MC 501a_3c0e
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is associated with the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene E is associated with the 'elephant' phenotype. A budding yeast that is homozygous recessive for Gene E cells absorb excessive amounts of liquid, resulting in giant, swollen cells.
- Gene M is affiliated with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 134 |
| 2 | | | | | 2,298 |
| 3 | | | | | 2,904 |
| 4 | | | | | 1,200 |
| 5 | | | | | 31 |
| 6 | | | | | 33 |
| TOTAL = | 6,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and E
A. distance = = = = 0.1050 = 10.50 cM Incorrect B. distance = | ½(1,200 + 2,298) + 3×(33) |
| 6,600 |
= = = 0.2800 = 28 cM Incorrect C. distance = | ½(2,298) + 3×(33 + 134) |
| 6,600 |
= = = 0.2500 = 25 cM Incorrect D. distance = | ½(1,200 + 2,298) + 3×(31 + 134) |
| 6,600 |
= = = 0.3400 = 34 cM Incorrect E. distance = | ½(1,200) + 3×(31 + 33) |
| 6,600 |
= = = 0.1200 = 12 cM Correct
MC 1b40_4556
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is linked with the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene K is affiliated with the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
- Gene T is affiliated with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 64 |
| 2 | | | | | 7 |
| 3 | | | | | 1,494 |
| 4 | | | | | 9,264 |
| 5 | | | | | 8,202 |
| 6 | | | | | 169 |
| TOTAL = | 19,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and T
A. distance = = = = 0.0400 = 4 cM Incorrect B. distance = | ½(1,494 + 8,202) + 3×(7 + 169) |
| 19,200 |
= = = 0.2800 = 28 cM Correct C. distance = | ½(1,494) + 3×(7 + 64) |
| 19,200 |
= = = 0.0500 = 5 cM Incorrect D. distance = | ½(1,494 + 8,202) + 3×(7 + 64 + 169) |
| 19,200 |
= = = 0.2900 = 29 cM Incorrect E. distance = | ½(8,202) + 3×(64 + 169) |
| 19,200 |
= = = 0.2500 = 25 cM Incorrect
MC 2032_5702
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene K is related to the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
- Gene N is analogous to the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
- Gene T is related to the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 250 |
| 2 | | | | | 5,112 |
| 3 | | | | | 75 |
| 4 | | | | | 3,030 |
| 5 | | | | | 8,875 |
| 6 | | | | | 58 |
| TOTAL = | 17,400 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes K and N
A. distance = | ½(3,030 + 5,112) + 3×(75 + 250) |
| 17,400 |
= = = 0.2900 = 29 cM Correct B. distance = = = = 0.1000 = 10 cM Incorrect C. distance = | ½(5,112) + 3×(58 + 250) |
| 17,400 |
= = = 0.2000 = 20 cM Incorrect D. distance = | ½(3,030 + 5,112) + 3×(58 + 75 + 250) |
| 17,400 |
= = = 0.3000 = 30 cM Incorrect E. distance = | ½(3,030) + 3×(58 + 75) |
| 17,400 |
= = = 0.1100 = 11 cM Incorrect
MC 3033_c514
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene D is correlated with the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene J is connected with the 'jeweled' phenotype. A budding yeast that is homozygous recessive for Gene J colonies appear dotted with tiny, iridescent spots that sparkle under light.
- Gene K is related to the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 310 |
| 2 | | | | | 4,782 |
| 3 | | | | | 8,488 |
| 4 | | | | | 4,368 |
| 5 | | | | | 246 |
| 6 | | | | | 256 |
| TOTAL = | 18,450 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes J and K
A. distance = | ½(4,782) + 3×(310) |
| 18,450 |
= = = 0.1800 = 18 cM Incorrect B. distance = | ½(4,368) + 3×(256) |
| 18,450 |
= = = 0.1600 = 16 cM Incorrect C. distance = | ½(4,368) + 3×(246 + 256) |
| 18,450 |
= = = 0.2000 = 20 cM Incorrect D. distance = | ½(4,368 + 4,782) + 3×(256 + 310) |
| 18,450 |
= = = 0.3400 = 34 cM Correct E. distance = | ½(4,782) + 3×(246 + 310) |
| 18,450 |
= = = 0.2200 = 22 cM Incorrect
MC c74e_df19
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene E is affiliated with the 'elephant' phenotype. A budding yeast that is homozygous recessive for Gene E cells absorb excessive amounts of liquid, resulting in giant, swollen cells.
- Gene M is correlated with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene T is linked with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 451 |
| 2 | | | | | 7,728 |
| 3 | | | | | 254 |
| 4 | | | | | 6,090 |
| 5 | | | | | 13,536 |
| 6 | | | | | 141 |
| TOTAL = | 28,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes M and T
A. distance = | ½(7,728) + 3×(141 + 451) |
| 28,200 |
= = = 0.2000 = 20 cM Correct B. distance = | ½(6,090 + 7,728) + 3×(141) |
| 28,200 |
= = = 0.2600 = 26 cM Incorrect C. distance = | ½(6,090 + 7,728) + 3×(254 + 451) |
| 28,200 |
= = = 0.3200 = 32 cM Incorrect D. distance = | ½(6,090) + 3×(254) |
| 28,200 |
= = = 0.1350 = 13.50 cM Incorrect E. distance = | ½(6,090) + 3×(141 + 254) |
| 28,200 |
= = = 0.1500 = 15 cM Incorrect
MC 5eb2_db71
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is related to the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene B is related to the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene M is correlated with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 299 |
| 2 | | | | | 8,418 |
| 3 | | | | | 14,905 |
| 4 | | | | | 3,780 |
| 5 | | | | | 138 |
| 6 | | | | | 60 |
| TOTAL = | 27,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and B
A. distance = | ½(3,780) + 3×(60 + 138) |
| 27,600 |
= = = 0.0900 = 9 cM Correct B. distance = | ½(8,418) + 3×(299) |
| 27,600 |
= = = 0.1850 = 18.50 cM Incorrect C. distance = | ½(8,418) + 3×(138 + 299) |
| 27,600 |
= = = 0.2000 = 20 cM Incorrect D. distance = | ½(3,780 + 8,418) + 3×(60 + 299) |
| 27,600 |
= = = 0.2600 = 26 cM Incorrect E. distance = | ½(3,780 + 8,418) + 3×(60 + 138 + 299) |
| 27,600 |
= = = 0.2750 = 27.50 cM Incorrect
MC 312c_4228
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene B is correlated with the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene N is connected with the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
- Gene X is associated with the 'xenon' phenotype. A budding yeast that is homozygous recessive for Gene X cells emit a faint glow under UV light, as if they were fluorescent.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 231 |
| 2 | | | | | 163 |
| 3 | | | | | 3,972 |
| 4 | | | | | 8,910 |
| 5 | | | | | 6,126 |
| 6 | | | | | 398 |
| TOTAL = | 19,800 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes B and N
A. distance = | ½(3,972 + 6,126) + 3×(163 + 398) |
| 19,800 |
= = = 0.3400 = 34 cM Incorrect B. distance = | ½(3,972 + 6,126) + 3×(231) |
| 19,800 |
= = = 0.2900 = 29 cM Incorrect C. distance = | ½(3,972) + 3×(163 + 231) |
| 19,800 |
= = = 0.1600 = 16 cM Correct D. distance = | ½(3,972 + 6,126) + 3×(163 + 231 + 398) |
| 19,800 |
= = = 0.3750 = 37.50 cM Incorrect E. distance = | ½(6,126) + 3×(231 + 398) |
| 19,800 |
= = = 0.2500 = 25 cM Incorrect
MC 7660_b96e
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene D is correlated with the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene X is correlated with the 'xenon' phenotype. A budding yeast that is homozygous recessive for Gene X cells emit a faint glow under UV light, as if they were fluorescent.
- Gene Y is analogous to the 'yolk' phenotype. A budding yeast that is homozygous recessive for Gene Y cells develop a dense, yellowish core that resembles an egg yolk when viewed under a microscope.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 22 |
| 2 | | | | | 8,383 |
| 3 | | | | | 1,374 |
| 4 | | | | | 3,324 |
| 5 | | | | | 13 |
| 6 | | | | | 84 |
| TOTAL = | 13,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes D and X
A. distance = | ½(1,374 + 3,324) + 3×(13 + 84) |
| 13,200 |
= = = 0.2000 = 20 cM Correct B. distance = | ½(1,374) + 3×(13 + 22) |
| 13,200 |
= = = 0.0600 = 6 cM Incorrect C. distance = = = = 0.0550 = 5.50 cM Incorrect D. distance = | ½(3,324) + 3×(22 + 84) |
| 13,200 |
= = = 0.1500 = 15 cM Incorrect E. distance = = = = 0.1450 = 14.50 cM Incorrect
MC c626_b2b1
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is associated with the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene J is related to the 'jeweled' phenotype. A budding yeast that is homozygous recessive for Gene J colonies appear dotted with tiny, iridescent spots that sparkle under light.
- Gene R is connected with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 6 |
| 2 | | | | | 1,644 |
| 3 | | | | | 40 |
| 4 | | | | | 13,680 |
| 5 | | | | | 8,484 |
| 6 | | | | | 146 |
| TOTAL = | 24,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes C and R
A. distance = | ½(1,644 + 8,484) + 3×(40) |
| 24,000 |
= = = 0.2160 = 21.60 cM Incorrect B. distance = = = = 0.0350 = 3.50 cM Incorrect C. distance = | ½(8,484) + 3×(40 + 146) |
| 24,000 |
= = = 0.2000 = 20 cM Correct D. distance = | ½(1,644 + 8,484) + 3×(6 + 146) |
| 24,000 |
= = = 0.2300 = 23 cM Incorrect E. distance = | ½(1,644) + 3×(6 + 40) |
| 24,000 |
= = = 0.0400 = 4 cM Incorrect
MC edba_d317
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene J is linked with the 'jeweled' phenotype. A budding yeast that is homozygous recessive for Gene J colonies appear dotted with tiny, iridescent spots that sparkle under light.
- Gene P is affiliated with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
- Gene R is associated with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 268 |
| 2 | | | | | 4,596 |
| 3 | | | | | 5,808 |
| 4 | | | | | 2,400 |
| 5 | | | | | 62 |
| 6 | | | | | 66 |
| TOTAL = | 13,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes J and R
A. distance = | ½(2,400) + 3×(62 + 66) |
| 13,200 |
= = = 0.1200 = 12 cM Incorrect B. distance = | ½(4,596) + 3×(268) |
| 13,200 |
= = = 0.2350 = 23.50 cM Incorrect C. distance = | ½(2,400 + 4,596) + 3×(66) |
| 13,200 |
= = = 0.2800 = 28 cM Incorrect D. distance = | ½(2,400 + 4,596) + 3×(62 + 268) |
| 13,200 |
= = = 0.3400 = 34 cM Correct E. distance = | ½(4,596) + 3×(66 + 268) |
| 13,200 |
= = = 0.2500 = 25 cM Incorrect
MC 9a28_56a7
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is affiliated with the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene F is associated with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene X is connected with the 'xenon' phenotype. A budding yeast that is homozygous recessive for Gene X cells emit a faint glow under UV light, as if they were fluorescent.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 69 |
| 2 | | | | | 3,096 |
| 3 | | | | | 130 |
| 4 | | | | | 10,335 |
| 5 | | | | | 5,640 |
| 6 | | | | | 230 |
| TOTAL = | 19,500 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and F
A. distance = | ½(5,640) + 3×(130 + 230) |
| 19,500 |
= = = 0.2000 = 20 cM Incorrect B. distance = | ½(3,096 + 5,640) + 3×(69 + 130 + 230) |
| 19,500 |
= = = 0.2900 = 29 cM Incorrect C. distance = | ½(3,096) + 3×(69 + 130) |
| 19,500 |
= = = 0.1100 = 11 cM Correct D. distance = = = = 0.0900 = 9 cM Incorrect E. distance = | ½(3,096 + 5,640) + 3×(69 + 230) |
| 19,500 |
= = = 0.2700 = 27 cM Incorrect
MC 14c9_0bfe
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene K is analogous to the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
- Gene P is associated with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
- Gene Y is related to the 'yolk' phenotype. A budding yeast that is homozygous recessive for Gene Y cells develop a dense, yellowish core that resembles an egg yolk when viewed under a microscope.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 315 |
| 2 | | | | | 9,054 |
| 3 | | | | | 15,553 |
| 4 | | | | | 3,732 |
| 5 | | | | | 96 |
| 6 | | | | | 50 |
| TOTAL = | 28,800 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes P and Y
A. distance = | ½(9,054) + 3×(315) |
| 28,800 |
= = = 0.1900 = 19 cM Incorrect B. distance = | ½(3,732) + 3×(50 + 96) |
| 28,800 |
= = = 0.0800 = 8 cM Incorrect C. distance = | ½(3,732 + 9,054) + 3×(50 + 96 + 315) |
| 28,800 |
= = = 0.2700 = 27 cM Incorrect D. distance = | ½(3,732 + 9,054) + 3×(50 + 315) |
| 28,800 |
= = = 0.2600 = 26 cM Correct E. distance = | ½(9,054) + 3×(96 + 315) |
| 28,800 |
= = = 0.2000 = 20 cM Incorrect
MC 992d_d781
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is correlated with the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene J is linked with the 'jeweled' phenotype. A budding yeast that is homozygous recessive for Gene J colonies appear dotted with tiny, iridescent spots that sparkle under light.
- Gene P is analogous to the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 350 |
| 2 | | | | | 364 |
| 3 | | | | | 6,216 |
| 4 | | | | | 441 |
| 5 | | | | | 6,804 |
| 6 | | | | | 12,075 |
| TOTAL = | 26,250 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and J
A. distance = | ½(6,804) + 3×(350 + 441) |
| 26,250 |
= = = 0.2200 = 22 cM Incorrect B. distance = | ½(6,216 + 6,804) + 3×(350 + 364 + 441) |
| 26,250 |
= = = 0.3800 = 38 cM Incorrect C. distance = | ½(6,804) + 3×(441) |
| 26,250 |
= = = 0.1800 = 18 cM Incorrect D. distance = | ½(6,216) + 3×(350 + 364) |
| 26,250 |
= = = 0.2000 = 20 cM Correct E. distance = | ½(6,216) + 3×(364) |
| 26,250 |
= = = 0.1600 = 16 cM Incorrect
MC 017e_c7cb
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene B is analogous to the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene C is associated with the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene F is affiliated with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 36 |
| 2 | | | | | 11 |
| 3 | | | | | 1,878 |
| 4 | | | | | 147 |
| 5 | | | | | 6,678 |
| 6 | | | | | 12,850 |
| TOTAL = | 21,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes B and C
A. distance = | ½(6,678) + 3×(147) |
| 21,600 |
= = = 0.1750 = 17.50 cM Incorrect B. distance = | ½(1,878) + 3×(11 + 36) |
| 21,600 |
= = = 0.0500 = 5 cM Correct C. distance = = = = 0.0450 = 4.50 cM Incorrect D. distance = | ½(1,878 + 6,678) + 3×(11 + 147) |
| 21,600 |
= = = 0.2200 = 22 cM Incorrect E. distance = | ½(6,678) + 3×(36 + 147) |
| 21,600 |
= = = 0.1800 = 18 cM Incorrect
MC 6585_f728
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene K is correlated with the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
- Gene T is connected with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
- Gene W is connected with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 213 |
| 2 | | | | | 6,114 |
| 3 | | | | | 15,708 |
| 4 | | | | | 4,182 |
| 5 | | | | | 88 |
| 6 | | | | | 95 |
| TOTAL = | 26,400 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes K and W
A. distance = | ½(4,182 + 6,114) + 3×(95 + 213) |
| 26,400 |
= = = 0.2300 = 23 cM Incorrect B. distance = | ½(6,114) + 3×(213) |
| 26,400 |
= = = 0.1400 = 14 cM Incorrect C. distance = | ½(6,114) + 3×(88 + 213) |
| 26,400 |
= = = 0.1500 = 15 cM Correct D. distance = | ½(4,182) + 3×(88 + 95) |
| 26,400 |
= = = 0.1000 = 10 cM Incorrect E. distance = | ½(4,182 + 6,114) + 3×(88 + 95 + 213) |
| 26,400 |
= = = 0.2400 = 24 cM Incorrect
MC 3c54_cd16
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is analogous to the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene C is related to the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene F is analogous to the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 417 |
| 2 | | | | | 4,980 |
| 3 | | | | | 7,180 |
| 4 | | | | | 4,326 |
| 5 | | | | | 203 |
| 6 | | | | | 294 |
| TOTAL = | 17,400 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and F
A. distance = | ½(4,326) + 3×(203 + 294) |
| 17,400 |
= = = 0.2100 = 21 cM Incorrect B. distance = | ½(4,326) + 3×(294) |
| 17,400 |
= = = 0.1750 = 17.50 cM Incorrect C. distance = | ½(4,326 + 4,980) + 3×(294 + 417) |
| 17,400 |
= = = 0.3900 = 39 cM Incorrect D. distance = | ½(4,980) + 3×(203 + 417) |
| 17,400 |
= = = 0.2500 = 25 cM Correct E. distance = | ½(4,980) + 3×(417) |
| 17,400 |
= = = 0.2150 = 21.50 cM Incorrect
MC a954_e920
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene F is affiliated with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene T is correlated with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
- Gene W is correlated with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 48 |
| 2 | | | | | 49 |
| 3 | | | | | 1,722 |
| 4 | | | | | 5,569 |
| 5 | | | | | 2,136 |
| 6 | | | | | 76 |
| TOTAL = | 9,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes F and W
A. distance = | ½(2,136) + 3×(48 + 76) |
| 9,600 |
= = = 0.1500 = 15 cM Incorrect B. distance = | ½(1,722 + 2,136) + 3×(49 + 76) |
| 9,600 |
= = = 0.2400 = 24 cM Correct C. distance = | ½(1,722) + 3×(48 + 49) |
| 9,600 |
= = = 0.1200 = 12 cM Incorrect D. distance = = = = 0.1350 = 13.50 cM Incorrect E. distance = | ½(1,722 + 2,136) + 3×(48 + 49 + 76) |
| 9,600 |
= = = 0.2550 = 25.50 cM Incorrect
MC 1847_61ba
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is affiliated with the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene R is linked with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
- Gene T is connected with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 26 |
| 2 | | | | | 1,902 |
| 3 | | | | | 49 |
| 4 | | | | | 160 |
| 5 | | | | | 4,626 |
| 6 | | | | | 7,937 |
| TOTAL = | 14,700 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes C and T
A. distance = | ½(4,626) + 3×(160) |
| 14,700 |
= = = 0.1900 = 19 cM Incorrect B. distance = | ½(4,626) + 3×(49 + 160) |
| 14,700 |
= = = 0.2000 = 20 cM Incorrect C. distance = | ½(1,902 + 4,626) + 3×(26 + 49 + 160) |
| 14,700 |
= = = 0.2700 = 27 cM Incorrect D. distance = | ½(1,902 + 4,626) + 3×(26 + 160) |
| 14,700 |
= = = 0.2600 = 26 cM Correct E. distance = = = = 0.0700 = 7 cM Incorrect
MC f166_e149
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is analogous to the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene K is related to the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
- Gene W is connected with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 38 |
| 2 | | | | | 1,224 |
| 3 | | | | | 59 |
| 4 | | | | | 1,494 |
| 5 | | | | | 22 |
| 6 | | | | | 3,763 |
| TOTAL = | 6,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and W
A. distance = = = = 0.1400 = 14 cM Incorrect B. distance = = = = 0.1100 = 11 cM Incorrect C. distance = | ½(1,224 + 1,494) + 3×(22 + 38 + 59) |
| 6,600 |
= = = 0.2600 = 26 cM Incorrect D. distance = | ½(1,224) + 3×(22 + 38) |
| 6,600 |
= = = 0.1200 = 12 cM Correct E. distance = | ½(1,494) + 3×(22 + 59) |
| 6,600 |
= = = 0.1500 = 15 cM Incorrect
MC 4d9e_564e
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene B is linked with the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene E is affiliated with the 'elephant' phenotype. A budding yeast that is homozygous recessive for Gene E cells absorb excessive amounts of liquid, resulting in giant, swollen cells.
- Gene H is linked with the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 115 |
| 2 | | | | | 116 |
| 3 | | | | | 1,650 |
| 4 | | | | | 149 |
| 5 | | | | | 1,866 |
| 6 | | | | | 3,004 |
| TOTAL = | 6,900 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes B and E
A. distance = | ½(1,650 + 1,866) + 3×(116 + 149) |
| 6,900 |
= = = 0.3700 = 37 cM Incorrect B. distance = | ½(1,650 + 1,866) + 3×(115 + 116 + 149) |
| 6,900 |
= = = 0.4200 = 42 cM Incorrect C. distance = | ½(1,650) + 3×(115 + 116) |
| 6,900 |
= = = 0.2200 = 22 cM Correct D. distance = = = = 0.1700 = 17 cM Incorrect E. distance = = = = 0.2000 = 20 cM Incorrect
MC 1cd0_c82c
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is linked with the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene P is associated with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
- Gene R is linked with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 152 |
| 2 | | | | | 2,322 |
| 3 | | | | | 2,772 |
| 4 | | | | | 1,308 |
| 5 | | | | | 35 |
| 6 | | | | | 11 |
| TOTAL = | 6,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and R
A. distance = | ½(1,308 + 2,322) + 3×(35 + 152) |
| 6,600 |
= = = 0.3600 = 36 cM Correct B. distance = | ½(1,308) + 3×(11 + 35) |
| 6,600 |
= = = 0.1200 = 12 cM Incorrect C. distance = | ½(1,308 + 2,322) + 3×(11) |
| 6,600 |
= = = 0.2800 = 28 cM Incorrect D. distance = | ½(2,322) + 3×(11 + 152) |
| 6,600 |
= = = 0.2500 = 25 cM Incorrect E. distance = = = = 0.2450 = 24.50 cM Incorrect
MC bac9_b219
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene D is correlated with the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene K is affiliated with the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
- Gene W is correlated with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 104 |
| 2 | | | | | 1,728 |
| 3 | | | | | 2,052 |
| 4 | | | | | 888 |
| 5 | | | | | 8 |
| 6 | | | | | 20 |
| TOTAL = | 4,800 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes D and W
A. distance = | ½(888 + 1,728) + 3×(20 + 104) |
| 4,800 |
= = = 0.3500 = 35 cM Incorrect B. distance = | ½(1,728) + 3×(8 + 104) |
| 4,800 |
= = = 0.2500 = 25 cM Correct C. distance = | ½(888) + 3×(20 + 104) |
| 4,800 |
= = = 0.1700 = 17 cM Incorrect D. distance = | ½(888) + 3×(8 + 20) |
| 4,800 |
= = = 0.1100 = 11 cM Incorrect E. distance = | ½(888 + 1,728) + 3×(8 + 20) |
| 4,800 |
= = = 0.2900 = 29 cM Incorrect
MC 97cf_a1df
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene B is linked with the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene E is connected with the 'elephant' phenotype. A budding yeast that is homozygous recessive for Gene E cells absorb excessive amounts of liquid, resulting in giant, swollen cells.
- Gene F is connected with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 280 |
| 2 | | | | | 273 |
| 3 | | | | | 5,322 |
| 4 | | | | | 527 |
| 5 | | | | | 7,158 |
| 6 | | | | | 10,440 |
| TOTAL = | 24,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes B and F
A. distance = | ½(5,322 + 7,158) + 3×(273 + 527) |
| 24,000 |
= = = 0.3600 = 36 cM Incorrect B. distance = | ½(5,322 + 7,158) + 3×(280) |
| 24,000 |
= = = 0.2950 = 29.50 cM Incorrect C. distance = | ½(7,158) + 3×(280 + 527) |
| 24,000 |
= = = 0.2500 = 25 cM Correct D. distance = | ½(5,322 + 7,158) + 3×(273 + 280 + 527) |
| 24,000 |
= = = 0.3950 = 39.50 cM Incorrect E. distance = | ½(5,322) + 3×(273 + 280) |
| 24,000 |
= = = 0.1800 = 18 cM Incorrect
MC 2f80_533c
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene H is linked with the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
- Gene K is correlated with the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
- Gene W is affiliated with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 100 |
| 2 | | | | | 1,920 |
| 3 | | | | | 3,600 |
| 4 | | | | | 1,464 |
| 5 | | | | | 60 |
| 6 | | | | | 56 |
| TOTAL = | 7,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes K and W
A. distance = | ½(1,464 + 1,920) + 3×(56 + 60 + 100) |
| 7,200 |
= = = 0.3250 = 32.50 cM Incorrect B. distance = | ½(1,464) + 3×(56 + 60) |
| 7,200 |
= = = 0.1500 = 15 cM Incorrect C. distance = | ½(1,464 + 1,920) + 3×(60) |
| 7,200 |
= = = 0.2600 = 26 cM Incorrect D. distance = | ½(1,920) + 3×(60 + 100) |
| 7,200 |
= = = 0.2000 = 20 cM Incorrect E. distance = | ½(1,464 + 1,920) + 3×(56 + 100) |
| 7,200 |
= = = 0.3000 = 30 cM Correct
MC 5475_1ec1
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene D is connected with the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene R is correlated with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
- Gene X is associated with the 'xenon' phenotype. A budding yeast that is homozygous recessive for Gene X cells emit a faint glow under UV light, as if they were fluorescent.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 488 |
| 2 | | | | | 5,844 |
| 3 | | | | | 345 |
| 4 | | | | | 5,070 |
| 5 | | | | | 8,415 |
| 6 | | | | | 238 |
| TOTAL = | 20,400 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes D and X
A. distance = | ½(5,844) + 3×(238 + 488) |
| 20,400 |
= = = 0.2500 = 25 cM Incorrect B. distance = | ½(5,070) + 3×(345) |
| 20,400 |
= = = 0.1750 = 17.50 cM Incorrect C. distance = | ½(5,070 + 5,844) + 3×(345 + 488) |
| 20,400 |
= = = 0.3900 = 39 cM Incorrect D. distance = | ½(5,070) + 3×(238 + 345) |
| 20,400 |
= = = 0.2100 = 21 cM Correct E. distance = | ½(5,844) + 3×(488) |
| 20,400 |
= = = 0.2150 = 21.50 cM Incorrect
MC e0e3_de8b
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene B is analogous to the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene C is affiliated with the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene E is associated with the 'elephant' phenotype. A budding yeast that is homozygous recessive for Gene E cells absorb excessive amounts of liquid, resulting in giant, swollen cells.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 47 |
| 2 | | | | | 5 |
| 3 | | | | | 1,098 |
| 4 | | | | | 7,049 |
| 5 | | | | | 5,784 |
| 6 | | | | | 117 |
| TOTAL = | 14,100 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes B and E
A. distance = | ½(5,784) + 3×(47 + 117) |
| 14,100 |
= = = 0.2400 = 24 cM Incorrect B. distance = = = = 0.0400 = 4 cM Incorrect C. distance = | ½(1,098) + 3×(5 + 47) |
| 14,100 |
= = = 0.0500 = 5 cM Incorrect D. distance = | ½(1,098 + 5,784) + 3×(5 + 47 + 117) |
| 14,100 |
= = = 0.2800 = 28 cM Incorrect E. distance = | ½(1,098 + 5,784) + 3×(5 + 117) |
| 14,100 |
= = = 0.2700 = 27 cM Correct
MC af23_d136
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene F is analogous to the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene T is analogous to the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
- Gene Y is connected with the 'yolk' phenotype. A budding yeast that is homozygous recessive for Gene Y cells develop a dense, yellowish core that resembles an egg yolk when viewed under a microscope.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 186 |
| 2 | | | | | 6,418 |
| 3 | | | | | 3,300 |
| 4 | | | | | 3,618 |
| 5 | | | | | 194 |
| 6 | | | | | 234 |
| TOTAL = | 13,950 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes T and Y
A. distance = | ½(3,300 + 3,618) + 3×(186 + 194 + 234) |
| 13,950 |
= = = 0.3800 = 38 cM Incorrect B. distance = | ½(3,618) + 3×(234) |
| 13,950 |
= = = 0.1800 = 18 cM Incorrect C. distance = | ½(3,618) + 3×(186 + 234) |
| 13,950 |
= = = 0.2200 = 22 cM Correct D. distance = | ½(3,300 + 3,618) + 3×(194 + 234) |
| 13,950 |
= = = 0.3400 = 34 cM Incorrect E. distance = | ½(3,300) + 3×(194) |
| 13,950 |
= = = 0.1600 = 16 cM Incorrect
MC 4e4b_41ce
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene H is associated with the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
- Gene M is affiliated with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene Y is related to the 'yolk' phenotype. A budding yeast that is homozygous recessive for Gene Y cells develop a dense, yellowish core that resembles an egg yolk when viewed under a microscope.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 51 |
| 2 | | | | | 44 |
| 3 | | | | | 1,062 |
| 4 | | | | | 2,244 |
| 5 | | | | | 1,590 |
| 6 | | | | | 109 |
| TOTAL = | 5,100 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes H and Y
A. distance = = = = 0.1300 = 13 cM Incorrect B. distance = | ½(1,062 + 1,590) + 3×(44 + 109) |
| 5,100 |
= = = 0.3500 = 35 cM Correct C. distance = | ½(1,062) + 3×(44 + 51) |
| 5,100 |
= = = 0.1600 = 16 cM Incorrect D. distance = | ½(1,062 + 1,590) + 3×(51) |
| 5,100 |
= = = 0.2900 = 29 cM Incorrect E. distance = | ½(1,590) + 3×(51 + 109) |
| 5,100 |
= = = 0.2500 = 25 cM Incorrect
MC 3ff7_ceeb
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene B is connected with the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene R is related to the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
- Gene T is linked with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 75 |
| 2 | | | | | 9,000 |
| 3 | | | | | 5,310 |
| 4 | | | | | 7,290 |
| 5 | | | | | 240 |
| 6 | | | | | 585 |
| TOTAL = | 22,500 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes R and T
A. distance = | ½(7,290) + 3×(585) |
| 22,500 |
= = = 0.2400 = 24 cM Incorrect B. distance = | ½(5,310) + 3×(240) |
| 22,500 |
= = = 0.1500 = 15 cM Incorrect C. distance = | ½(5,310) + 3×(75 + 240) |
| 22,500 |
= = = 0.1600 = 16 cM Incorrect D. distance = | ½(7,290) + 3×(75 + 585) |
| 22,500 |
= = = 0.2500 = 25 cM Correct E. distance = | ½(5,310 + 7,290) + 3×(240 + 585) |
| 22,500 |
= = = 0.3900 = 39 cM Incorrect
MC 10af_aa3c
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene B is connected with the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene J is connected with the 'jeweled' phenotype. A budding yeast that is homozygous recessive for Gene J colonies appear dotted with tiny, iridescent spots that sparkle under light.
- Gene W is analogous to the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 127 |
| 2 | | | | | 1,878 |
| 3 | | | | | 2,640 |
| 4 | | | | | 1,266 |
| 5 | | | | | 49 |
| 6 | | | | | 40 |
| TOTAL = | 6,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes B and W
A. distance = = = = 0.2200 = 22 cM Incorrect B. distance = | ½(1,266 + 1,878) + 3×(40 + 49 + 127) |
| 6,000 |
= = = 0.3700 = 37 cM Incorrect C. distance = | ½(1,878) + 3×(40 + 127) |
| 6,000 |
= = = 0.2400 = 24 cM Incorrect D. distance = = = = 0.1300 = 13 cM Incorrect E. distance = | ½(1,266 + 1,878) + 3×(49 + 127) |
| 6,000 |
= = = 0.3500 = 35 cM Correct
MC ffb1_13b4
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene M is linked with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene P is associated with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
- Gene T is analogous to the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 20 |
| 2 | | | | | 2,760 |
| 3 | | | | | 972 |
| 4 | | | | | 2,124 |
| 5 | | | | | 18 |
| 6 | | | | | 106 |
| TOTAL = | 6,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes P and T
A. distance = | ½(972 + 2,124) + 3×(18 + 106) |
| 6,000 |
= = = 0.3200 = 32 cM Incorrect B. distance = = = = 0.0900 = 9 cM Incorrect C. distance = | ½(2,124) + 3×(20 + 106) |
| 6,000 |
= = = 0.2400 = 24 cM Correct D. distance = = = = 0.2300 = 23 cM Incorrect E. distance = | ½(972) + 3×(18 + 20) |
| 6,000 |
= = = 0.1000 = 10 cM Incorrect
MC 7fa6_a588
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is analogous to the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene P is linked with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
- Gene X is linked with the 'xenon' phenotype. A budding yeast that is homozygous recessive for Gene X cells emit a faint glow under UV light, as if they were fluorescent.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 362 |
| 2 | | | | | 4,578 |
| 3 | | | | | 188 |
| 4 | | | | | 3,522 |
| 5 | | | | | 6,225 |
| 6 | | | | | 125 |
| TOTAL = | 15,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes P and X
A. distance = | ½(3,522 + 4,578) + 3×(125) |
| 15,000 |
= = = 0.2950 = 29.50 cM Incorrect B. distance = | ½(4,578) + 3×(125 + 362) |
| 15,000 |
= = = 0.2500 = 25 cM Correct C. distance = | ½(3,522 + 4,578) + 3×(188 + 362) |
| 15,000 |
= = = 0.3800 = 38 cM Incorrect D. distance = | ½(3,522) + 3×(125 + 188) |
| 15,000 |
= = = 0.1800 = 18 cM Incorrect E. distance = | ½(3,522 + 4,578) + 3×(125 + 188 + 362) |
| 15,000 |
= = = 0.4050 = 40.50 cM Incorrect
MC 6c5b_8e6d
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene B is associated with the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene R is linked with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
- Gene W is linked with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 78 |
| 2 | | | | | 134 |
| 3 | | | | | 4,344 |
| 4 | | | | | 209 |
| 5 | | | | | 5,298 |
| 6 | | | | | 13,337 |
| TOTAL = | 23,400 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes B and W
A. distance = | ½(4,344) + 3×(78 + 134) |
| 23,400 |
= = = 0.1200 = 12 cM Incorrect B. distance = | ½(5,298) + 3×(78 + 209) |
| 23,400 |
= = = 0.1500 = 15 cM Correct C. distance = | ½(4,344) + 3×(134) |
| 23,400 |
= = = 0.1100 = 11 cM Incorrect D. distance = | ½(4,344 + 5,298) + 3×(134 + 209) |
| 23,400 |
= = = 0.2500 = 25 cM Incorrect E. distance = | ½(5,298) + 3×(209) |
| 23,400 |
= = = 0.1400 = 14 cM Incorrect
MC 9f42_8f91
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene N is correlated with the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
- Gene P is linked with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
- Gene R is analogous to the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 315 |
| 2 | | | | | 10,395 |
| 3 | | | | | 5,544 |
| 4 | | | | | 6,552 |
| 5 | | | | | 336 |
| 6 | | | | | 483 |
| TOTAL = | 23,625 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes N and P
A. distance = | ½(5,544 + 6,552) + 3×(336 + 483) |
| 23,625 |
= = = 0.3600 = 36 cM Correct B. distance = | ½(5,544 + 6,552) + 3×(315 + 336 + 483) |
| 23,625 |
= = = 0.4000 = 40 cM Incorrect C. distance = | ½(5,544) + 3×(336) |
| 23,625 |
= = = 0.1600 = 16 cM Incorrect D. distance = | ½(6,552) + 3×(315 + 483) |
| 23,625 |
= = = 0.2400 = 24 cM Incorrect E. distance = | ½(6,552) + 3×(483) |
| 23,625 |
= = = 0.2000 = 20 cM Incorrect
MC 1204_360f
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene E is correlated with the 'elephant' phenotype. A budding yeast that is homozygous recessive for Gene E cells absorb excessive amounts of liquid, resulting in giant, swollen cells.
- Gene K is linked with the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
- Gene M is connected with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 45 |
| 2 | | | | | 1,350 |
| 3 | | | | | 3,555 |
| 4 | | | | | 996 |
| 5 | | | | | 30 |
| 6 | | | | | 24 |
| TOTAL = | 6,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes E and M
A. distance = | ½(996 + 1,350) + 3×(24 + 45) |
| 6,000 |
= = = 0.2300 = 23 cM Incorrect B. distance = | ½(996 + 1,350) + 3×(24 + 30 + 45) |
| 6,000 |
= = = 0.2450 = 24.50 cM Incorrect C. distance = | ½(996) + 3×(24 + 30) |
| 6,000 |
= = = 0.1100 = 11 cM Incorrect D. distance = | ½(1,350) + 3×(24 + 45) |
| 6,000 |
= = = 0.1470 = 14.70 cM Incorrect E. distance = | ½(1,350) + 3×(30 + 45) |
| 6,000 |
= = = 0.1500 = 15 cM Correct
MC 28bf_c72c
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene K is associated with the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
- Gene M is linked with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene W is linked with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 51 |
| 2 | | | | | 89 |
| 3 | | | | | 2,220 |
| 4 | | | | | 227 |
| 5 | | | | | 3,228 |
| 6 | | | | | 4,385 |
| TOTAL = | 10,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes K and M
A. distance = | ½(3,228) + 3×(51 + 227) |
| 10,200 |
= = = 0.2400 = 24 cM Incorrect B. distance = | ½(2,220) + 3×(51 + 89) |
| 10,200 |
= = = 0.1500 = 15 cM Correct C. distance = | ½(3,228) + 3×(227) |
| 10,200 |
= = = 0.2250 = 22.50 cM Incorrect D. distance = = = = 0.1350 = 13.50 cM Incorrect E. distance = | ½(2,220 + 3,228) + 3×(89 + 227) |
| 10,200 |
= = = 0.3600 = 36 cM Incorrect
MC 0012_8e6c
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene R is linked with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
- Gene T is associated with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
- Gene X is analogous to the 'xenon' phenotype. A budding yeast that is homozygous recessive for Gene X cells emit a faint glow under UV light, as if they were fluorescent.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 314 |
| 2 | | | | | 5,046 |
| 3 | | | | | 491 |
| 4 | | | | | 6,084 |
| 5 | | | | | 245 |
| 6 | | | | | 8,820 |
| TOTAL = | 21,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes R and T
A. distance = | ½(5,046 + 6,084) + 3×(314 + 491) |
| 21,000 |
= = = 0.3800 = 38 cM Correct B. distance = | ½(5,046 + 6,084) + 3×(245) |
| 21,000 |
= = = 0.3000 = 30 cM Incorrect C. distance = | ½(5,046) + 3×(314) |
| 21,000 |
= = = 0.1650 = 16.50 cM Incorrect D. distance = | ½(6,084) + 3×(245 + 491) |
| 21,000 |
= = = 0.2500 = 25 cM Incorrect E. distance = | ½(5,046) + 3×(245 + 314) |
| 21,000 |
= = = 0.2000 = 20 cM Incorrect
MC 13db_231c
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene D is correlated with the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene R is correlated with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
- Gene Y is correlated with the 'yolk' phenotype. A budding yeast that is homozygous recessive for Gene Y cells develop a dense, yellowish core that resembles an egg yolk when viewed under a microscope.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 295 |
| 2 | | | | | 3,888 |
| 3 | | | | | 6,072 |
| 4 | | | | | 3,150 |
| 5 | | | | | 207 |
| 6 | | | | | 188 |
| TOTAL = | 13,800 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes D and R
A. distance = | ½(3,150 + 3,888) + 3×(188 + 207 + 295) |
| 13,800 |
= = = 0.4050 = 40.50 cM Incorrect B. distance = | ½(3,150 + 3,888) + 3×(207) |
| 13,800 |
= = = 0.3000 = 30 cM Incorrect C. distance = | ½(3,150) + 3×(188 + 207) |
| 13,800 |
= = = 0.2000 = 20 cM Correct D. distance = | ½(3,888) + 3×(207 + 295) |
| 13,800 |
= = = 0.2500 = 25 cM Incorrect E. distance = | ½(3,150 + 3,888) + 3×(188 + 295) |
| 13,800 |
= = = 0.3600 = 36 cM Incorrect
MC bede_b86d
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene J is associated with the 'jeweled' phenotype. A budding yeast that is homozygous recessive for Gene J colonies appear dotted with tiny, iridescent spots that sparkle under light.
- Gene P is associated with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
- Gene X is connected with the 'xenon' phenotype. A budding yeast that is homozygous recessive for Gene X cells emit a faint glow under UV light, as if they were fluorescent.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 128 |
| 2 | | | | | 8,352 |
| 3 | | | | | 3,828 |
| 4 | | | | | 6,348 |
| 5 | | | | | 130 |
| 6 | | | | | 414 |
| TOTAL = | 19,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes J and P
A. distance = | ½(3,828 + 6,348) + 3×(128 + 130 + 414) |
| 19,200 |
= = = 0.3700 = 37 cM Incorrect B. distance = | ½(6,348) + 3×(128 + 414) |
| 19,200 |
= = = 0.2500 = 25 cM Incorrect C. distance = | ½(3,828 + 6,348) + 3×(130 + 414) |
| 19,200 |
= = = 0.3500 = 35 cM Correct D. distance = | ½(6,348) + 3×(414) |
| 19,200 |
= = = 0.2300 = 23 cM Incorrect E. distance = | ½(3,828) + 3×(128 + 130) |
| 19,200 |
= = = 0.1400 = 14 cM Incorrect
MC 8138_87de
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene E is affiliated with the 'elephant' phenotype. A budding yeast that is homozygous recessive for Gene E cells absorb excessive amounts of liquid, resulting in giant, swollen cells.
- Gene J is correlated with the 'jeweled' phenotype. A budding yeast that is homozygous recessive for Gene J colonies appear dotted with tiny, iridescent spots that sparkle under light.
- Gene N is linked with the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 87 |
| 2 | | | | | 3,582 |
| 3 | | | | | 108 |
| 4 | | | | | 161 |
| 5 | | | | | 4,866 |
| 6 | | | | | 12,796 |
| TOTAL = | 21,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes J and N
A. distance = | ½(4,866) + 3×(108 + 161) |
| 21,600 |
= = = 0.1500 = 15 cM Correct B. distance = | ½(3,582 + 4,866) + 3×(87 + 161) |
| 21,600 |
= = = 0.2300 = 23 cM Incorrect C. distance = | ½(3,582) + 3×(87 + 108) |
| 21,600 |
= = = 0.1100 = 11 cM Incorrect D. distance = | ½(4,866) + 3×(161) |
| 21,600 |
= = = 0.1350 = 13.50 cM Incorrect E. distance = = = = 0.0950 = 9.50 cM Incorrect
MC 5936_b2cf
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene N is correlated with the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
- Gene P is correlated with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
- Gene T is associated with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 228 |
| 2 | | | | | 311 |
| 3 | | | | | 4,290 |
| 4 | | | | | 7,099 |
| 5 | | | | | 4,770 |
| 6 | | | | | 402 |
| TOTAL = | 17,100 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes N and P
A. distance = | ½(4,290) + 3×(311) |
| 17,100 |
= = = 0.1800 = 18 cM Incorrect B. distance = | ½(4,290 + 4,770) + 3×(311 + 402) |
| 17,100 |
= = = 0.3900 = 39 cM Incorrect C. distance = | ½(4,770) + 3×(228 + 402) |
| 17,100 |
= = = 0.2500 = 25 cM Incorrect D. distance = | ½(4,770) + 3×(402) |
| 17,100 |
= = = 0.2100 = 21 cM Incorrect E. distance = | ½(4,290) + 3×(228 + 311) |
| 17,100 |
= = = 0.2200 = 22 cM Correct
MC 1f35_1f54
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene J is connected with the 'jeweled' phenotype. A budding yeast that is homozygous recessive for Gene J colonies appear dotted with tiny, iridescent spots that sparkle under light.
- Gene M is correlated with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene Y is linked with the 'yolk' phenotype. A budding yeast that is homozygous recessive for Gene Y cells develop a dense, yellowish core that resembles an egg yolk when viewed under a microscope.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 150 |
| 2 | | | | | 3,090 |
| 3 | | | | | 46 |
| 4 | | | | | 1,824 |
| 5 | | | | | 5,355 |
| 6 | | | | | 35 |
| TOTAL = | 10,500 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes J and Y
A. distance = | ½(1,824 + 3,090) + 3×(35 + 46 + 150) |
| 10,500 |
= = = 0.3000 = 30 cM Incorrect B. distance = | ½(3,090) + 3×(35 + 150) |
| 10,500 |
= = = 0.2000 = 20 cM Incorrect C. distance = = = = 0.1000 = 10 cM Incorrect D. distance = | ½(1,824 + 3,090) + 3×(46 + 150) |
| 10,500 |
= = = 0.2900 = 29 cM Incorrect E. distance = | ½(1,824) + 3×(35 + 46) |
| 10,500 |
= = = 0.1100 = 11 cM Correct
MC 9c4b_c736
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is affiliated with the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene C is analogous to the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene W is associated with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 285 |
| 2 | | | | | 5,580 |
| 3 | | | | | 66 |
| 4 | | | | | 2,682 |
| 5 | | | | | 7,452 |
| 6 | | | | | 135 |
| TOTAL = | 16,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and W
A. distance = | ½(2,682 + 5,580) + 3×(135) |
| 16,200 |
= = = 0.2800 = 28 cM Incorrect B. distance = | ½(2,682) + 3×(66 + 135) |
| 16,200 |
= = = 0.1200 = 12 cM Correct C. distance = | ½(2,682 + 5,580) + 3×(66 + 135 + 285) |
| 16,200 |
= = = 0.3450 = 34.50 cM Incorrect D. distance = | ½(2,682 + 5,580) + 3×(66 + 285) |
| 16,200 |
= = = 0.3200 = 32 cM Incorrect E. distance = | ½(5,580) + 3×(135 + 285) |
| 16,200 |
= = = 0.2500 = 25 cM Incorrect
MC e661_b807
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is affiliated with the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene C is linked with the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene J is associated with the 'jeweled' phenotype. A budding yeast that is homozygous recessive for Gene J colonies appear dotted with tiny, iridescent spots that sparkle under light.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 60 |
| 2 | | | | | 44 |
| 3 | | | | | 1,176 |
| 4 | | | | | 121 |
| 5 | | | | | 1,914 |
| 6 | | | | | 2,685 |
| TOTAL = | 6,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and J
A. distance = | ½(1,176 + 1,914) + 3×(44 + 121) |
| 6,000 |
= = = 0.3400 = 34 cM Incorrect B. distance = | ½(1,914) + 3×(60 + 121) |
| 6,000 |
= = = 0.2500 = 25 cM Correct C. distance = = = = 0.1200 = 12 cM Incorrect D. distance = = = = 0.2200 = 22 cM Incorrect E. distance = | ½(1,176 + 1,914) + 3×(44 + 60 + 121) |
| 6,000 |
= = = 0.3700 = 37 cM Incorrect
MC 9a67_c6f9
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is analogous to the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene F is associated with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene M is related to the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 16 |
| 2 | | | | | 2,016 |
| 3 | | | | | 182 |
| 4 | | | | | 6,396 |
| 5 | | | | | 32 |
| 6 | | | | | 10,558 |
| TOTAL = | 19,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes C and F
A. distance = | ½(2,016 + 6,396) + 3×(16 + 182) |
| 19,200 |
= = = 0.2500 = 25 cM Correct B. distance = | ½(2,016 + 6,396) + 3×(16 + 32 + 182) |
| 19,200 |
= = = 0.2550 = 25.50 cM Incorrect C. distance = | ½(6,396) + 3×(32 + 182) |
| 19,200 |
= = = 0.2000 = 20 cM Incorrect D. distance = | ½(6,396) + 3×(182) |
| 19,200 |
= = = 0.1950 = 19.50 cM Incorrect E. distance = | ½(2,016) + 3×(16 + 32) |
| 19,200 |
= = = 0.0600 = 6 cM Incorrect
MC c33a_90b8
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene E is affiliated with the 'elephant' phenotype. A budding yeast that is homozygous recessive for Gene E cells absorb excessive amounts of liquid, resulting in giant, swollen cells.
- Gene W is analogous to the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
- Gene Y is linked with the 'yolk' phenotype. A budding yeast that is homozygous recessive for Gene Y cells develop a dense, yellowish core that resembles an egg yolk when viewed under a microscope.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 467 |
| 2 | | | | | 7,782 |
| 3 | | | | | 9,234 |
| 4 | | | | | 3,990 |
| 5 | | | | | 91 |
| 6 | | | | | 36 |
| TOTAL = | 21,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes E and W
A. distance = | ½(7,782) + 3×(36 + 467) |
| 21,600 |
= = = 0.2500 = 25 cM Incorrect B. distance = | ½(3,990) + 3×(36 + 91) |
| 21,600 |
= = = 0.1100 = 11 cM Correct C. distance = | ½(7,782) + 3×(467) |
| 21,600 |
= = = 0.2450 = 24.50 cM Incorrect D. distance = | ½(3,990 + 7,782) + 3×(91 + 467) |
| 21,600 |
= = = 0.3500 = 35 cM Incorrect E. distance = = = = 0.1050 = 10.50 cM Incorrect
MC c189_508a
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene D is connected with the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene J is linked with the 'jeweled' phenotype. A budding yeast that is homozygous recessive for Gene J colonies appear dotted with tiny, iridescent spots that sparkle under light.
- Gene X is related to the 'xenon' phenotype. A budding yeast that is homozygous recessive for Gene X cells emit a faint glow under UV light, as if they were fluorescent.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 246 |
| 2 | | | | | 286 |
| 3 | | | | | 5,664 |
| 4 | | | | | 11,810 |
| 5 | | | | | 6,240 |
| 6 | | | | | 354 |
| TOTAL = | 24,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes D and X
A. distance = | ½(6,240) + 3×(354) |
| 24,600 |
= = = 0.1700 = 17 cM Incorrect B. distance = | ½(5,664) + 3×(286) |
| 24,600 |
= = = 0.1500 = 15 cM Incorrect C. distance = | ½(5,664 + 6,240) + 3×(286 + 354) |
| 24,600 |
= = = 0.3200 = 32 cM Correct D. distance = | ½(5,664) + 3×(246 + 286) |
| 24,600 |
= = = 0.1800 = 18 cM Incorrect E. distance = | ½(5,664 + 6,240) + 3×(246 + 286 + 354) |
| 24,600 |
= = = 0.3500 = 35 cM Incorrect
MC 890c_d89b
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is connected with the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene N is affiliated with the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
- Gene P is correlated with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 47 |
| 2 | | | | | 1,446 |
| 3 | | | | | 1,683 |
| 4 | | | | | 408 |
| 5 | | | | | 12 |
| 6 | | | | | 4 |
| TOTAL = | 3,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes C and N
A. distance = | ½(408 + 1,446) + 3×(4 + 12 + 47) |
| 3,600 |
= = = 0.3100 = 31 cM Incorrect B. distance = | ½(408) + 3×(4 + 12) |
| 3,600 |
= = = 0.0700 = 7 cM Correct C. distance = = = = 0.2400 = 24 cM Incorrect D. distance = | ½(408 + 1,446) + 3×(4 + 47) |
| 3,600 |
= = = 0.3000 = 30 cM Incorrect E. distance = = = = 0.0600 = 6 cM Incorrect
MC 77d6_afce
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is affiliated with the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene B is linked with the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene P is linked with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 236 |
| 2 | | | | | 253 |
| 3 | | | | | 4,146 |
| 4 | | | | | 396 |
| 5 | | | | | 5,058 |
| 6 | | | | | 7,611 |
| TOTAL = | 17,700 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes B and P
A. distance = | ½(4,146 + 5,058) + 3×(236 + 253 + 396) |
| 17,700 |
= = = 0.4100 = 41 cM Incorrect B. distance = | ½(5,058) + 3×(236 + 396) |
| 17,700 |
= = = 0.2500 = 25 cM Incorrect C. distance = | ½(4,146) + 3×(253) |
| 17,700 |
= = = 0.1600 = 16 cM Incorrect D. distance = | ½(4,146) + 3×(236 + 253) |
| 17,700 |
= = = 0.2000 = 20 cM Incorrect E. distance = | ½(4,146 + 5,058) + 3×(253 + 396) |
| 17,700 |
= = = 0.3700 = 37 cM Correct
MC 45e0_45bb
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is analogous to the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene F is connected with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene K is affiliated with the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 258 |
| 2 | | | | | 4,932 |
| 3 | | | | | 9,000 |
| 4 | | | | | 3,564 |
| 5 | | | | | 120 |
| 6 | | | | | 126 |
| TOTAL = | 18,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes C and K
A. distance = | ½(4,932) + 3×(258) |
| 18,000 |
= = = 0.1800 = 18 cM Incorrect B. distance = | ½(3,564 + 4,932) + 3×(126 + 258) |
| 18,000 |
= = = 0.3000 = 30 cM Incorrect C. distance = | ½(3,564 + 4,932) + 3×(120 + 126 + 258) |
| 18,000 |
= = = 0.3200 = 32 cM Incorrect D. distance = | ½(4,932) + 3×(120 + 258) |
| 18,000 |
= = = 0.2000 = 20 cM Correct E. distance = | ½(3,564) + 3×(120 + 126) |
| 18,000 |
= = = 0.1400 = 14 cM Incorrect
MC 02f0_0b40
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene B is connected with the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene F is related to the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene H is related to the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 589 |
| 2 | | | | | 8,592 |
| 3 | | | | | 185 |
| 4 | | | | | 5,340 |
| 5 | | | | | 10,965 |
| 6 | | | | | 129 |
| TOTAL = | 25,800 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes F and H
A. distance = | ½(5,340) + 3×(129 + 185) |
| 25,800 |
= = = 0.1400 = 14 cM Incorrect B. distance = | ½(5,340 + 8,592) + 3×(185 + 589) |
| 25,800 |
= = = 0.3600 = 36 cM Incorrect C. distance = | ½(5,340 + 8,592) + 3×(129 + 185 + 589) |
| 25,800 |
= = = 0.3750 = 37.50 cM Incorrect D. distance = | ½(5,340) + 3×(185) |
| 25,800 |
= = = 0.1250 = 12.50 cM Incorrect E. distance = | ½(8,592) + 3×(129 + 589) |
| 25,800 |
= = = 0.2500 = 25 cM Correct
MC 695e_707c
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is connected with the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene D is connected with the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene J is affiliated with the 'jeweled' phenotype. A budding yeast that is homozygous recessive for Gene J colonies appear dotted with tiny, iridescent spots that sparkle under light.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 19 |
| 2 | | | | | 48 |
| 3 | | | | | 2,106 |
| 4 | | | | | 4,876 |
| 5 | | | | | 4,104 |
| 6 | | | | | 247 |
| TOTAL = | 11,400 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes C and D
A. distance = | ½(4,104) + 3×(19 + 247) |
| 11,400 |
= = = 0.2500 = 25 cM Incorrect B. distance = | ½(2,106) + 3×(19 + 48) |
| 11,400 |
= = = 0.1100 = 11 cM Correct C. distance = | ½(2,106 + 4,104) + 3×(48 + 247) |
| 11,400 |
= = = 0.3500 = 35 cM Incorrect D. distance = | ½(2,106) + 3×(48 + 247) |
| 11,400 |
= = = 0.1700 = 17 cM Incorrect E. distance = | ½(2,106 + 4,104) + 3×(19 + 48) |
| 11,400 |
= = = 0.2900 = 29 cM Incorrect
MC fc2f_4ddd
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene F is correlated with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene M is associated with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene X is affiliated with the 'xenon' phenotype. A budding yeast that is homozygous recessive for Gene X cells emit a faint glow under UV light, as if they were fluorescent.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 76 |
| 2 | | | | | 996 |
| 3 | | | | | 40 |
| 4 | | | | | 750 |
| 5 | | | | | 1,405 |
| 6 | | | | | 33 |
| TOTAL = | 3,300 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes F and X
A. distance = = = = 0.1500 = 15 cM Incorrect B. distance = | ½(750 + 996) + 3×(33 + 40 + 76) |
| 3,300 |
= = = 0.4000 = 40 cM Incorrect C. distance = = = = 0.2200 = 22 cM Incorrect D. distance = | ½(750) + 3×(33 + 40) |
| 3,300 |
= = = 0.1800 = 18 cM Correct E. distance = | ½(996) + 3×(33 + 76) |
| 3,300 |
= = = 0.2500 = 25 cM Incorrect
MC 84fa_7e2e
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene K is analogous to the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
- Gene T is linked with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
- Gene W is associated with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 9 |
| 2 | | | | | 1,176 |
| 3 | | | | | 41 |
| 4 | | | | | 6,890 |
| 5 | | | | | 4,086 |
| 6 | | | | | 98 |
| TOTAL = | 12,300 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes K and T
A. distance = | ½(1,176 + 4,086) + 3×(9 + 41 + 98) |
| 12,300 |
= = = 0.2500 = 25 cM Incorrect B. distance = | ½(1,176 + 4,086) + 3×(9 + 98) |
| 12,300 |
= = = 0.2400 = 24 cM Incorrect C. distance = | ½(1,176) + 3×(9 + 41) |
| 12,300 |
= = = 0.0600 = 6 cM Correct D. distance = = = = 0.1900 = 19 cM Incorrect E. distance = | ½(4,086) + 3×(41 + 98) |
| 12,300 |
= = = 0.2000 = 20 cM Incorrect
MC 79ba_68d6
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is correlated with the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene R is linked with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
- Gene Y is related to the 'yolk' phenotype. A budding yeast that is homozygous recessive for Gene Y cells develop a dense, yellowish core that resembles an egg yolk when viewed under a microscope.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 546 |
| 2 | | | | | 6,528 |
| 3 | | | | | 9,405 |
| 4 | | | | | 5,670 |
| 5 | | | | | 266 |
| 6 | | | | | 385 |
| TOTAL = | 22,800 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and Y
A. distance = | ½(5,670 + 6,528) + 3×(385 + 546) |
| 22,800 |
= = = 0.3900 = 39 cM Incorrect B. distance = | ½(5,670) + 3×(266 + 385) |
| 22,800 |
= = = 0.2100 = 21 cM Incorrect C. distance = | ½(5,670 + 6,528) + 3×(266 + 385 + 546) |
| 22,800 |
= = = 0.4250 = 42.50 cM Incorrect D. distance = | ½(6,528) + 3×(546) |
| 22,800 |
= = = 0.2150 = 21.50 cM Incorrect E. distance = | ½(6,528) + 3×(266 + 546) |
| 22,800 |
= = = 0.2500 = 25 cM Correct
MC a89e_a2da
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is affiliated with the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene F is correlated with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene K is affiliated with the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 133 |
| 2 | | | | | 3,594 |
| 3 | | | | | 183 |
| 4 | | | | | 8,188 |
| 5 | | | | | 5,832 |
| 6 | | | | | 370 |
| TOTAL = | 18,300 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes C and K
A. distance = | ½(3,594 + 5,832) + 3×(133 + 183 + 370) |
| 18,300 |
= = = 0.3700 = 37 cM Incorrect B. distance = | ½(3,594 + 5,832) + 3×(133 + 370) |
| 18,300 |
= = = 0.3400 = 34 cM Incorrect C. distance = | ½(5,832) + 3×(183 + 370) |
| 18,300 |
= = = 0.2500 = 25 cM Correct D. distance = | ½(3,594) + 3×(133 + 183) |
| 18,300 |
= = = 0.1500 = 15 cM Incorrect E. distance = | ½(5,832) + 3×(370) |
| 18,300 |
= = = 0.2200 = 22 cM Incorrect
MC 2042_f680
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene H is associated with the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
- Gene M is linked with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene X is linked with the 'xenon' phenotype. A budding yeast that is homozygous recessive for Gene X cells emit a faint glow under UV light, as if they were fluorescent.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 26 |
| 2 | | | | | 27 |
| 3 | | | | | 2,178 |
| 4 | | | | | 7,020 |
| 5 | | | | | 6,090 |
| 6 | | | | | 259 |
| TOTAL = | 15,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes M and X
A. distance = | ½(2,178) + 3×(26 + 27) |
| 15,600 |
= = = 0.0800 = 8 cM Incorrect B. distance = | ½(2,178 + 6,090) + 3×(27 + 259) |
| 15,600 |
= = = 0.3200 = 32 cM Incorrect C. distance = | ½(6,090) + 3×(26 + 259) |
| 15,600 |
= = = 0.2500 = 25 cM Correct D. distance = | ½(2,178 + 6,090) + 3×(26) |
| 15,600 |
= = = 0.2700 = 27 cM Incorrect E. distance = | ½(2,178 + 6,090) + 3×(26 + 27 + 259) |
| 15,600 |
= = = 0.3250 = 32.50 cM Incorrect
MC f22b_ff5d
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene D is related to the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene E is affiliated with the 'elephant' phenotype. A budding yeast that is homozygous recessive for Gene E cells absorb excessive amounts of liquid, resulting in giant, swollen cells.
- Gene P is correlated with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 259 |
| 2 | | | | | 5,496 |
| 3 | | | | | 6,825 |
| 4 | | | | | 2,304 |
| 5 | | | | | 41 |
| 6 | | | | | 75 |
| TOTAL = | 15,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes E and P
A. distance = | ½(2,304 + 5,496) + 3×(41 + 259) |
| 15,000 |
= = = 0.3200 = 32 cM Incorrect B. distance = | ½(5,496) + 3×(75 + 259) |
| 15,000 |
= = = 0.2500 = 25 cM Correct C. distance = | ½(2,304 + 5,496) + 3×(75) |
| 15,000 |
= = = 0.2750 = 27.50 cM Incorrect D. distance = = = = 0.0850 = 8.50 cM Incorrect E. distance = | ½(2,304) + 3×(41 + 75) |
| 15,000 |
= = = 0.1000 = 10 cM Incorrect
MC 2635_5b76
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene B is correlated with the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene T is linked with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
- Gene Y is associated with the 'yolk' phenotype. A budding yeast that is homozygous recessive for Gene Y cells develop a dense, yellowish core that resembles an egg yolk when viewed under a microscope.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 11 |
| 2 | | | | | 3 |
| 3 | | | | | 576 |
| 4 | | | | | 52 |
| 5 | | | | | 2,262 |
| 6 | | | | | 3,696 |
| TOTAL = | 6,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes B and T
A. distance = | ½(576 + 2,262) + 3×(11) |
| 6,600 |
= = = 0.2200 = 22 cM Incorrect B. distance = | ½(576 + 2,262) + 3×(3 + 52) |
| 6,600 |
= = = 0.2400 = 24 cM Incorrect C. distance = | ½(576 + 2,262) + 3×(3 + 11 + 52) |
| 6,600 |
= = = 0.2450 = 24.50 cM Incorrect D. distance = | ½(576) + 3×(3 + 11) |
| 6,600 |
= = = 0.0500 = 5 cM Correct E. distance = | ½(2,262) + 3×(11 + 52) |
| 6,600 |
= = = 0.2000 = 20 cM Incorrect
MC b403_15bf
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene M is connected with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene N is connected with the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
- Gene P is connected with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 100 |
| 2 | | | | | 4,104 |
| 3 | | | | | 4,536 |
| 4 | | | | | 840 |
| 5 | | | | | 4 |
| 6 | | | | | 16 |
| TOTAL = | 9,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes M and P
A. distance = | ½(840 + 4,104) + 3×(4 + 100) |
| 9,600 |
= = = 0.2900 = 29 cM Correct B. distance = | ½(840) + 3×(4 + 16) |
| 9,600 |
= = = 0.0500 = 5 cM Incorrect C. distance = | ½(4,104) + 3×(16 + 100) |
| 9,600 |
= = = 0.2500 = 25 cM Incorrect D. distance = | ½(840) + 3×(16 + 100) |
| 9,600 |
= = = 0.0800 = 8 cM Incorrect E. distance = | ½(4,104) + 3×(4 + 16) |
| 9,600 |
= = = 0.2200 = 22 cM Incorrect
MC cab8_4907
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is related to the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene R is correlated with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
- Gene X is affiliated with the 'xenon' phenotype. A budding yeast that is homozygous recessive for Gene X cells emit a faint glow under UV light, as if they were fluorescent.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 134 |
| 2 | | | | | 2,862 |
| 3 | | | | | 22 |
| 4 | | | | | 1,194 |
| 5 | | | | | 3,549 |
| 6 | | | | | 39 |
| TOTAL = | 7,800 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes C and X
A. distance = | ½(1,194 + 2,862) + 3×(22 + 134) |
| 7,800 |
= = = 0.3200 = 32 cM Incorrect B. distance = = = = 0.0850 = 8.50 cM Incorrect C. distance = | ½(1,194) + 3×(22 + 39) |
| 7,800 |
= = = 0.1000 = 10 cM Correct D. distance = | ½(2,862) + 3×(39 + 134) |
| 7,800 |
= = = 0.2500 = 25 cM Incorrect E. distance = | ½(1,194 + 2,862) + 3×(22 + 39 + 134) |
| 7,800 |
= = = 0.3350 = 33.50 cM Incorrect
MC 974f_3312
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is associated with the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene M is connected with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene T is analogous to the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 144 |
| 2 | | | | | 1,716 |
| 3 | | | | | 2,475 |
| 4 | | | | | 1,494 |
| 5 | | | | | 70 |
| 6 | | | | | 101 |
| TOTAL = | 6,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and T
A. distance = | ½(1,494 + 1,716) + 3×(101 + 144) |
| 6,000 |
= = = 0.3900 = 39 cM Incorrect B. distance = | ½(1,716) + 3×(70 + 144) |
| 6,000 |
= = = 0.2500 = 25 cM Correct C. distance = | ½(1,494) + 3×(70 + 101) |
| 6,000 |
= = = 0.2100 = 21 cM Incorrect D. distance = | ½(1,494) + 3×(101 + 144) |
| 6,000 |
= = = 0.2470 = 24.70 cM Incorrect E. distance = | ½(1,494) + 3×(70 + 101 + 144) |
| 6,000 |
= = = 0.2820 = 28.20 cM Incorrect
MC 4213_85d4
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene B is analogous to the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene H is correlated with the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
- Gene Y is associated with the 'yolk' phenotype. A budding yeast that is homozygous recessive for Gene Y cells develop a dense, yellowish core that resembles an egg yolk when viewed under a microscope.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 97 |
| 2 | | | | | 1,476 |
| 3 | | | | | 1,764 |
| 4 | | | | | 834 |
| 5 | | | | | 22 |
| 6 | | | | | 7 |
| TOTAL = | 4,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes B and H
A. distance = = = = 0.1150 = 11.50 cM Incorrect B. distance = | ½(834 + 1,476) + 3×(22 + 97) |
| 4,200 |
= = = 0.3600 = 36 cM Incorrect C. distance = | ½(1,476) + 3×(7 + 97) |
| 4,200 |
= = = 0.2500 = 25 cM Incorrect D. distance = | ½(834 + 1,476) + 3×(7) |
| 4,200 |
= = = 0.2800 = 28 cM Incorrect E. distance = | ½(834) + 3×(7 + 22) |
| 4,200 |
= = = 0.1200 = 12 cM Correct
MC ccea_77a7
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is related to the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene J is linked with the 'jeweled' phenotype. A budding yeast that is homozygous recessive for Gene J colonies appear dotted with tiny, iridescent spots that sparkle under light.
- Gene N is connected with the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 526 |
| 2 | | | | | 6,876 |
| 3 | | | | | 9,690 |
| 4 | | | | | 5,208 |
| 5 | | | | | 272 |
| 6 | | | | | 228 |
| TOTAL = | 22,800 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes J and N
A. distance = | ½(5,208) + 3×(228 + 272) |
| 22,800 |
= = = 0.1800 = 18 cM Incorrect B. distance = | ½(5,208) + 3×(272) |
| 22,800 |
= = = 0.1500 = 15 cM Incorrect C. distance = | ½(6,876) + 3×(526) |
| 22,800 |
= = = 0.2200 = 22 cM Incorrect D. distance = | ½(5,208 + 6,876) + 3×(272 + 526) |
| 22,800 |
= = = 0.3700 = 37 cM Incorrect E. distance = | ½(6,876) + 3×(228 + 526) |
| 22,800 |
= = = 0.2500 = 25 cM Correct
MC 9d22_383c
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is correlated with the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene H is connected with the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
- Gene P is related to the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 204 |
| 2 | | | | | 5,160 |
| 3 | | | | | 12,994 |
| 4 | | | | | 4,236 |
| 5 | | | | | 130 |
| 6 | | | | | 76 |
| TOTAL = | 22,800 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes C and H
A. distance = | ½(4,236 + 5,160) + 3×(76 + 130 + 204) |
| 22,800 |
= = = 0.2600 = 26 cM Incorrect B. distance = | ½(5,160) + 3×(76 + 204) |
| 22,800 |
= = = 0.1500 = 15 cM Incorrect C. distance = | ½(5,160) + 3×(204) |
| 22,800 |
= = = 0.1400 = 14 cM Incorrect D. distance = | ½(4,236 + 5,160) + 3×(130 + 204) |
| 22,800 |
= = = 0.2500 = 25 cM Incorrect E. distance = | ½(4,236) + 3×(76 + 130) |
| 22,800 |
= = = 0.1200 = 12 cM Correct
MC 623c_8df6
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene N is analogous to the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
- Gene P is correlated with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
- Gene Y is connected with the 'yolk' phenotype. A budding yeast that is homozygous recessive for Gene Y cells develop a dense, yellowish core that resembles an egg yolk when viewed under a microscope.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 47 |
| 2 | | | | | 2,364 |
| 3 | | | | | 49 |
| 4 | | | | | 227 |
| 5 | | | | | 4,812 |
| 6 | | | | | 7,201 |
| TOTAL = | 14,700 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes P and Y
A. distance = | ½(2,364 + 4,812) + 3×(47 + 227) |
| 14,700 |
= = = 0.3000 = 30 cM Incorrect B. distance = = = = 0.0900 = 9 cM Incorrect C. distance = | ½(4,812) + 3×(49 + 227) |
| 14,700 |
= = = 0.2200 = 22 cM Correct D. distance = | ½(2,364 + 4,812) + 3×(47 + 49 + 227) |
| 14,700 |
= = = 0.3100 = 31 cM Incorrect E. distance = | ½(2,364) + 3×(47 + 49) |
| 14,700 |
= = = 0.1000 = 10 cM Incorrect
MC 1df8_f791
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene E is analogous to the 'elephant' phenotype. A budding yeast that is homozygous recessive for Gene E cells absorb excessive amounts of liquid, resulting in giant, swollen cells.
- Gene H is related to the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
- Gene P is analogous to the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 87 |
| 2 | | | | | 2,976 |
| 3 | | | | | 53 |
| 4 | | | | | 242 |
| 5 | | | | | 4,590 |
| 6 | | | | | 7,952 |
| TOTAL = | 15,900 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes H and P
A. distance = | ½(2,976 + 4,590) + 3×(53 + 87 + 242) |
| 15,900 |
= = = 0.3100 = 31 cM Incorrect B. distance = | ½(2,976 + 4,590) + 3×(87 + 242) |
| 15,900 |
= = = 0.3000 = 30 cM Incorrect C. distance = | ½(4,590) + 3×(242) |
| 15,900 |
= = = 0.1900 = 19 cM Incorrect D. distance = = = = 0.1100 = 11 cM Incorrect E. distance = | ½(4,590) + 3×(53 + 242) |
| 15,900 |
= = = 0.2000 = 20 cM Correct
MC e174_d1cf
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene H is linked with the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
- Gene K is related to the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
- Gene M is correlated with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 105 |
| 2 | | | | | 77 |
| 3 | | | | | 3,528 |
| 4 | | | | | 9,400 |
| 5 | | | | | 7,494 |
| 6 | | | | | 396 |
| TOTAL = | 21,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes H and M
A. distance = | ½(3,528) + 3×(77 + 105) |
| 21,000 |
= = = 0.1100 = 11 cM Incorrect B. distance = | ½(3,528 + 7,494) + 3×(77 + 396) |
| 21,000 |
= = = 0.3300 = 33 cM Correct C. distance = | ½(7,494) + 3×(105 + 396) |
| 21,000 |
= = = 0.2500 = 25 cM Incorrect D. distance = | ½(7,494) + 3×(396) |
| 21,000 |
= = = 0.2350 = 23.50 cM Incorrect E. distance = = = = 0.0950 = 9.50 cM Incorrect
MC 23bb_b146
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is linked with the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene D is analogous to the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene R is correlated with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 6 |
| 2 | | | | | 1,854 |
| 3 | | | | | 45 |
| 4 | | | | | 12,960 |
| 5 | | | | | 11,916 |
| 6 | | | | | 219 |
| TOTAL = | 27,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes C and R
A. distance = | ½(1,854 + 11,916) + 3×(45) |
| 27,000 |
= = = 0.2600 = 26 cM Incorrect B. distance = = = = 0.0350 = 3.50 cM Incorrect C. distance = | ½(11,916) + 3×(45 + 219) |
| 27,000 |
= = = 0.2500 = 25 cM Correct D. distance = | ½(1,854) + 3×(6 + 45) |
| 27,000 |
= = = 0.0400 = 4 cM Incorrect E. distance = | ½(1,854 + 11,916) + 3×(6 + 219) |
| 27,000 |
= = = 0.2800 = 28 cM Incorrect
MC 00d9_58a4
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene F is linked with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene M is affiliated with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene N is correlated with the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 47 |
| 2 | | | | | 1,176 |
| 3 | | | | | 27 |
| 4 | | | | | 2,320 |
| 5 | | | | | 1,710 |
| 6 | | | | | 120 |
| TOTAL = | 5,400 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes M and N
A. distance = | ½(1,176 + 1,710) + 3×(27 + 47 + 120) |
| 5,400 |
= = = 0.3750 = 37.50 cM Incorrect B. distance = | ½(1,176 + 1,710) + 3×(47 + 120) |
| 5,400 |
= = = 0.3600 = 36 cM Correct C. distance = = = = 0.1350 = 13.50 cM Incorrect D. distance = | ½(1,176) + 3×(27 + 47) |
| 5,400 |
= = = 0.1500 = 15 cM Incorrect E. distance = | ½(1,710) + 3×(27 + 120) |
| 5,400 |
= = = 0.2400 = 24 cM Incorrect
MC 5787_855e
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene B is correlated with the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene K is correlated with the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
- Gene P is analogous to the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 21 |
| 2 | | | | | 7,183 |
| 3 | | | | | 864 |
| 4 | | | | | 4,452 |
| 5 | | | | | 3 |
| 6 | | | | | 77 |
| TOTAL = | 12,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes B and K
A. distance = | ½(864) + 3×(3 + 21) |
| 12,600 |
= = = 0.0400 = 4 cM Incorrect B. distance = = = = 0.0350 = 3.50 cM Incorrect C. distance = | ½(864 + 4,452) + 3×(3 + 77) |
| 12,600 |
= = = 0.2300 = 23 cM Correct D. distance = | ½(864 + 4,452) + 3×(3 + 21 + 77) |
| 12,600 |
= = = 0.2350 = 23.50 cM Incorrect E. distance = | ½(4,452) + 3×(21 + 77) |
| 12,600 |
= = = 0.2000 = 20 cM Incorrect
MC cc3a_e583
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is connected with the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene D is associated with the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene R is related to the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 437 |
| 2 | | | | | 5,568 |
| 3 | | | | | 8,190 |
| 4 | | | | | 4,806 |
| 5 | | | | | 304 |
| 6 | | | | | 195 |
| TOTAL = | 19,500 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes D and R
A. distance = | ½(4,806 + 5,568) + 3×(304 + 437) |
| 19,500 |
= = = 0.3800 = 38 cM Incorrect B. distance = | ½(5,568) + 3×(437) |
| 19,500 |
= = = 0.2100 = 21 cM Incorrect C. distance = | ½(4,806) + 3×(195 + 304) |
| 19,500 |
= = = 0.2000 = 20 cM Incorrect D. distance = | ½(4,806) + 3×(304) |
| 19,500 |
= = = 0.1700 = 17 cM Incorrect E. distance = | ½(5,568) + 3×(195 + 437) |
| 19,500 |
= = = 0.2400 = 24 cM Correct
MC 2a40_0462
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene D is connected with the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene E is affiliated with the 'elephant' phenotype. A budding yeast that is homozygous recessive for Gene E cells absorb excessive amounts of liquid, resulting in giant, swollen cells.
- Gene T is linked with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 16 |
| 2 | | | | | 98 |
| 3 | | | | | 2,196 |
| 4 | | | | | 4,416 |
| 5 | | | | | 2,700 |
| 6 | | | | | 174 |
| TOTAL = | 9,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes E and T
A. distance = | ½(2,196) + 3×(16 + 98) |
| 9,600 |
= = = 0.1500 = 15 cM Incorrect B. distance = | ½(2,196 + 2,700) + 3×(16) |
| 9,600 |
= = = 0.2600 = 26 cM Incorrect C. distance = | ½(2,196 + 2,700) + 3×(98 + 174) |
| 9,600 |
= = = 0.3400 = 34 cM Incorrect D. distance = | ½(2,700) + 3×(16 + 174) |
| 9,600 |
= = = 0.2000 = 20 cM Correct E. distance = = = = 0.1450 = 14.50 cM Incorrect
MC 4522_2116
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene D is correlated with the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene E is associated with the 'elephant' phenotype. A budding yeast that is homozygous recessive for Gene E cells absorb excessive amounts of liquid, resulting in giant, swollen cells.
- Gene R is affiliated with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 53 |
| 2 | | | | | 56 |
| 3 | | | | | 1,572 |
| 4 | | | | | 114 |
| 5 | | | | | 2,178 |
| 6 | | | | | 3,977 |
| TOTAL = | 7,950 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes D and R
A. distance = | ½(1,572 + 2,178) + 3×(56 + 114) |
| 7,950 |
= = = 0.3000 = 30 cM Incorrect B. distance = = = = 0.1800 = 18 cM Incorrect C. distance = | ½(1,572 + 2,178) + 3×(53 + 56 + 114) |
| 7,950 |
= = = 0.3200 = 32 cM Incorrect D. distance = | ½(2,178) + 3×(53 + 114) |
| 7,950 |
= = = 0.2000 = 20 cM Correct E. distance = = = = 0.1200 = 12 cM Incorrect
MC 8b61_ba1d
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is related to the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene F is connected with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene T is associated with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 46 |
| 2 | | | | | 66 |
| 3 | | | | | 4,296 |
| 4 | | | | | 12,213 |
| 5 | | | | | 10,470 |
| 6 | | | | | 509 |
| TOTAL = | 27,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and T
A. distance = | ½(4,296 + 10,470) + 3×(66 + 509) |
| 27,600 |
= = = 0.3300 = 33 cM Correct B. distance = | ½(10,470) + 3×(509) |
| 27,600 |
= = = 0.2450 = 24.50 cM Incorrect C. distance = | ½(4,296 + 10,470) + 3×(46 + 66 + 509) |
| 27,600 |
= = = 0.3350 = 33.50 cM Incorrect D. distance = | ½(4,296) + 3×(46 + 66) |
| 27,600 |
= = = 0.0900 = 9 cM Incorrect E. distance = | ½(10,470) + 3×(46 + 509) |
| 27,600 |
= = = 0.2500 = 25 cM Incorrect
MC 0de8_160d
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene E is correlated with the 'elephant' phenotype. A budding yeast that is homozygous recessive for Gene E cells absorb excessive amounts of liquid, resulting in giant, swollen cells.
- Gene R is correlated with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
- Gene W is linked with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 602 |
| 2 | | | | | 8,232 |
| 3 | | | | | 10,521 |
| 4 | | | | | 5,502 |
| 5 | | | | | 126 |
| 6 | | | | | 217 |
| TOTAL = | 25,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes R and W
A. distance = | ½(8,232) + 3×(602) |
| 25,200 |
= = = 0.2350 = 23.50 cM Incorrect B. distance = | ½(5,502) + 3×(126 + 217) |
| 25,200 |
= = = 0.1500 = 15 cM Incorrect C. distance = | ½(5,502 + 8,232) + 3×(217 + 602) |
| 25,200 |
= = = 0.3700 = 37 cM Correct D. distance = | ½(8,232) + 3×(126 + 602) |
| 25,200 |
= = = 0.2500 = 25 cM Incorrect E. distance = | ½(5,502 + 8,232) + 3×(126 + 217 + 602) |
| 25,200 |
= = = 0.3850 = 38.50 cM Incorrect
MC de05_76f0
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene F is related to the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene M is associated with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene P is analogous to the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 45 |
| 2 | | | | | 14 |
| 3 | | | | | 2,346 |
| 4 | | | | | 16,065 |
| 5 | | | | | 8,346 |
| 6 | | | | | 184 |
| TOTAL = | 27,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes M and P
A. distance = | ½(2,346) + 3×(14 + 45) |
| 27,000 |
= = = 0.0500 = 5 cM Incorrect B. distance = | ½(2,346 + 8,346) + 3×(14 + 184) |
| 27,000 |
= = = 0.2200 = 22 cM Incorrect C. distance = | ½(2,346 + 8,346) + 3×(14 + 45 + 184) |
| 27,000 |
= = = 0.2250 = 22.50 cM Incorrect D. distance = | ½(2,346 + 8,346) + 3×(45) |
| 27,000 |
= = = 0.2030 = 20.30 cM Incorrect E. distance = | ½(8,346) + 3×(45 + 184) |
| 27,000 |
= = = 0.1800 = 18 cM Correct
MC 09ca_2dc5
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is associated with the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene E is connected with the 'elephant' phenotype. A budding yeast that is homozygous recessive for Gene E cells absorb excessive amounts of liquid, resulting in giant, swollen cells.
- Gene H is affiliated with the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 47 |
| 2 | | | | | 3,120 |
| 3 | | | | | 164 |
| 4 | | | | | 5,820 |
| 5 | | | | | 81 |
| 6 | | | | | 15,068 |
| TOTAL = | 24,300 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes C and H
A. distance = | ½(3,120 + 5,820) + 3×(47 + 164) |
| 24,300 |
= = = 0.2100 = 21 cM Incorrect B. distance = | ½(5,820) + 3×(81 + 164) |
| 24,300 |
= = = 0.1500 = 15 cM Incorrect C. distance = = = = 0.0700 = 7 cM Incorrect D. distance = | ½(3,120) + 3×(47 + 81) |
| 24,300 |
= = = 0.0800 = 8 cM Correct E. distance = | ½(3,120 + 5,820) + 3×(47 + 81 + 164) |
| 24,300 |
= = = 0.2200 = 22 cM Incorrect
MC 0457_d637
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene D is correlated with the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene F is affiliated with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene N is linked with the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 36 |
| 2 | | | | | 13,609 |
| 3 | | | | | 1,872 |
| 4 | | | | | 5,946 |
| 5 | | | | | 12 |
| 6 | | | | | 125 |
| TOTAL = | 21,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes D and F
A. distance = | ½(1,872) + 3×(12 + 36) |
| 21,600 |
= = = 0.0500 = 5 cM Incorrect B. distance = | ½(5,946) + 3×(36 + 125) |
| 21,600 |
= = = 0.1600 = 16 cM Incorrect C. distance = | ½(1,872 + 5,946) + 3×(12 + 125) |
| 21,600 |
= = = 0.2000 = 20 cM Correct D. distance = | ½(5,946) + 3×(125) |
| 21,600 |
= = = 0.1550 = 15.50 cM Incorrect E. distance = = = = 0.0450 = 4.50 cM Incorrect
MC 0ad0_630e
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene B is connected with the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene F is affiliated with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene R is connected with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 150 |
| 2 | | | | | 234 |
| 3 | | | | | 3,696 |
| 4 | | | | | 366 |
| 5 | | | | | 4,404 |
| 6 | | | | | 6,150 |
| TOTAL = | 15,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes B and R
A. distance = | ½(4,404) + 3×(150 + 366) |
| 15,000 |
= = = 0.2500 = 25 cM Correct B. distance = | ½(3,696 + 4,404) + 3×(234 + 366) |
| 15,000 |
= = = 0.3900 = 39 cM Incorrect C. distance = | ½(3,696) + 3×(150 + 234) |
| 15,000 |
= = = 0.2000 = 20 cM Incorrect D. distance = | ½(3,696 + 4,404) + 3×(150) |
| 15,000 |
= = = 0.3000 = 30 cM Incorrect E. distance = | ½(3,696 + 4,404) + 3×(150 + 234 + 366) |
| 15,000 |
= = = 0.4200 = 42 cM Incorrect
MC 1840_fc36
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene J is related to the 'jeweled' phenotype. A budding yeast that is homozygous recessive for Gene J colonies appear dotted with tiny, iridescent spots that sparkle under light.
- Gene K is linked with the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
- Gene R is related to the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 46 |
| 2 | | | | | 32 |
| 3 | | | | | 3,396 |
| 4 | | | | | 405 |
| 5 | | | | | 11,094 |
| 6 | | | | | 12,627 |
| TOTAL = | 27,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes J and K
A. distance = | ½(3,396) + 3×(32 + 46) |
| 27,600 |
= = = 0.0700 = 7 cM Correct B. distance = | ½(3,396 + 11,094) + 3×(32 + 405) |
| 27,600 |
= = = 0.3100 = 31 cM Incorrect C. distance = = = = 0.0650 = 6.50 cM Incorrect D. distance = | ½(11,094) + 3×(46 + 405) |
| 27,600 |
= = = 0.2500 = 25 cM Incorrect E. distance = | ½(11,094) + 3×(405) |
| 27,600 |
= = = 0.2450 = 24.50 cM Incorrect
MC 4b3d_779f
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is connected with the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene N is connected with the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
- Gene P is connected with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 42 |
| 2 | | | | | 29 |
| 3 | | | | | 3,102 |
| 4 | | | | | 370 |
| 5 | | | | | 10,128 |
| 6 | | | | | 11,529 |
| TOTAL = | 25,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and N
A. distance = | ½(3,102) + 3×(29 + 42) |
| 25,200 |
= = = 0.0700 = 7 cM Correct B. distance = | ½(3,102 + 10,128) + 3×(29 + 42 + 370) |
| 25,200 |
= = = 0.3150 = 31.50 cM Incorrect C. distance = | ½(10,128) + 3×(42 + 370) |
| 25,200 |
= = = 0.2500 = 25 cM Incorrect D. distance = | ½(3,102 + 10,128) + 3×(29 + 370) |
| 25,200 |
= = = 0.3100 = 31 cM Incorrect E. distance = = = = 0.0650 = 6.50 cM Incorrect
MC 3572_1c34
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is related to the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene T is affiliated with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
- Gene Y is affiliated with the 'yolk' phenotype. A budding yeast that is homozygous recessive for Gene Y cells develop a dense, yellowish core that resembles an egg yolk when viewed under a microscope.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 218 |
| 2 | | | | | 2,874 |
| 3 | | | | | 139 |
| 4 | | | | | 2,328 |
| 5 | | | | | 4,488 |
| 6 | | | | | 153 |
| TOTAL = | 10,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and T
A. distance = | ½(2,328) + 3×(139) |
| 10,200 |
= = = 0.1550 = 15.50 cM Incorrect B. distance = | ½(2,328 + 2,874) + 3×(153) |
| 10,200 |
= = = 0.3000 = 30 cM Incorrect C. distance = | ½(2,874) + 3×(153 + 218) |
| 10,200 |
= = = 0.2500 = 25 cM Incorrect D. distance = | ½(2,328) + 3×(139 + 153) |
| 10,200 |
= = = 0.2000 = 20 cM Incorrect E. distance = | ½(2,328 + 2,874) + 3×(139 + 218) |
| 10,200 |
= = = 0.3600 = 36 cM Correct
MC fc5a_a19b
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene A is correlated with the 'amber' phenotype. A budding yeast that is homozygous recessive for Gene A cells develop a rich yellow-orange pigmentation, giving the colony a warm, amber hue.
- Gene D is analogous to the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene F is affiliated with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 323 |
| 2 | | | | | 4,542 |
| 3 | | | | | 6,192 |
| 4 | | | | | 3,090 |
| 5 | | | | | 120 |
| 6 | | | | | 133 |
| TOTAL = | 14,400 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes A and F
A. distance = | ½(3,090 + 4,542) + 3×(120) |
| 14,400 |
= = = 0.2900 = 29 cM Incorrect B. distance = | ½(4,542) + 3×(120 + 323) |
| 14,400 |
= = = 0.2500 = 25 cM Correct C. distance = | ½(3,090 + 4,542) + 3×(133 + 323) |
| 14,400 |
= = = 0.3600 = 36 cM Incorrect D. distance = | ½(3,090) + 3×(133) |
| 14,400 |
= = = 0.1350 = 13.50 cM Incorrect E. distance = | ½(3,090) + 3×(120 + 133) |
| 14,400 |
= = = 0.1600 = 16 cM Incorrect
MC e607_318b
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is connected with the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene M is connected with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene T is affiliated with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 90 |
| 2 | | | | | 259 |
| 3 | | | | | 6,006 |
| 4 | | | | | 461 |
| 5 | | | | | 7,494 |
| 6 | | | | | 12,690 |
| TOTAL = | 27,000 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes M and T
A. distance = | ½(6,006) + 3×(259) |
| 27,000 |
= = = 0.1400 = 14 cM Incorrect B. distance = | ½(6,006 + 7,494) + 3×(259 + 461) |
| 27,000 |
= = = 0.3300 = 33 cM Correct C. distance = | ½(6,006 + 7,494) + 3×(90) |
| 27,000 |
= = = 0.2600 = 26 cM Incorrect D. distance = | ½(6,006) + 3×(90 + 259) |
| 27,000 |
= = = 0.1500 = 15 cM Incorrect E. distance = | ½(7,494) + 3×(461) |
| 27,000 |
= = = 0.1900 = 19 cM Incorrect
MC 7f14_878c
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene P is linked with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
- Gene T is related to the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
- Gene W is affiliated with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 31 |
| 2 | | | | | 1,686 |
| 3 | | | | | 148 |
| 4 | | | | | 3,792 |
| 5 | | | | | 78 |
| 6 | | | | | 5,965 |
| TOTAL = | 11,700 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes P and W
A. distance = = = = 0.0800 = 8 cM Incorrect B. distance = | ½(1,686) + 3×(31 + 78) |
| 11,700 |
= = = 0.1000 = 10 cM Correct C. distance = | ½(3,792) + 3×(31 + 78) |
| 11,700 |
= = = 0.1900 = 19 cM Incorrect D. distance = | ½(1,686) + 3×(148) |
| 11,700 |
= = = 0.1100 = 11 cM Incorrect E. distance = | ½(3,792) + 3×(148) |
| 11,700 |
= = = 0.2000 = 20 cM Incorrect
MC 8d9e_3672
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene W is correlated with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
- Gene X is related to the 'xenon' phenotype. A budding yeast that is homozygous recessive for Gene X cells emit a faint glow under UV light, as if they were fluorescent.
- Gene Y is affiliated with the 'yolk' phenotype. A budding yeast that is homozygous recessive for Gene Y cells develop a dense, yellowish core that resembles an egg yolk when viewed under a microscope.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 350 |
| 2 | | | | | 6,216 |
| 3 | | | | | 431 |
| 4 | | | | | 6,738 |
| 5 | | | | | 126 |
| 6 | | | | | 11,339 |
| TOTAL = | 25,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes X and Y
A. distance = | ½(6,216 + 6,738) + 3×(126 + 350 + 431) |
| 25,200 |
= = = 0.3650 = 36.50 cM Incorrect B. distance = | ½(6,216 + 6,738) + 3×(350 + 431) |
| 25,200 |
= = = 0.3500 = 35 cM Incorrect C. distance = | ½(6,738) + 3×(431) |
| 25,200 |
= = = 0.1850 = 18.50 cM Incorrect D. distance = | ½(6,216) + 3×(126 + 350) |
| 25,200 |
= = = 0.1800 = 18 cM Incorrect E. distance = | ½(6,738) + 3×(126 + 431) |
| 25,200 |
= = = 0.2000 = 20 cM Correct
MC b761_a1c0
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene D is associated with the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene N is correlated with the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
- Gene Y is correlated with the 'yolk' phenotype. A budding yeast that is homozygous recessive for Gene Y cells develop a dense, yellowish core that resembles an egg yolk when viewed under a microscope.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 84 |
| 2 | | | | | 3,372 |
| 3 | | | | | 188 |
| 4 | | | | | 4,788 |
| 5 | | | | | 34 |
| 6 | | | | | 11,934 |
| TOTAL = | 20,400 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes N and Y
A. distance = | ½(4,788) + 3×(34 + 188) |
| 20,400 |
= = = 0.1500 = 15 cM Correct B. distance = | ½(3,372) + 3×(34 + 84) |
| 20,400 |
= = = 0.1000 = 10 cM Incorrect C. distance = = = = 0.0950 = 9.50 cM Incorrect D. distance = | ½(3,372 + 4,788) + 3×(84 + 188) |
| 20,400 |
= = = 0.2400 = 24 cM Incorrect E. distance = | ½(3,372 + 4,788) + 3×(34) |
| 20,400 |
= = = 0.2050 = 20.50 cM Incorrect
MC 5afe_4978
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene N is analogous to the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
- Gene R is correlated with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
- Gene W is associated with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 50 |
| 2 | | | | | 3,900 |
| 3 | | | | | 1,296 |
| 4 | | | | | 2,124 |
| 5 | | | | | 34 |
| 6 | | | | | 96 |
| TOTAL = | 7,500 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes N and W
A. distance = = = = 0.1000 = 10 cM Incorrect B. distance = = = = 0.1800 = 18 cM Incorrect C. distance = | ½(2,124) + 3×(50 + 96) |
| 7,500 |
= = = 0.2000 = 20 cM Incorrect D. distance = | ½(1,296) + 3×(34 + 50) |
| 7,500 |
= = = 0.1200 = 12 cM Correct E. distance = | ½(1,296 + 2,124) + 3×(34 + 96) |
| 7,500 |
= = = 0.2800 = 28 cM Incorrect
MC 3bb4_8291
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene H is related to the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
- Gene M is related to the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene N is linked with the 'nude' phenotype. A budding yeast that is homozygous recessive for Gene N cells have an unusually smooth surface with no visible external features or textures.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 81 |
| 2 | | | | | 3,732 |
| 3 | | | | | 419 |
| 4 | | | | | 7,920 |
| 5 | | | | | 111 |
| 6 | | | | | 9,937 |
| TOTAL = | 22,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes M and N
A. distance = | ½(7,920) + 3×(419) |
| 22,200 |
= = = 0.2350 = 23.50 cM Incorrect B. distance = | ½(3,732) + 3×(81 + 111) |
| 22,200 |
= = = 0.1100 = 11 cM Incorrect C. distance = | ½(3,732 + 7,920) + 3×(81 + 419) |
| 22,200 |
= = = 0.3300 = 33 cM Incorrect D. distance = | ½(3,732 + 7,920) + 3×(81 + 111 + 419) |
| 22,200 |
= = = 0.3450 = 34.50 cM Incorrect E. distance = | ½(7,920) + 3×(111 + 419) |
| 22,200 |
= = = 0.2500 = 25 cM Correct
MC a16f_4ffa
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene D is linked with the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene M is connected with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene T is associated with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 126 |
| 2 | | | | | 11,655 |
| 3 | | | | | 3,474 |
| 4 | | | | | 9,504 |
| 5 | | | | | 51 |
| 6 | | | | | 390 |
| TOTAL = | 25,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes D and T
A. distance = | ½(9,504) + 3×(126 + 390) |
| 25,200 |
= = = 0.2500 = 25 cM Incorrect B. distance = | ½(3,474 + 9,504) + 3×(51 + 390) |
| 25,200 |
= = = 0.3100 = 31 cM Incorrect C. distance = | ½(3,474 + 9,504) + 3×(51 + 126 + 390) |
| 25,200 |
= = = 0.3250 = 32.50 cM Incorrect D. distance = = = = 0.0750 = 7.50 cM Incorrect E. distance = | ½(3,474) + 3×(51 + 126) |
| 25,200 |
= = = 0.0900 = 9 cM Correct
MC a624_5fc5
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is affiliated with the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene R is correlated with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
- Gene W is linked with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 29 |
| 2 | | | | | 5 |
| 3 | | | | | 1,188 |
| 4 | | | | | 11,482 |
| 5 | | | | | 4,626 |
| 6 | | | | | 70 |
| TOTAL = | 17,400 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes R and W
A. distance = = = = 0.1450 = 14.50 cM Incorrect B. distance = | ½(1,188 + 4,626) + 3×(5 + 70) |
| 17,400 |
= = = 0.1800 = 18 cM Incorrect C. distance = | ½(4,626) + 3×(29 + 70) |
| 17,400 |
= = = 0.1500 = 15 cM Correct D. distance = | ½(1,188) + 3×(5 + 29) |
| 17,400 |
= = = 0.0400 = 4 cM Incorrect E. distance = | ½(1,188 + 4,626) + 3×(5 + 29 + 70) |
| 17,400 |
= = = 0.1850 = 18.50 cM Incorrect
MC 72a6_37af
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene B is affiliated with the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
- Gene H is analogous to the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
- Gene P is associated with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 576 |
| 2 | | | | | 8,388 |
| 3 | | | | | 180 |
| 4 | | | | | 5,220 |
| 5 | | | | | 10,710 |
| 6 | | | | | 126 |
| TOTAL = | 25,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes B and P
A. distance = | ½(8,388) + 3×(126 + 576) |
| 25,200 |
= = = 0.2500 = 25 cM Incorrect B. distance = | ½(5,220) + 3×(180) |
| 25,200 |
= = = 0.1250 = 12.50 cM Incorrect C. distance = | ½(5,220) + 3×(126 + 180) |
| 25,200 |
= = = 0.1400 = 14 cM Correct D. distance = | ½(5,220 + 8,388) + 3×(180 + 576) |
| 25,200 |
= = = 0.3600 = 36 cM Incorrect E. distance = | ½(5,220 + 8,388) + 3×(126 + 180 + 576) |
| 25,200 |
= = = 0.3750 = 37.50 cM Incorrect
MC 88d3_6801
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene F is correlated with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene W is correlated with the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
- Gene X is related to the 'xenon' phenotype. A budding yeast that is homozygous recessive for Gene X cells emit a faint glow under UV light, as if they were fluorescent.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 42 |
| 2 | | | | | 10,773 |
| 3 | | | | | 4,656 |
| 4 | | | | | 9,078 |
| 5 | | | | | 106 |
| 6 | | | | | 545 |
| TOTAL = | 25,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes W and X
A. distance = | ½(4,656) + 3×(42 + 106) |
| 25,200 |
= = = 0.1100 = 11 cM Incorrect B. distance = | ½(9,078) + 3×(42 + 545) |
| 25,200 |
= = = 0.2500 = 25 cM Correct C. distance = | ½(4,656 + 9,078) + 3×(42 + 106 + 545) |
| 25,200 |
= = = 0.3550 = 35.50 cM Incorrect D. distance = | ½(4,656) + 3×(106) |
| 25,200 |
= = = 0.1050 = 10.50 cM Incorrect E. distance = | ½(4,656 + 9,078) + 3×(106 + 545) |
| 25,200 |
= = = 0.3500 = 35 cM Incorrect
MC 601d_01cb
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene H is linked with the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
- Gene M is analogous to the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
- Gene P is linked with the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 338 |
| 2 | | | | | 6,612 |
| 3 | | | | | 8,832 |
| 4 | | | | | 3,180 |
| 5 | | | | | 78 |
| 6 | | | | | 160 |
| TOTAL = | 19,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes M and P
A. distance = | ½(3,180 + 6,612) + 3×(78 + 338) |
| 19,200 |
= = = 0.3200 = 32 cM Incorrect B. distance = | ½(3,180 + 6,612) + 3×(160) |
| 19,200 |
= = = 0.2800 = 28 cM Incorrect C. distance = | ½(6,612) + 3×(160 + 338) |
| 19,200 |
= = = 0.2500 = 25 cM Correct D. distance = | ½(3,180) + 3×(78 + 160) |
| 19,200 |
= = = 0.1200 = 12 cM Incorrect E. distance = | ½(6,612) + 3×(338) |
| 19,200 |
= = = 0.2250 = 22.50 cM Incorrect
MC 20af_293f
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene D is analogous to the 'doubled' phenotype. A budding yeast that is homozygous recessive for Gene D cells display double or multiple budding, with several buds emerging simultaneously.
- Gene H is linked with the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
- Gene P is related to the 'pebble' phenotype. A budding yeast that is homozygous recessive for Gene P produces colonies with a rough, uneven surface that resembles a collection of tiny pebbles.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 18 |
| 2 | | | | | 31 |
| 3 | | | | | 786 |
| 4 | | | | | 86 |
| 5 | | | | | 1,176 |
| 6 | | | | | 1,503 |
| TOTAL = | 3,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes H and P
A. distance = | ½(1,176) + 3×(18 + 86) |
| 3,600 |
= = = 0.2500 = 25 cM Incorrect B. distance = | ½(786 + 1,176) + 3×(18 + 31 + 86) |
| 3,600 |
= = = 0.3850 = 38.50 cM Incorrect C. distance = | ½(786) + 3×(18 + 31) |
| 3,600 |
= = = 0.1500 = 15 cM Incorrect D. distance = = = = 0.2350 = 23.50 cM Incorrect E. distance = | ½(786 + 1,176) + 3×(31 + 86) |
| 3,600 |
= = = 0.3700 = 37 cM Correct
MC e4af_1151
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene F is linked with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
- Gene H is analogous to the 'hairy' phenotype. A budding yeast that is homozygous recessive for Gene H cells develop long, thread-like filaments that extend outward, creating a hairy, shaggy texture on the colony.
- Gene R is linked with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 48 |
| 2 | | | | | 2,364 |
| 3 | | | | | 78 |
| 4 | | | | | 8,737 |
| 5 | | | | | 4,218 |
| 6 | | | | | 155 |
| TOTAL = | 15,600 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes H and R
A. distance = | ½(2,364 + 4,218) + 3×(48 + 155) |
| 15,600 |
= = = 0.2500 = 25 cM Correct B. distance = | ½(4,218) + 3×(78 + 155) |
| 15,600 |
= = = 0.1800 = 18 cM Incorrect C. distance = | ½(2,364) + 3×(48 + 78) |
| 15,600 |
= = = 0.1000 = 10 cM Incorrect D. distance = | ½(2,364 + 4,218) + 3×(48 + 78 + 155) |
| 15,600 |
= = = 0.2650 = 26.50 cM Incorrect E. distance = | ½(4,218) + 3×(155) |
| 15,600 |
= = = 0.1650 = 16.50 cM Incorrect
MC af75_2775
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene T is analogous to the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.
- Gene W is related to the 'webbed' phenotype. A budding yeast that is homozygous recessive for Gene W colonies produce delicate, web-like strands that connect neighboring colonies in a cobweb pattern.
- Gene X is affiliated with the 'xenon' phenotype. A budding yeast that is homozygous recessive for Gene X cells emit a faint glow under UV light, as if they were fluorescent.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 42 |
| 2 | | | | | 15 |
| 3 | | | | | 2,178 |
| 4 | | | | | 16,756 |
| 5 | | | | | 6,090 |
| 6 | | | | | 119 |
| TOTAL = | 25,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes T and X
A. distance = | ½(6,090) + 3×(42 + 119) |
| 25,200 |
= = = 0.1400 = 14 cM Incorrect B. distance = | ½(2,178) + 3×(15 + 42) |
| 25,200 |
= = = 0.0500 = 5 cM Incorrect C. distance = | ½(2,178 + 6,090) + 3×(15 + 42 + 119) |
| 25,200 |
= = = 0.1850 = 18.50 cM Incorrect D. distance = | ½(6,090) + 3×(119) |
| 25,200 |
= = = 0.1350 = 13.50 cM Incorrect E. distance = | ½(2,178 + 6,090) + 3×(15 + 119) |
| 25,200 |
= = = 0.1800 = 18 cM Correct
MC 4177_2585
Unordered Tetrad Three Gene Mapping
In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.
Characteristics of Recessive Phenotypes
- Gene C is analogous to the 'clumpy' phenotype. A budding yeast that is homozygous recessive for Gene C grows in dense, irregular clusters, with cells clumping together rather than spreading smoothly.
- Gene K is related to the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
- Gene M is affiliated with the 'militant' phenotype. A budding yeast that is homozygous recessive for Gene M colonies are small, dense, and secrete compounds that inhibit the growth of nearby colonies.
| Set # | Tetrad Genotypes | Progeny
Count |
| 1 | | | | | 12 |
| 2 | | | | | 2 |
| 3 | | | | | 492 |
| 4 | | | | | 29 |
| 5 | | | | | 1,914 |
| 6 | | | | | 4,751 |
| TOTAL = | 7,200 |
The resulting phenotypes are summarized in the table above.
Step-by-Step Instructions
- Step 1: Find the row for the Parental Type for all three genes.
- Step 2: Looking at only your two genes, assign PD, NPD, TT.
- Step 3: Determine if the two genes are linked.
- PD >> NPD → linked; PD ≈ NPD → unlinked
- Step 4: Determine the map distance between the two genes.
- D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total
Determine the distance between the two genes C and K
A. distance = | ½(492 + 1,914) + 3×(2 + 29) |
| 7,200 |
= = = 0.1800 = 18 cM Incorrect B. distance = = = = 0.0350 = 3.50 cM Incorrect C. distance = | ½(492) + 3×(2 + 12) |
| 7,200 |
= = = 0.0400 = 4 cM Correct D. distance = | ½(1,914) + 3×(12 + 29) |
| 7,200 |
= = = 0.1500 = 15 cM Incorrect E. distance = = = = 0.1450 = 14.50 cM Incorrect