fbb3_6490

Unordered Tetrad Two 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 two 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 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.

Set #

Tetrad Genotypes

Progeny
Count

868

886

TOTAL =

1,800

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 E

A. distance =

½(886) + 3×(0)

1,800

443 + 0

1,800

443

1,800

= 0.2461 = 24.61 cM Incorrect B. distance =

½(46) + 3×(46)

1,800

23 + 138

1,800

161

1,800

= 0.0894 = 8.94 cM Incorrect C. distance =

½(868) + 3×(0)

1,800

434 + 0

1,800

434

1,800

= 0.2411 = 24.11 cM Incorrect D. distance =

½(868) + 3×(46)

1,800

434 + 138

1,800

572

1,800

= 0.3178 = 31.78 cM Correct E. distance =

½(0) + 3×(46)

1,800

0 + 138

1,800

138

1,800

= 0.0767 = 7.67 cM Incorrect MC

ceeb_ce8b

Unordered Tetrad Two Gene Mapping

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 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

1,161

808

TOTAL =

2,000

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 F and J

A. distance =

½(808 + 1,161) + 3×(31)

2,000

984.5 + 93

2,000

1077.5

2,000

= 0.5387 = 53.87 cM Incorrect B. distance =

½(808) + 3×(31)

2,000

404 + 93

2,000

497

2,000

= 0.2485 = 24.85 cM Correct C. distance =

½(1,161) + 3×(0)

2,000

580.5 + 0

2,000

580.5

2,000

= 0.2903 = 29.03 cM Incorrect D. distance =

½(808) + 3×(0)

2,000

404 + 0

2,000

404

2,000

= 0.2020 = 20.20 cM Incorrect E. distance =

½(31) + 3×(31)

2,000

15.5 + 93

2,000

108.5

2,000

= 0.0542 = 5.42 cM Incorrect MC

a062_35a4

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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.
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

169

3,862

4,769

TOTAL =

8,800

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 W and X

A. distance =

½(3,862) + 3×(0)

8,800

1,931 + 0

8,800

1,931

8,800

= 0.2194 = 21.94 cM Incorrect B. distance =

½(169) + 3×(0)

8,800

84.5 + 0

8,800

84.5

8,800

= 0.0096 = 0.96 cM Incorrect C. distance =

½(3,862 + 4,769) + 3×(169)

8,800

4315.5 + 507

8,800

4822.5

8,800

= 0.5480 = 54.80 cM Incorrect D. distance =

½(0) + 3×(169)

8,800

0 + 507

8,800

507

8,800

= 0.0576 = 5.76 cM Incorrect E. distance =

½(3,862) + 3×(169)

8,800

1,931 + 507

8,800

2,438

8,800

= 0.2770 = 27.70 cM Correct MC

b81d_ed7c

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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.
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

153

3,454

4,193

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 W and Y

A. distance =

½(153) + 3×(0)

7,800

76.5 + 0

7,800

76.5

7,800

= 0.0098 = 0.98 cM Incorrect B. distance =

½(4,193) + 3×(0)

7,800

2096.5 + 0

7,800

2096.5

7,800

= 0.2688 = 26.88 cM Incorrect C. distance =

½(0) + 3×(153)

7,800

0 + 459

7,800

459

7,800

= 0.0588 = 5.88 cM Incorrect D. distance =

½(153) + 3×(153)

7,800

76.5 + 459

7,800

535.5

7,800

= 0.0687 = 6.87 cM Incorrect E. distance =

½(3,454) + 3×(153)

7,800

1,727 + 459

7,800

2,186

7,800

= 0.2803 = 28.03 cM Correct MC

5161_261c

Unordered Tetrad Two Gene Mapping

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 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

1,362

1,575

TOTAL =

3,000

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 D and P

A. distance =

½(1,362 + 1,575) + 3×(63)

3,000

1468.5 + 189

3,000

1657.5

3,000

= 0.5525 = 55.25 cM Incorrect B. distance =

½(1,362) + 3×(63)

3,000

681 + 189

3,000

870

3,000

= 0.2900 = 29 cM Correct C. distance =

½(63) + 3×(0)

3,000

31.5 + 0

3,000

31.5

3,000

= 0.0105 = 1.05 cM Incorrect D. distance =

½(1,575) + 3×(63)

3,000

787.5 + 189

3,000

976.5

3,000

= 0.3255 = 32.55 cM Incorrect E. distance =

½(1,575) + 3×(0)

3,000

787.5 + 0

3,000

787.5

3,000

= 0.2625 = 26.25 cM Incorrect MC

11c9_ccee

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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.
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

4,856

2,662

TOTAL =

7,600

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 N and W

A. distance =

½(2,662) + 3×(82)

7,600

1,331 + 246

7,600

1,577

7,600

= 0.2075 = 20.75 cM Correct B. distance =

½(82) + 3×(0)

7,600

41 + 0

7,600

= 0.0054 = 0.54 cM Incorrect C. distance =

½(2,662) + 3×(0)

7,600

1,331 + 0

7,600

1,331

7,600

= 0.1751 = 17.51 cM Incorrect D. distance =

½(82) + 3×(82)

7,600

41 + 246

7,600

287

7,600

= 0.0378 = 3.78 cM Incorrect E. distance =

½(4,856) + 3×(0)

7,600

2,428 + 0

7,600

2,428

7,600

= 0.3195 = 31.95 cM Incorrect MC

06ac_4c76

Unordered Tetrad Two Gene Mapping

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 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

4,025

2,302

TOTAL =

6,400

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 H and W

A. distance =

½(2,302 + 4,025) + 3×(73)

6,400

3163.5 + 219

6,400

3382.5

6,400

= 0.5285 = 52.85 cM Incorrect B. distance =

½(4,025) + 3×(73)

6,400

2012.5 + 219

6,400

2231.5

6,400

= 0.3487 = 34.87 cM Incorrect C. distance =

½(2,302) + 3×(73)

6,400

1,151 + 219

6,400

1,370

6,400

= 0.2141 = 21.41 cM Correct D. distance =

½(0) + 3×(73)

6,400

0 + 219

6,400

219

6,400

= 0.0342 = 3.42 cM Incorrect E. distance =

½(73) + 3×(73)

6,400

36.5 + 219

6,400

255.5

6,400

= 0.0399 = 3.99 cM Incorrect MC

29d8_2ed4

Unordered Tetrad Two Gene Mapping

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 K is connected 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

1,972

1,928

100

TOTAL =

4,000

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(1,972) + 3×(100)

4,000

986 + 300

4,000

1,286

4,000

= 0.3215 = 32.15 cM Incorrect B. distance =

½(0) + 3×(100)

4,000

0 + 300

4,000

300

4,000

= 0.0750 = 7.50 cM Incorrect C. distance =

½(1,972) + 3×(0)

4,000

986 + 0

4,000

986

4,000

= 0.2465 = 24.65 cM Incorrect D. distance =

½(100) + 3×(100)

4,000

50 + 300

4,000

350

4,000

= 0.0875 = 8.75 cM Incorrect E. distance =

½(1,928) + 3×(100)

4,000

964 + 300

4,000

1,264

4,000

= 0.3160 = 31.60 cM Correct MC

7369_9577

Unordered Tetrad Two Gene Mapping

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 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

2,144

3,584

TOTAL =

5,800

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 T

A. distance =

½(72) + 3×(0)

5,800

36 + 0

5,800

= 0.0062 = 0.62 cM Incorrect B. distance =

½(2,144 + 3,584) + 3×(72)

5,800

2,864 + 216

5,800

3,080

5,800

= 0.5310 = 53.10 cM Incorrect C. distance =

½(3,584) + 3×(0)

5,800

1,792 + 0

5,800

1,792

5,800

= 0.3090 = 30.90 cM Incorrect D. distance =

½(72) + 3×(72)

5,800

36 + 216

5,800

252

5,800

= 0.0434 = 4.34 cM Incorrect E. distance =

½(2,144) + 3×(72)

5,800

1,072 + 216

5,800

1,288

5,800

= 0.2221 = 22.21 cM Correct MC

6974_ee3f

Unordered Tetrad Two Gene Mapping

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 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

980

970

TOTAL =

2,000

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 F and X

A. distance =

½(980) + 3×(0)

2,000

490 + 0

2,000

490

2,000

= 0.2450 = 24.50 cM Incorrect B. distance =

½(970) + 3×(50)

2,000

485 + 150

2,000

635

2,000

= 0.3175 = 31.75 cM Correct C. distance =

½(0) + 3×(50)

2,000

0 + 150

2,000

150

2,000

= 0.0750 = 7.50 cM Incorrect D. distance =

½(50) + 3×(0)

2,000

25 + 0

2,000

= 0.0125 = 1.25 cM Incorrect E. distance =

½(970 + 980) + 3×(50)

2,000

975 + 150

2,000

1,125

2,000

= 0.5625 = 56.25 cM Incorrect MC

d2da_8767

Unordered Tetrad Two Gene Mapping

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 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

3,034

2,642

124

TOTAL =

5,800

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(2,642) + 3×(124)

5,800

1,321 + 372

5,800

1,693

5,800

= 0.2919 = 29.19 cM Correct B. distance =

½(124) + 3×(0)

5,800

62 + 0

5,800

= 0.0107 = 1.07 cM Incorrect C. distance =

½(3,034) + 3×(124)

5,800

1,517 + 372

5,800

1,889

5,800

= 0.3257 = 32.57 cM Incorrect D. distance =

½(0) + 3×(124)

5,800

0 + 372

5,800

372

5,800

= 0.0641 = 6.41 cM Incorrect E. distance =

½(2,642 + 3,034) + 3×(124)

5,800

2,838 + 372

5,800

3,210

5,800

= 0.5534 = 55.34 cM Incorrect MC

ff43_eeaf

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

182

3,918

4,500

TOTAL =

8,600

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 R

A. distance =

½(3,918 + 4,500) + 3×(182)

8,600

4,209 + 546

8,600

4,755

8,600

= 0.5529 = 55.29 cM Incorrect B. distance =

½(4,500) + 3×(0)

8,600

2,250 + 0

8,600

2,250

8,600

= 0.2616 = 26.16 cM Incorrect C. distance =

½(4,500) + 3×(182)

8,600

2,250 + 546

8,600

2,796

8,600

= 0.3251 = 32.51 cM Incorrect D. distance =

½(182) + 3×(182)

8,600

91 + 546

8,600

637

8,600

= 0.0741 = 7.41 cM Incorrect E. distance =

½(3,918) + 3×(182)

8,600

1,959 + 546

8,600

2,505

8,600

= 0.2913 = 29.13 cM Correct MC

17bc_c5be

Unordered Tetrad Two Gene Mapping

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 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

2,362

2,318

120

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 A and X

A. distance =

½(0) + 3×(120)

4,800

0 + 360

4,800

360

4,800

= 0.0750 = 7.50 cM Incorrect B. distance =

½(2,318 + 2,362) + 3×(120)

4,800

2,340 + 360

4,800

2,700

4,800

= 0.5625 = 56.25 cM Incorrect C. distance =

½(2,318) + 3×(0)

4,800

1,159 + 0

4,800

1,159

4,800

= 0.2415 = 24.15 cM Incorrect D. distance =

½(2,362) + 3×(0)

4,800

1,181 + 0

4,800

1,181

4,800

= 0.2460 = 24.60 cM Incorrect E. distance =

½(2,318) + 3×(120)

4,800

1,159 + 360

4,800

1,519

4,800

= 0.3165 = 31.65 cM Correct MC

0c35_a318

Unordered Tetrad Two Gene Mapping

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 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

3,116

2,570

114

TOTAL =

5,800

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 X

A. distance =

½(2,570) + 3×(114)

5,800

1,285 + 342

5,800

1,627

5,800

= 0.2805 = 28.05 cM Correct B. distance =

½(114) + 3×(114)

5,800

57 + 342

5,800

399

5,800

= 0.0688 = 6.88 cM Incorrect C. distance =

½(3,116) + 3×(114)

5,800

1,558 + 342

5,800

1,900

5,800

= 0.3276 = 32.76 cM Incorrect D. distance =

½(0) + 3×(114)

5,800

0 + 342

5,800

342

5,800

= 0.0590 = 5.90 cM Incorrect E. distance =

½(2,570 + 3,116) + 3×(114)

5,800

2,843 + 342

5,800

3,185

5,800

= 0.5491 = 54.91 cM Incorrect MC

4f56_ab50

Unordered Tetrad Two Gene Mapping

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 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

2,276

3,236

TOTAL =

5,600

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 N and R

A. distance =

½(3,236) + 3×(88)

5,600

1,618 + 264

5,600

1,882

5,600

= 0.3361 = 33.61 cM Incorrect B. distance =

½(3,236) + 3×(0)

5,600

1,618 + 0

5,600

1,618

5,600

= 0.2889 = 28.89 cM Incorrect C. distance =

½(0) + 3×(88)

5,600

0 + 264

5,600

264

5,600

= 0.0471 = 4.71 cM Incorrect D. distance =

½(88) + 3×(0)

5,600

44 + 0

5,600

= 0.0079 = 0.79 cM Incorrect E. distance =

½(2,276) + 3×(88)

5,600

1,138 + 264

5,600

1,402

5,600

= 0.2504 = 25.04 cM Correct MC

daa3_0ca7

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

4,181

2,908

111

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 R and Y

A. distance =

½(2,908) + 3×(111)

7,200

1,454 + 333

7,200

1,787

7,200

= 0.2482 = 24.82 cM Correct B. distance =

½(2,908 + 4,181) + 3×(111)

7,200

3544.5 + 333

7,200

3877.5

7,200

= 0.5385 = 53.85 cM Incorrect C. distance =

½(4,181) + 3×(0)

7,200

2090.5 + 0

7,200

2090.5

7,200

= 0.2903 = 29.03 cM Incorrect D. distance =

½(111) + 3×(0)

7,200

55.5 + 0

7,200

55.5

7,200

= 0.0077 = 0.77 cM Incorrect E. distance =

½(111) + 3×(111)

7,200

55.5 + 333

7,200

388.5

7,200

= 0.0540 = 5.40 cM Incorrect MC

a5e7_3148

Unordered Tetrad Two Gene Mapping

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 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

722

1,051

TOTAL =

1,800

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 R

A. distance =

½(27) + 3×(27)

1,800

13.5 + 81

1,800

94.5

1,800

= 0.0525 = 5.25 cM Incorrect B. distance =

½(0) + 3×(27)

1,800

0 + 81

1,800

= 0.0450 = 4.50 cM Incorrect C. distance =

½(722 + 1,051) + 3×(27)

1,800

886.5 + 81

1,800

967.5

1,800

= 0.5375 = 53.75 cM Incorrect D. distance =

½(722) + 3×(27)

1,800

361 + 81

1,800

442

1,800

= 0.2456 = 24.56 cM Correct E. distance =

½(722) + 3×(0)

1,800

361 + 0

1,800

361

1,800

= 0.2006 = 20.06 cM Incorrect MC

ead5_5e52

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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.
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

118

2,776

3,506

TOTAL =

6,400

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 X and Y

A. distance =

½(2,776) + 3×(118)

6,400

1,388 + 354

6,400

1,742

6,400

= 0.2722 = 27.22 cM Correct B. distance =

½(3,506) + 3×(118)

6,400

1,753 + 354

6,400

2,107

6,400

= 0.3292 = 32.92 cM Incorrect C. distance =

½(3,506) + 3×(0)

6,400

1,753 + 0

6,400

1,753

6,400

= 0.2739 = 27.39 cM Incorrect D. distance =

½(2,776) + 3×(0)

6,400

1,388 + 0

6,400

1,388

6,400

= 0.2169 = 21.69 cM Incorrect E. distance =

½(2,776 + 3,506) + 3×(118)

6,400

3,141 + 354

6,400

3,495

6,400

= 0.5461 = 54.61 cM Incorrect MC

892c_3e9b

Unordered Tetrad Two Gene Mapping

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 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

149

3,480

4,371

TOTAL =

8,000

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 W

A. distance =

½(3,480) + 3×(0)

8,000

1,740 + 0

8,000

1,740

8,000

= 0.2175 = 21.75 cM Incorrect B. distance =

½(149) + 3×(0)

8,000

74.5 + 0

8,000

74.5

8,000

= 0.0093 = 0.93 cM Incorrect C. distance =

½(149) + 3×(149)

8,000

74.5 + 447

8,000

521.5

8,000

= 0.0652 = 6.52 cM Incorrect D. distance =

½(4,371) + 3×(0)

8,000

2185.5 + 0

8,000

2185.5

8,000

= 0.2732 = 27.32 cM Incorrect E. distance =

½(3,480) + 3×(149)

8,000

1,740 + 447

8,000

2,187

8,000

= 0.2734 = 27.34 cM Correct MC

86ae_186c

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

183

3,680

3,937

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 M and R

A. distance =

½(3,680) + 3×(183)

7,800

1,840 + 549

7,800

2,389

7,800

= 0.3063 = 30.63 cM Correct B. distance =

½(3,680) + 3×(0)

7,800

1,840 + 0

7,800

1,840

7,800

= 0.2359 = 23.59 cM Incorrect C. distance =

½(183) + 3×(0)

7,800

91.5 + 0

7,800

91.5

7,800

= 0.0117 = 1.17 cM Incorrect D. distance =

½(3,937) + 3×(0)

7,800

1968.5 + 0

7,800

1968.5

7,800

= 0.2524 = 25.24 cM Incorrect E. distance =

½(3,937) + 3×(183)

7,800

1968.5 + 549

7,800

2517.5

7,800

= 0.3228 = 32.28 cM Incorrect MC

51dc_f711

Unordered Tetrad Two Gene Mapping

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 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.

Set #

Tetrad Genotypes

Progeny
Count

3,846

3,026

128

TOTAL =

7,000

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 H

A. distance =

½(3,846) + 3×(128)

7,000

1,923 + 384

7,000

2,307

7,000

= 0.3296 = 32.96 cM Incorrect B. distance =

½(3,026) + 3×(0)

7,000

1,513 + 0

7,000

1,513

7,000

= 0.2161 = 21.61 cM Incorrect C. distance =

½(3,026) + 3×(128)

7,000

1,513 + 384

7,000

1,897

7,000

= 0.2710 = 27.10 cM Correct D. distance =

½(0) + 3×(128)

7,000

0 + 384

7,000

384

7,000

= 0.0549 = 5.49 cM Incorrect E. distance =

½(3,026 + 3,846) + 3×(128)

7,000

3,436 + 384

7,000

3,820

7,000

= 0.5457 = 54.57 cM Incorrect MC

818e_ca16

Unordered Tetrad Two Gene Mapping

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 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

1,678

3,071

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 F and T

A. distance =

½(3,071) + 3×(51)

4,800

1535.5 + 153

4,800

1688.5

4,800

= 0.3518 = 35.18 cM Incorrect B. distance =

½(1,678) + 3×(51)

4,800

839 + 153

4,800

992

4,800

= 0.2067 = 20.67 cM Correct C. distance =

½(1,678) + 3×(0)

4,800

839 + 0

4,800

839

4,800

= 0.1748 = 17.48 cM Incorrect D. distance =

½(1,678 + 3,071) + 3×(51)

4,800

2374.5 + 153

4,800

2527.5

4,800

= 0.5266 = 52.66 cM Incorrect E. distance =

½(0) + 3×(51)

4,800

0 + 153

4,800

153

4,800

= 0.0319 = 3.19 cM Incorrect MC

c90b_be52

Unordered Tetrad Two Gene Mapping

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 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

1,122

1,028

TOTAL =

2,200

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(1,028) + 3×(0)

2,200

514 + 0

2,200

514

2,200

= 0.2336 = 23.36 cM Incorrect B. distance =

½(1,028) + 3×(50)

2,200

514 + 150

2,200

664

2,200

= 0.3018 = 30.18 cM Correct C. distance =

½(50) + 3×(0)

2,200

25 + 0

2,200

= 0.0114 = 1.14 cM Incorrect D. distance =

½(0) + 3×(50)

2,200

0 + 150

2,200

150

2,200

= 0.0682 = 6.82 cM Incorrect E. distance =

½(1,028 + 1,122) + 3×(50)

2,200

1,075 + 150

2,200

1,225

2,200

= 0.5568 = 55.68 cM Incorrect MC

0278_18c0

Unordered Tetrad Two Gene Mapping

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 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.

Set #

Tetrad Genotypes

Progeny
Count

117

3,456

5,627

TOTAL =

9,200

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 N

A. distance =

½(3,456) + 3×(117)

9,200

1,728 + 351

9,200

2,079

9,200

= 0.2260 = 22.60 cM Correct B. distance =

½(5,627) + 3×(0)

9,200

2813.5 + 0

9,200

2813.5

9,200

= 0.3058 = 30.58 cM Incorrect C. distance =

½(3,456 + 5,627) + 3×(117)

9,200

4541.5 + 351

9,200

4892.5

9,200

= 0.5318 = 53.18 cM Incorrect D. distance =

½(117) + 3×(0)

9,200

58.5 + 0

9,200

58.5

9,200

= 0.0064 = 0.64 cM Incorrect E. distance =

½(3,456) + 3×(0)

9,200

1,728 + 0

9,200

1,728

9,200

= 0.1878 = 18.78 cM Incorrect MC

3c4c_77ce

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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.
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

2,452

1,684

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 P and Y

A. distance =

½(0) + 3×(64)

4,200

0 + 192

4,200

192

4,200

= 0.0457 = 4.57 cM Incorrect B. distance =

½(1,684) + 3×(64)

4,200

842 + 192

4,200

1,034

4,200

= 0.2462 = 24.62 cM Correct C. distance =

½(2,452) + 3×(0)

4,200

1,226 + 0

4,200

1,226

4,200

= 0.2919 = 29.19 cM Incorrect D. distance =

½(2,452) + 3×(64)

4,200

1,226 + 192

4,200

1,418

4,200

= 0.3376 = 33.76 cM Incorrect E. distance =

½(64) + 3×(0)

4,200

32 + 0

4,200

= 0.0076 = 0.76 cM Incorrect MC

06ed_90ee

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

3,042

1,704

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 K and R

A. distance =

½(1,704) + 3×(54)

4,800

852 + 162

4,800

1,014

4,800

= 0.2112 = 21.12 cM Correct B. distance =

½(54) + 3×(54)

4,800

27 + 162

4,800

189

4,800

= 0.0394 = 3.94 cM Incorrect C. distance =

½(3,042) + 3×(54)

4,800

1,521 + 162

4,800

1,683

4,800

= 0.3506 = 35.06 cM Incorrect D. distance =

½(3,042) + 3×(0)

4,800

1,521 + 0

4,800

1,521

4,800

= 0.3169 = 31.69 cM Incorrect E. distance =

½(0) + 3×(54)

4,800

0 + 162

4,800

162

4,800

= 0.0338 = 3.38 cM Incorrect MC

632a_98e9

Unordered Tetrad Two Gene Mapping

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 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

806

1,163

TOTAL =

2,000

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 J and N

A. distance =

½(806) + 3×(31)

2,000

403 + 93

2,000

496

2,000

= 0.2480 = 24.80 cM Correct B. distance =

½(806 + 1,163) + 3×(31)

2,000

984.5 + 93

2,000

1077.5

2,000

= 0.5387 = 53.87 cM Incorrect C. distance =

½(806) + 3×(0)

2,000

403 + 0

2,000

403

2,000

= 0.2015 = 20.15 cM Incorrect D. distance =

½(1,163) + 3×(0)

2,000

581.5 + 0

2,000

581.5

2,000

= 0.2908 = 29.07 cM Incorrect E. distance =

½(0) + 3×(31)

2,000

0 + 93

2,000

= 0.0465 = 4.65 cM Incorrect MC

5c18_5fbb

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

125

2,934

3,741

TOTAL =

6,800

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 R

A. distance =

½(2,934) + 3×(125)

6,800

1,467 + 375

6,800

1,842

6,800

= 0.2709 = 27.09 cM Correct B. distance =

½(3,741) + 3×(0)

6,800

1870.5 + 0

6,800

1870.5

6,800

= 0.2751 = 27.51 cM Incorrect C. distance =

½(2,934) + 3×(0)

6,800

1,467 + 0

6,800

1,467

6,800

= 0.2157 = 21.57 cM Incorrect D. distance =

½(2,934 + 3,741) + 3×(125)

6,800

3337.5 + 375

6,800

3712.5

6,800

= 0.5460 = 54.60 cM Incorrect E. distance =

½(3,741) + 3×(125)

6,800

1870.5 + 375

6,800

2245.5

6,800

= 0.3302 = 33.02 cM Incorrect MC

e7d2_711c

Unordered Tetrad Two Gene Mapping

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 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

5,288

4,136

176

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 D and Y

A. distance =

½(176) + 3×(176)

9,600

88 + 528

9,600

616

9,600

= 0.0642 = 6.42 cM Incorrect B. distance =

½(5,288) + 3×(0)

9,600

2,644 + 0

9,600

2,644

9,600

= 0.2754 = 27.54 cM Incorrect C. distance =

½(0) + 3×(176)

9,600

0 + 528

9,600

528

9,600

= 0.0550 = 5.50 cM Incorrect D. distance =

½(5,288) + 3×(176)

9,600

2,644 + 528

9,600

3,172

9,600

= 0.3304 = 33.04 cM Incorrect E. distance =

½(4,136) + 3×(176)

9,600

2,068 + 528

9,600

2,596

9,600

= 0.2704 = 27.04 cM Correct MC

9e13_f95c

Unordered Tetrad Two Gene Mapping

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 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

1,932

2,383

TOTAL =

4,400

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 F and M

A. distance =

½(85) + 3×(85)

4,400

42.5 + 255

4,400

297.5

4,400

= 0.0676 = 6.76 cM Incorrect B. distance =

½(1,932 + 2,383) + 3×(85)

4,400

2157.5 + 255

4,400

2412.5

4,400

= 0.5483 = 54.83 cM Incorrect C. distance =

½(2,383) + 3×(85)

4,400

1191.5 + 255

4,400

1446.5

4,400

= 0.3287 = 32.88 cM Incorrect D. distance =

½(1,932) + 3×(85)

4,400

966 + 255

4,400

1,221

4,400

= 0.2775 = 27.75 cM Correct E. distance =

½(0) + 3×(85)

4,400

0 + 255

4,400

255

4,400

= 0.0580 = 5.80 cM Incorrect MC

2a34_bf14

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

4,646

3,972

182

TOTAL =

8,800

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 K and Y

A. distance =

½(182) + 3×(182)

8,800

91 + 546

8,800

637

8,800

= 0.0724 = 7.24 cM Incorrect B. distance =

½(3,972) + 3×(182)

8,800

1,986 + 546

8,800

2,532

8,800

= 0.2877 = 28.77 cM Correct C. distance =

½(0) + 3×(182)

8,800

0 + 546

8,800

546

8,800

= 0.0620 = 6.20 cM Incorrect D. distance =

½(3,972 + 4,646) + 3×(182)

8,800

4,309 + 546

8,800

4,855

8,800

= 0.5517 = 55.17 cM Incorrect E. distance =

½(3,972) + 3×(0)

8,800

1,986 + 0

8,800

1,986

8,800

= 0.2257 = 22.57 cM Incorrect MC

d775_3114

Unordered Tetrad Two Gene Mapping

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.

Set #

Tetrad Genotypes

Progeny
Count

131

3,604

5,465

TOTAL =

9,200

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(0) + 3×(131)

9,200

0 + 393

9,200

393

9,200

= 0.0427 = 4.27 cM Incorrect B. distance =

½(3,604 + 5,465) + 3×(131)

9,200

4534.5 + 393

9,200

4927.5

9,200

= 0.5356 = 53.56 cM Incorrect C. distance =

½(5,465) + 3×(0)

9,200

2732.5 + 0

9,200

2732.5

9,200

= 0.2970 = 29.70 cM Incorrect D. distance =

½(131) + 3×(131)

9,200

65.5 + 393

9,200

458.5

9,200

= 0.0498 = 4.98 cM Incorrect E. distance =

½(3,604) + 3×(131)

9,200

1,802 + 393

9,200

2,195

9,200

= 0.2386 = 23.86 cM Correct MC

a7f2_4973

Unordered Tetrad Two Gene Mapping

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 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

3,751

1,990

TOTAL =

5,800

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 E and R

A. distance =

½(0) + 3×(59)

5,800

0 + 177

5,800

177

5,800

= 0.0305 = 3.05 cM Incorrect B. distance =

½(59) + 3×(0)

5,800

29.5 + 0

5,800

29.5

5,800

= 0.0051 = 0.51 cM Incorrect C. distance =

½(1,990) + 3×(59)

5,800

995 + 177

5,800

1,172

5,800

= 0.2021 = 20.21 cM Correct D. distance =

½(59) + 3×(59)

5,800

29.5 + 177

5,800

206.5

5,800

= 0.0356 = 3.56 cM Incorrect E. distance =

½(1,990) + 3×(0)

5,800

995 + 0

5,800

995

5,800

= 0.1716 = 17.16 cM Incorrect MC

d56e_4b33

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

118

3,380

5,302

TOTAL =

8,800

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 R and T

A. distance =

½(0) + 3×(118)

8,800

0 + 354

8,800

354

8,800

= 0.0402 = 4.02 cM Incorrect B. distance =

½(3,380) + 3×(118)

8,800

1,690 + 354

8,800

2,044

8,800

= 0.2323 = 23.23 cM Correct C. distance =

½(5,302) + 3×(0)

8,800

2,651 + 0

8,800

2,651

8,800

= 0.3013 = 30.12 cM Incorrect D. distance =

½(3,380) + 3×(0)

8,800

1,690 + 0

8,800

1,690

8,800

= 0.1920 = 19.20 cM Incorrect E. distance =

½(5,302) + 3×(118)

8,800

2,651 + 354

8,800

3,005

8,800

= 0.3415 = 34.15 cM Incorrect MC

0ca3_8d9e

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

Gene N is associated 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 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

142

2,776

2,882

TOTAL =

5,800

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 N and W

A. distance =

½(2,882) + 3×(142)

5,800

1,441 + 426

5,800

1,867

5,800

= 0.3219 = 32.19 cM Incorrect B. distance =

½(2,776) + 3×(142)

5,800

1,388 + 426

5,800

1,814

5,800

= 0.3128 = 31.28 cM Correct C. distance =

½(0) + 3×(142)

5,800

0 + 426

5,800

426

5,800

= 0.0734 = 7.34 cM Incorrect D. distance =

½(2,776 + 2,882) + 3×(142)

5,800

2,829 + 426

5,800

3,255

5,800

= 0.5612 = 56.12 cM Incorrect E. distance =

½(142) + 3×(142)

5,800

71 + 426

5,800

497

5,800

= 0.0857 = 8.57 cM Incorrect MC

1110_fc92

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

173

3,724

4,303

TOTAL =

8,200

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(3,724) + 3×(0)

8,200

1,862 + 0

8,200

1,862

8,200

= 0.2271 = 22.71 cM Incorrect B. distance =

½(173) + 3×(173)

8,200

86.5 + 519

8,200

605.5

8,200

= 0.0738 = 7.38 cM Incorrect C. distance =

½(4,303) + 3×(173)

8,200

2151.5 + 519

8,200

2670.5

8,200

= 0.3257 = 32.57 cM Incorrect D. distance =

½(4,303) + 3×(0)

8,200

2151.5 + 0

8,200

2151.5

8,200

= 0.2624 = 26.24 cM Incorrect E. distance =

½(3,724) + 3×(173)

8,200

1,862 + 519

8,200

2,381

8,200

= 0.2904 = 29.04 cM Correct MC

e3e0_396e

Unordered Tetrad Two Gene Mapping

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 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

2,138

1,598

TOTAL =

3,800

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 Y

A. distance =

½(1,598) + 3×(64)

3,800

799 + 192

3,800

991

3,800

= 0.2608 = 26.08 cM Correct B. distance =

½(2,138) + 3×(0)

3,800

1,069 + 0

3,800

1,069

3,800

= 0.2813 = 28.13 cM Incorrect C. distance =

½(1,598) + 3×(0)

3,800

799 + 0

3,800

799

3,800

= 0.2103 = 21.03 cM Incorrect D. distance =

½(64) + 3×(0)

3,800

32 + 0

3,800

= 0.0084 = 0.84 cM Incorrect E. distance =

½(64) + 3×(64)

3,800

32 + 192

3,800

224

3,800

= 0.0589 = 5.89 cM Incorrect MC

39d9_323e

Unordered Tetrad Two Gene Mapping

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 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

5,410

3,280

110

TOTAL =

8,800

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 K and X

A. distance =

½(5,410) + 3×(110)

8,800

2,705 + 330

8,800

3,035

8,800

= 0.3449 = 34.49 cM Incorrect B. distance =

½(0) + 3×(110)

8,800

0 + 330

8,800

330

8,800

= 0.0375 = 3.75 cM Incorrect C. distance =

½(5,410) + 3×(0)

8,800

2,705 + 0

8,800

2,705

8,800

= 0.3074 = 30.74 cM Incorrect D. distance =

½(3,280) + 3×(110)

8,800

1,640 + 330

8,800

1,970

8,800

= 0.2239 = 22.39 cM Correct E. distance =

½(3,280) + 3×(0)

8,800

1,640 + 0

8,800

1,640

8,800

= 0.1864 = 18.64 cM Incorrect MC

a979_be10

Unordered Tetrad Two Gene Mapping

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 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

2,673

2,594

133

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 E and R

A. distance =

½(2,673) + 3×(0)

5,400

1336.5 + 0

5,400

1336.5

5,400

= 0.2475 = 24.75 cM Incorrect B. distance =

½(2,594) + 3×(0)

5,400

1,297 + 0

5,400

1,297

5,400

= 0.2402 = 24.02 cM Incorrect C. distance =

½(2,594) + 3×(133)

5,400

1,297 + 399

5,400

1,696

5,400

= 0.3141 = 31.41 cM Correct D. distance =

½(0) + 3×(133)

5,400

0 + 399

5,400

399

5,400

= 0.0739 = 7.39 cM Incorrect E. distance =

½(2,594 + 2,673) + 3×(133)

5,400

2633.5 + 399

5,400

3032.5

5,400

= 0.5616 = 56.16 cM Incorrect MC

a47e_d040

Unordered Tetrad Two Gene Mapping

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 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

102

3,172

5,526

TOTAL =

8,800

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 N

A. distance =

½(102) + 3×(102)

8,800

51 + 306

8,800

357

8,800

= 0.0406 = 4.06 cM Incorrect B. distance =

½(0) + 3×(102)

8,800

0 + 306

8,800

306

8,800

= 0.0348 = 3.48 cM Incorrect C. distance =

½(5,526) + 3×(0)

8,800

2,763 + 0

8,800

2,763

8,800

= 0.3140 = 31.40 cM Incorrect D. distance =

½(3,172) + 3×(102)

8,800

1,586 + 306

8,800

1,892

8,800

= 0.2150 = 21.50 cM Correct E. distance =

½(3,172 + 5,526) + 3×(102)

8,800

4,349 + 306

8,800

4,655

8,800

= 0.5290 = 52.90 cM Incorrect MC

51d5_b77b

Unordered Tetrad Two Gene Mapping

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 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

1,280

1,666

TOTAL =

3,000

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 H

A. distance =

½(54) + 3×(0)

3,000

27 + 0

3,000

= 0.0090 = 0.90 cM Incorrect B. distance =

½(1,666) + 3×(0)

3,000

833 + 0

3,000

833

3,000

= 0.2777 = 27.77 cM Incorrect C. distance =

½(1,280 + 1,666) + 3×(54)

3,000

1,473 + 162

3,000

1,635

3,000

= 0.5450 = 54.50 cM Incorrect D. distance =

½(1,280) + 3×(54)

3,000

640 + 162

3,000

802

3,000

= 0.2673 = 26.73 cM Correct E. distance =

½(1,280) + 3×(0)

3,000

640 + 0

3,000

640

3,000

= 0.2133 = 21.33 cM Incorrect MC

c208_50af

Unordered Tetrad Two Gene Mapping

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 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

142

3,130

3,728

TOTAL =

7,000

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 N

A. distance =

½(142) + 3×(142)

7,000

71 + 426

7,000

497

7,000

= 0.0710 = 7.10 cM Incorrect B. distance =

½(0) + 3×(142)

7,000

0 + 426

7,000

426

7,000

= 0.0609 = 6.09 cM Incorrect C. distance =

½(3,130) + 3×(142)

7,000

1,565 + 426

7,000

1,991

7,000

= 0.2844 = 28.44 cM Correct D. distance =

½(3,728) + 3×(142)

7,000

1,864 + 426

7,000

2,290

7,000

= 0.3271 = 32.71 cM Incorrect E. distance =

½(3,130 + 3,728) + 3×(142)

7,000

3,429 + 426

7,000

3,855

7,000

= 0.5507 = 55.07 cM Incorrect MC

26f9_c4bf

Unordered Tetrad Two Gene Mapping

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 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.

Set #

Tetrad Genotypes

Progeny
Count

136

2,854

3,210

TOTAL =

6,200

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 K

A. distance =

½(3,210) + 3×(136)

6,200

1,605 + 408

6,200

2,013

6,200

= 0.3247 = 32.47 cM Incorrect B. distance =

½(3,210) + 3×(0)

6,200

1,605 + 0

6,200

1,605

6,200

= 0.2589 = 25.89 cM Incorrect C. distance =

½(2,854) + 3×(136)

6,200

1,427 + 408

6,200

1,835

6,200

= 0.2960 = 29.60 cM Correct D. distance =

½(136) + 3×(0)

6,200

68 + 0

6,200

= 0.0110 = 1.10 cM Incorrect E. distance =

½(2,854 + 3,210) + 3×(136)

6,200

3,032 + 408

6,200

3,440

6,200

= 0.5548 = 55.48 cM Incorrect MC

82d5_f332

Unordered Tetrad Two Gene Mapping

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 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

118

2,816

3,666

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 J and N

A. distance =

½(2,816 + 3,666) + 3×(118)

6,600

3,241 + 354

6,600

3,595

6,600

= 0.5447 = 54.47 cM Incorrect B. distance =

½(0) + 3×(118)

6,600

0 + 354

6,600

354

6,600

= 0.0536 = 5.36 cM Incorrect C. distance =

½(3,666) + 3×(0)

6,600

1,833 + 0

6,600

1,833

6,600

= 0.2777 = 27.77 cM Incorrect D. distance =

½(2,816) + 3×(118)

6,600

1,408 + 354

6,600

1,762

6,600

= 0.2670 = 26.70 cM Correct E. distance =

½(118) + 3×(0)

6,600

59 + 0

6,600

= 0.0089 = 0.89 cM Incorrect MC

feaa_8de1

Unordered Tetrad Two Gene Mapping

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 K is connected 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

1,914

1,989

TOTAL =

4,000

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 K

A. distance =

½(1,914 + 1,989) + 3×(97)

4,000

1951.5 + 291

4,000

2242.5

4,000

= 0.5606 = 56.06 cM Incorrect B. distance =

½(1,989) + 3×(97)

4,000

994.5 + 291

4,000

1285.5

4,000

= 0.3214 = 32.14 cM Incorrect C. distance =

½(97) + 3×(0)

4,000

48.5 + 0

4,000

48.5

4,000

= 0.0121 = 1.21 cM Incorrect D. distance =

½(0) + 3×(97)

4,000

0 + 291

4,000

291

4,000

= 0.0727 = 7.27 cM Incorrect E. distance =

½(1,914) + 3×(97)

4,000

957 + 291

4,000

1,248

4,000

= 0.3120 = 31.20 cM Correct MC

1115_361e

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

Gene D is affiliated 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 related 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.

Set #

Tetrad Genotypes

Progeny
Count

3,664

3,372

164

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 D and N

A. distance =

½(3,664) + 3×(0)

7,200

1,832 + 0

7,200

1,832

7,200

= 0.2544 = 25.44 cM Incorrect B. distance =

½(3,372 + 3,664) + 3×(164)

7,200

3,518 + 492

7,200

4,010

7,200

= 0.5569 = 55.69 cM Incorrect C. distance =

½(0) + 3×(164)

7,200

0 + 492

7,200

492

7,200

= 0.0683 = 6.83 cM Incorrect D. distance =

½(3,372) + 3×(164)

7,200

1,686 + 492

7,200

2,178

7,200

= 0.3025 = 30.25 cM Correct E. distance =

½(3,664) + 3×(164)

7,200

1,832 + 492

7,200

2,324

7,200

= 0.3228 = 32.28 cM Incorrect MC

96c9_da26

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

2,365

1,390

TOTAL =

3,800

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 K and P

A. distance =

½(2,365) + 3×(0)

3,800

1182.5 + 0

3,800

1182.5

3,800

= 0.3112 = 31.12 cM Incorrect B. distance =

½(1,390) + 3×(0)

3,800

695 + 0

3,800

695

3,800

= 0.1829 = 18.29 cM Incorrect C. distance =

½(1,390) + 3×(45)

3,800

695 + 135

3,800

830

3,800

= 0.2184 = 21.84 cM Correct D. distance =

½(1,390 + 2,365) + 3×(45)

3,800

1877.5 + 135

3,800

2012.5

3,800

= 0.5296 = 52.96 cM Incorrect E. distance =

½(45) + 3×(45)

3,800

22.5 + 135

3,800

157.5

3,800

= 0.0414 = 4.14 cM Incorrect MC

cd34_56d4

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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.
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

1,708

1,806

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 J and T

A. distance =

½(86) + 3×(86)

3,600

43 + 258

3,600

301

3,600

= 0.0836 = 8.36 cM Incorrect B. distance =

½(1,708) + 3×(86)

3,600

854 + 258

3,600

1,112

3,600

= 0.3089 = 30.89 cM Correct C. distance =

½(86) + 3×(0)

3,600

43 + 0

3,600

= 0.0119 = 1.19 cM Incorrect D. distance =

½(1,806) + 3×(86)

3,600

903 + 258

3,600

1,161

3,600

= 0.3225 = 32.25 cM Incorrect E. distance =

½(1,708 + 1,806) + 3×(86)

3,600

1,757 + 258

3,600

2,015

3,600

= 0.5597 = 55.97 cM Incorrect MC

b1ab_1786

Unordered Tetrad Two Gene Mapping

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 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.

Set #

Tetrad Genotypes

Progeny
Count

1,734

2,397

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 C and M

A. distance =

½(69) + 3×(69)

4,200

34.5 + 207

4,200

241.5

4,200

= 0.0575 = 5.75 cM Incorrect B. distance =

½(1,734) + 3×(69)

4,200

867 + 207

4,200

1,074

4,200

= 0.2557 = 25.57 cM Correct C. distance =

½(1,734) + 3×(0)

4,200

867 + 0

4,200

867

4,200

= 0.2064 = 20.64 cM Incorrect D. distance =

½(2,397) + 3×(0)

4,200

1198.5 + 0

4,200

1198.5

4,200

= 0.2854 = 28.54 cM Incorrect E. distance =

½(2,397) + 3×(69)

4,200

1198.5 + 207

4,200

1405.5

4,200

= 0.3346 = 33.46 cM Incorrect MC

983d_43f5

Unordered Tetrad Two Gene Mapping

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 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

144

3,782

5,474

TOTAL =

9,400

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 X

A. distance =

½(144) + 3×(0)

9,400

72 + 0

9,400

= 0.0077 = 0.77 cM Incorrect B. distance =

½(5,474) + 3×(0)

9,400

2,737 + 0

9,400

2,737

9,400

= 0.2912 = 29.12 cM Incorrect C. distance =

½(3,782) + 3×(0)

9,400

1,891 + 0

9,400

1,891

9,400

= 0.2012 = 20.12 cM Incorrect D. distance =

½(144) + 3×(144)

9,400

72 + 432

9,400

504

9,400

= 0.0536 = 5.36 cM Incorrect E. distance =

½(3,782) + 3×(144)

9,400

1,891 + 432

9,400

2,323

9,400

= 0.2471 = 24.71 cM Correct MC

ba5c_9523

Unordered Tetrad Two Gene Mapping

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 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

2,046

2,463

TOTAL =

4,600

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 X

A. distance =

½(0) + 3×(91)

4,600

0 + 273

4,600

273

4,600

= 0.0593 = 5.93 cM Incorrect B. distance =

½(2,046) + 3×(91)

4,600

1,023 + 273

4,600

1,296

4,600

= 0.2817 = 28.17 cM Correct C. distance =

½(91) + 3×(91)

4,600

45.5 + 273

4,600

318.5

4,600

= 0.0692 = 6.92 cM Incorrect D. distance =

½(2,463) + 3×(0)

4,600

1231.5 + 0

4,600

1231.5

4,600

= 0.2677 = 26.77 cM Incorrect E. distance =

½(2,463) + 3×(91)

4,600

1231.5 + 273

4,600

1504.5

4,600

= 0.3271 = 32.71 cM Incorrect MC

61a4_3127

Unordered Tetrad Two Gene Mapping

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 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

796

969

TOTAL =

1,800

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 W

A. distance =

½(796) + 3×(35)

1,800

398 + 105

1,800

503

1,800

= 0.2794 = 27.94 cM Correct B. distance =

½(0) + 3×(35)

1,800

0 + 105

1,800

105

1,800

= 0.0583 = 5.83 cM Incorrect C. distance =

½(35) + 3×(35)

1,800

17.5 + 105

1,800

122.5

1,800

= 0.0681 = 6.81 cM Incorrect D. distance =

½(35) + 3×(0)

1,800

17.5 + 0

1,800

17.5

1,800

= 0.0097 = 0.97 cM Incorrect E. distance =

½(969) + 3×(0)

1,800

484.5 + 0

1,800

484.5

1,800

= 0.2692 = 26.92 cM Incorrect MC

0767_1842

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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.
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

2,294

3,216

TOTAL =

5,600

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 R and T

A. distance =

½(3,216) + 3×(0)

5,600

1,608 + 0

5,600

1,608

5,600

= 0.2871 = 28.71 cM Incorrect B. distance =

½(2,294 + 3,216) + 3×(90)

5,600

2,755 + 270

5,600

3,025

5,600

= 0.5402 = 54.02 cM Incorrect C. distance =

½(2,294) + 3×(90)

5,600

1,147 + 270

5,600

1,417

5,600

= 0.2530 = 25.30 cM Correct D. distance =

½(3,216) + 3×(90)

5,600

1,608 + 270

5,600

1,878

5,600

= 0.3354 = 33.54 cM Incorrect E. distance =

½(90) + 3×(0)

5,600

45 + 0

5,600

= 0.0080 = 0.80 cM Incorrect MC

2821_ef54

Unordered Tetrad Two Gene Mapping

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 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

1,642

2,702

TOTAL =

4,400

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 M

A. distance =

½(2,702) + 3×(0)

4,400

1,351 + 0

4,400

1,351

4,400

= 0.3070 = 30.70 cM Incorrect B. distance =

½(0) + 3×(56)

4,400

0 + 168

4,400

168

4,400

= 0.0382 = 3.82 cM Incorrect C. distance =

½(1,642) + 3×(0)

4,400

821 + 0

4,400

821

4,400

= 0.1866 = 18.66 cM Incorrect D. distance =

½(1,642) + 3×(56)

4,400

821 + 168

4,400

989

4,400

= 0.2248 = 22.48 cM Correct E. distance =

½(1,642 + 2,702) + 3×(56)

4,400

2,172 + 168

4,400

2,340

4,400

= 0.5318 = 53.18 cM Incorrect MC

a439_af0b

Unordered Tetrad Two Gene Mapping

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 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

940

822

TOTAL =

1,800

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 F

A. distance =

½(940) + 3×(38)

1,800

470 + 114

1,800

584

1,800

= 0.3244 = 32.44 cM Incorrect B. distance =

½(822 + 940) + 3×(38)

1,800

881 + 114

1,800

995

1,800

= 0.5528 = 55.28 cM Incorrect C. distance =

½(822) + 3×(38)

1,800

411 + 114

1,800

525

1,800

= 0.2917 = 29.17 cM Correct D. distance =

½(0) + 3×(38)

1,800

0 + 114

1,800

114

1,800

= 0.0633 = 6.33 cM Incorrect E. distance =

½(38) + 3×(0)

1,800

19 + 0

1,800

= 0.0106 = 1.06 cM Incorrect MC

c6ae_2384

Unordered Tetrad Two Gene Mapping

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 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.

Set #

Tetrad Genotypes

Progeny
Count

138

3,538

4,924

TOTAL =

8,600

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 J

A. distance =

½(3,538 + 4,924) + 3×(138)

8,600

4,231 + 414

8,600

4,645

8,600

= 0.5401 = 54.01 cM Incorrect B. distance =

½(3,538) + 3×(0)

8,600

1,769 + 0

8,600

1,769

8,600

= 0.2057 = 20.57 cM Incorrect C. distance =

½(138) + 3×(0)

8,600

69 + 0

8,600

= 0.0080 = 0.80 cM Incorrect D. distance =

½(3,538) + 3×(138)

8,600

1,769 + 414

8,600

2,183

8,600

= 0.2538 = 25.38 cM Correct E. distance =

½(0) + 3×(138)

8,600

0 + 414

8,600

414

8,600

= 0.0481 = 4.81 cM Incorrect MC

48ec_d97e

Unordered Tetrad Two Gene Mapping

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 N is related 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.

Set #

Tetrad Genotypes

Progeny
Count

1,805

966

TOTAL =

2,800

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(1,805) + 3×(0)

2,800

902.5 + 0

2,800

902.5

2,800

= 0.3223 = 32.23 cM Incorrect B. distance =

½(0) + 3×(29)

2,800

0 + 87

2,800

= 0.0311 = 3.11 cM Incorrect C. distance =

½(966) + 3×(29)

2,800

483 + 87

2,800

570

2,800

= 0.2036 = 20.36 cM Correct D. distance =

½(29) + 3×(0)

2,800

14.5 + 0

2,800

14.5

2,800

= 0.0052 = 0.52 cM Incorrect E. distance =

½(29) + 3×(29)

2,800

14.5 + 87

2,800

101.5

2,800

= 0.0362 = 3.62 cM Incorrect MC

73fd_a3d5

Unordered Tetrad Two Gene Mapping

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 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

1,802

1,344

TOTAL =

3,200

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 Y

A. distance =

½(1,802) + 3×(54)

3,200

901 + 162

3,200

1,063

3,200

= 0.3322 = 33.22 cM Incorrect B. distance =

½(1,344) + 3×(0)

3,200

672 + 0

3,200

672

3,200

= 0.2100 = 21 cM Incorrect C. distance =

½(1,344) + 3×(54)

3,200

672 + 162

3,200

834

3,200

= 0.2606 = 26.06 cM Correct D. distance =

½(1,344 + 1,802) + 3×(54)

3,200

1,573 + 162

3,200

1,735

3,200

= 0.5422 = 54.22 cM Incorrect E. distance =

½(54) + 3×(54)

3,200

27 + 162

3,200

189

3,200

= 0.0591 = 5.91 cM Incorrect MC

66bf_9f1b

Unordered Tetrad Two Gene Mapping

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 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

5,599

3,856

145

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 A and X

A. distance =

½(5,599) + 3×(0)

9,600

2799.5 + 0

9,600

2799.5

9,600

= 0.2916 = 29.16 cM Incorrect B. distance =

½(5,599) + 3×(145)

9,600

2799.5 + 435

9,600

3234.5

9,600

= 0.3369 = 33.69 cM Incorrect C. distance =

½(3,856) + 3×(0)

9,600

1,928 + 0

9,600

1,928

9,600

= 0.2008 = 20.08 cM Incorrect D. distance =

½(145) + 3×(145)

9,600

72.5 + 435

9,600

507.5

9,600

= 0.0529 = 5.29 cM Incorrect E. distance =

½(3,856) + 3×(145)

9,600

1,928 + 435

9,600

2,363

9,600

= 0.2461 = 24.61 cM Correct MC

1080_40bb

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

1,815

1,700

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 M and R

A. distance =

½(1,815) + 3×(85)

3,600

907.5 + 255

3,600

1162.5

3,600

= 0.3229 = 32.29 cM Incorrect B. distance =

½(1,700) + 3×(85)

3,600

850 + 255

3,600

1,105

3,600

= 0.3069 = 30.69 cM Correct C. distance =

½(1,700) + 3×(0)

3,600

850 + 0

3,600

850

3,600

= 0.2361 = 23.61 cM Incorrect D. distance =

½(85) + 3×(85)

3,600

42.5 + 255

3,600

297.5

3,600

= 0.0826 = 8.26 cM Incorrect E. distance =

½(1,815) + 3×(0)

3,600

907.5 + 0

3,600

907.5

3,600

= 0.2521 = 25.21 cM Incorrect MC

b4cc_90f0

Unordered Tetrad Two Gene Mapping

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 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

2,223

1,520

TOTAL =

3,800

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 K

A. distance =

½(1,520) + 3×(57)

3,800

760 + 171

3,800

931

3,800

= 0.2450 = 24.50 cM Correct B. distance =

½(0) + 3×(57)

3,800

0 + 171

3,800

171

3,800

= 0.0450 = 4.50 cM Incorrect C. distance =

½(57) + 3×(0)

3,800

28.5 + 0

3,800

28.5

3,800

= 0.0075 = 0.75 cM Incorrect D. distance =

½(1,520 + 2,223) + 3×(57)

3,800

1871.5 + 171

3,800

2042.5

3,800

= 0.5375 = 53.75 cM Incorrect E. distance =

½(2,223) + 3×(57)

3,800

1111.5 + 171

3,800

1282.5

3,800

= 0.3375 = 33.75 cM Incorrect MC

cffe_2c01

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

Gene N is related 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 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

2,251

2,050

TOTAL =

4,400

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 R

A. distance =

½(99) + 3×(0)

4,400

49.5 + 0

4,400

49.5

4,400

= 0.0112 = 1.12 cM Incorrect B. distance =

½(99) + 3×(99)

4,400

49.5 + 297

4,400

346.5

4,400

= 0.0788 = 7.88 cM Incorrect C. distance =

½(2,251) + 3×(99)

4,400

1125.5 + 297

4,400

1422.5

4,400

= 0.3233 = 32.33 cM Incorrect D. distance =

½(2,251) + 3×(0)

4,400

1125.5 + 0

4,400

1125.5

4,400

= 0.2558 = 25.58 cM Incorrect E. distance =

½(2,050) + 3×(99)

4,400

1,025 + 297

4,400

1,322

4,400

= 0.3005 = 30.05 cM Correct MC

b02d_7bd9

Unordered Tetrad Two Gene Mapping

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 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.

Set #

Tetrad Genotypes

Progeny
Count

1,976

2,334

TOTAL =

4,400

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 K

A. distance =

½(90) + 3×(0)

4,400

45 + 0

4,400

= 0.0102 = 1.02 cM Incorrect B. distance =

½(2,334) + 3×(0)

4,400

1,167 + 0

4,400

1,167

4,400

= 0.2652 = 26.52 cM Incorrect C. distance =

½(1,976) + 3×(90)

4,400

988 + 270

4,400

1,258

4,400

= 0.2859 = 28.59 cM Correct D. distance =

½(1,976 + 2,334) + 3×(90)

4,400

2,155 + 270

4,400

2,425

4,400

= 0.5511 = 55.11 cM Incorrect E. distance =

½(0) + 3×(90)

4,400

0 + 270

4,400

270

4,400

= 0.0614 = 6.14 cM Incorrect MC

825e_ad41

Unordered Tetrad Two Gene Mapping

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 N is related 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.

Set #

Tetrad Genotypes

Progeny
Count

2,248

3,267

TOTAL =

5,600

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 N

A. distance =

½(3,267) + 3×(0)

5,600

1633.5 + 0

5,600

1633.5

5,600

= 0.2917 = 29.17 cM Incorrect B. distance =

½(85) + 3×(85)

5,600

42.5 + 255

5,600

297.5

5,600

= 0.0531 = 5.31 cM Incorrect C. distance =

½(2,248) + 3×(0)

5,600

1,124 + 0

5,600

1,124

5,600

= 0.2007 = 20.07 cM Incorrect D. distance =

½(2,248) + 3×(85)

5,600

1,124 + 255

5,600

1,379

5,600

= 0.2462 = 24.62 cM Correct E. distance =

½(0) + 3×(85)

5,600

0 + 255

5,600

255

5,600

= 0.0455 = 4.55 cM Incorrect MC

c8bf_08f2

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

Gene N is associated 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 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

889

866

TOTAL =

1,800

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 N and W

A. distance =

½(889) + 3×(0)

1,800

444.5 + 0

1,800

444.5

1,800

= 0.2469 = 24.69 cM Incorrect B. distance =

½(866) + 3×(45)

1,800

433 + 135

1,800

568

1,800

= 0.3156 = 31.56 cM Correct C. distance =

½(866 + 889) + 3×(45)

1,800

877.5 + 135

1,800

1012.5

1,800

= 0.5625 = 56.25 cM Incorrect D. distance =

½(0) + 3×(45)

1,800

0 + 135

1,800

135

1,800

= 0.0750 = 7.50 cM Incorrect E. distance =

½(45) + 3×(0)

1,800

22.5 + 0

1,800

22.5

1,800

= 0.0125 = 1.25 cM Incorrect MC

e125_e586

Unordered Tetrad Two Gene Mapping

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 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

3,468

3,360

172

TOTAL =

7,000

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 P

A. distance =

½(3,468) + 3×(0)

7,000

1,734 + 0

7,000

1,734

7,000

= 0.2477 = 24.77 cM Incorrect B. distance =

½(3,360 + 3,468) + 3×(172)

7,000

3,414 + 516

7,000

3,930

7,000

= 0.5614 = 56.14 cM Incorrect C. distance =

½(0) + 3×(172)

7,000

0 + 516

7,000

516

7,000

= 0.0737 = 7.37 cM Incorrect D. distance =

½(172) + 3×(0)

7,000

86 + 0

7,000

= 0.0123 = 1.23 cM Incorrect E. distance =

½(3,360) + 3×(172)

7,000

1,680 + 516

7,000

2,196

7,000

= 0.3137 = 31.37 cM Correct MC

a900_6582

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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.
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

121

3,358

5,121

TOTAL =

8,600

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 W and X

A. distance =

½(3,358) + 3×(121)

8,600

1,679 + 363

8,600

2,042

8,600

= 0.2374 = 23.74 cM Correct B. distance =

½(121) + 3×(0)

8,600

60.5 + 0

8,600

60.5

8,600

= 0.0070 = 0.70 cM Incorrect C. distance =

½(5,121) + 3×(121)

8,600

2560.5 + 363

8,600

2923.5

8,600

= 0.3399 = 33.99 cM Incorrect D. distance =

½(5,121) + 3×(0)

8,600

2560.5 + 0

8,600

2560.5

8,600

= 0.2977 = 29.77 cM Incorrect E. distance =

½(3,358) + 3×(0)

8,600

1,679 + 0

8,600

1,679

8,600

= 0.1952 = 19.52 cM Incorrect MC

c1e7_ec6c

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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.

Set #

Tetrad Genotypes

Progeny
Count

1,238

1,110

TOTAL =

2,400

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 H and N

A. distance =

½(1,110 + 1,238) + 3×(52)

2,400

1,174 + 156

2,400

1,330

2,400

= 0.5542 = 55.42 cM Incorrect B. distance =

½(52) + 3×(0)

2,400

26 + 0

2,400

= 0.0108 = 1.08 cM Incorrect C. distance =

½(1,110) + 3×(52)

2,400

555 + 156

2,400

711

2,400

= 0.2963 = 29.62 cM Correct D. distance =

½(1,238) + 3×(0)

2,400

619 + 0

2,400

619

2,400

= 0.2579 = 25.79 cM Incorrect E. distance =

½(1,110) + 3×(0)

2,400

555 + 0

2,400

555

2,400

= 0.2313 = 23.12 cM Incorrect MC

f9c0_8522

Unordered Tetrad Two Gene Mapping

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 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

1,166

1,173

TOTAL =

2,400

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 E and Y

A. distance =

½(61) + 3×(61)

2,400

30.5 + 183

2,400

213.5

2,400

= 0.0890 = 8.90 cM Incorrect B. distance =

½(0) + 3×(61)

2,400

0 + 183

2,400

183

2,400

= 0.0762 = 7.62 cM Incorrect C. distance =

½(1,166 + 1,173) + 3×(61)

2,400

1169.5 + 183

2,400

1352.5

2,400

= 0.5635 = 56.35 cM Incorrect D. distance =

½(1,173) + 3×(0)

2,400

586.5 + 0

2,400

586.5

2,400

= 0.2444 = 24.44 cM Incorrect E. distance =

½(1,166) + 3×(61)

2,400

583 + 183

2,400

766

2,400

= 0.3192 = 31.92 cM Correct MC

53af_732d

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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.

Set #

Tetrad Genotypes

Progeny
Count

2,217

1,526

TOTAL =

3,800

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 H and X

A. distance =

½(1,526 + 2,217) + 3×(57)

3,800

1871.5 + 171

3,800

2042.5

3,800

= 0.5375 = 53.75 cM Incorrect B. distance =

½(57) + 3×(57)

3,800

28.5 + 171

3,800

199.5

3,800

= 0.0525 = 5.25 cM Incorrect C. distance =

½(1,526) + 3×(57)

3,800

763 + 171

3,800

934

3,800

= 0.2458 = 24.58 cM Correct D. distance =

½(2,217) + 3×(57)

3,800

1108.5 + 171

3,800

1279.5

3,800

= 0.3367 = 33.67 cM Incorrect E. distance =

½(57) + 3×(0)

3,800

28.5 + 0

3,800

28.5

3,800

= 0.0075 = 0.75 cM Incorrect MC

0803_2388

Unordered Tetrad Two Gene Mapping

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 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

1,210

950

TOTAL =

2,200

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 W

A. distance =

½(1,210) + 3×(40)

2,200

605 + 120

2,200

725

2,200

= 0.3295 = 32.95 cM Incorrect B. distance =

½(950) + 3×(40)

2,200

475 + 120

2,200

595

2,200

= 0.2705 = 27.05 cM Correct C. distance =

½(950) + 3×(0)

2,200

475 + 0

2,200

475

2,200

= 0.2159 = 21.59 cM Incorrect D. distance =

½(950 + 1,210) + 3×(40)

2,200

1,080 + 120

2,200

1,200

2,200

= 0.5455 = 54.55 cM Incorrect E. distance =

½(1,210) + 3×(0)

2,200

605 + 0

2,200

605

2,200

= 0.2750 = 27.50 cM Incorrect MC

8858_395d

Unordered Tetrad Two Gene Mapping

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 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.

Set #

Tetrad Genotypes

Progeny
Count

100

2,222

2,678

TOTAL =

5,000

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 W

A. distance =

½(2,222) + 3×(100)

5,000

1,111 + 300

5,000

1,411

5,000

= 0.2822 = 28.22 cM Correct B. distance =

½(100) + 3×(0)

5,000

50 + 0

5,000

= 0.0100 = 1 cM Incorrect C. distance =

½(2,678) + 3×(100)

5,000

1,339 + 300

5,000

1,639

5,000

= 0.3278 = 32.78 cM Incorrect D. distance =

½(2,222 + 2,678) + 3×(100)

5,000

2,450 + 300

5,000

2,750

5,000

= 0.5500 = 55 cM Incorrect E. distance =

½(2,222) + 3×(0)

5,000

1,111 + 0

5,000

1,111

5,000

= 0.2222 = 22.22 cM Incorrect MC

54b9_7ad4

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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.

Set #

Tetrad Genotypes

Progeny
Count

4,563

3,306

131

TOTAL =

8,000

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 F and J

A. distance =

½(4,563) + 3×(0)

8,000

2281.5 + 0

8,000

2281.5

8,000

= 0.2852 = 28.52 cM Incorrect B. distance =

½(4,563) + 3×(131)

8,000

2281.5 + 393

8,000

2674.5

8,000

= 0.3343 = 33.43 cM Incorrect C. distance =

½(3,306 + 4,563) + 3×(131)

8,000

3934.5 + 393

8,000

4327.5

8,000

= 0.5409 = 54.09 cM Incorrect D. distance =

½(131) + 3×(0)

8,000

65.5 + 0

8,000

65.5

8,000

= 0.0082 = 0.82 cM Incorrect E. distance =

½(3,306) + 3×(131)

8,000

1,653 + 393

8,000

2,046

8,000

= 0.2557 = 25.57 cM Correct MC

96d3_564c

Unordered Tetrad Two Gene Mapping

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 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.

Set #

Tetrad Genotypes

Progeny
Count

2,720

4,794

TOTAL =

7,600

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 J

A. distance =

½(4,794) + 3×(0)

7,600

2,397 + 0

7,600

2,397

7,600

= 0.3154 = 31.54 cM Incorrect B. distance =

½(0) + 3×(86)

7,600

0 + 258

7,600

258

7,600

= 0.0339 = 3.39 cM Incorrect C. distance =

½(86) + 3×(0)

7,600

43 + 0

7,600

= 0.0057 = 0.57 cM Incorrect D. distance =

½(2,720 + 4,794) + 3×(86)

7,600

3,757 + 258

7,600

4,015

7,600

= 0.5283 = 52.83 cM Incorrect E. distance =

½(2,720) + 3×(86)

7,600

1,360 + 258

7,600

1,618

7,600

= 0.2129 = 21.29 cM Correct MC

b81b_d79c

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

2,810

2,656

134

TOTAL =

5,600

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 Y

A. distance =

½(2,656) + 3×(134)

5,600

1,328 + 402

5,600

1,730

5,600

= 0.3089 = 30.89 cM Correct B. distance =

½(2,656) + 3×(0)

5,600

1,328 + 0

5,600

1,328

5,600

= 0.2371 = 23.71 cM Incorrect C. distance =

½(2,656 + 2,810) + 3×(134)

5,600

2,733 + 402

5,600

3,135

5,600

= 0.5598 = 55.98 cM Incorrect D. distance =

½(134) + 3×(0)

5,600

67 + 0

5,600

= 0.0120 = 1.20 cM Incorrect E. distance =

½(0) + 3×(134)

5,600

0 + 402

5,600

402

5,600

= 0.0718 = 7.18 cM Incorrect MC

8873_c112

Unordered Tetrad Two Gene Mapping

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 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

4,276

3,380

144

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 D and X

A. distance =

½(3,380) + 3×(144)

7,800

1,690 + 432

7,800

2,122

7,800

= 0.2721 = 27.21 cM Correct B. distance =

½(3,380 + 4,276) + 3×(144)

7,800

3,828 + 432

7,800

4,260

7,800

= 0.5462 = 54.62 cM Incorrect C. distance =

½(144) + 3×(144)

7,800

72 + 432

7,800

504

7,800

= 0.0646 = 6.46 cM Incorrect D. distance =

½(4,276) + 3×(144)

7,800

2,138 + 432

7,800

2,570

7,800

= 0.3295 = 32.95 cM Incorrect E. distance =

½(3,380) + 3×(0)

7,800

1,690 + 0

7,800

1,690

7,800

= 0.2167 = 21.67 cM Incorrect MC

2c29_d144

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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.

Set #

Tetrad Genotypes

Progeny
Count

4,396

2,334

TOTAL =

6,800

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 T and W

A. distance =

½(4,396) + 3×(0)

6,800

2,198 + 0

6,800

2,198

6,800

= 0.3232 = 32.32 cM Incorrect B. distance =

½(2,334) + 3×(0)

6,800

1,167 + 0

6,800

1,167

6,800

= 0.1716 = 17.16 cM Incorrect C. distance =

½(2,334) + 3×(70)

6,800

1,167 + 210

6,800

1,377

6,800

= 0.2025 = 20.25 cM Correct D. distance =

½(2,334 + 4,396) + 3×(70)

6,800

3,365 + 210

6,800

3,575

6,800

= 0.5257 = 52.57 cM Incorrect E. distance =

½(0) + 3×(70)

6,800

0 + 210

6,800

210

6,800

= 0.0309 = 3.09 cM Incorrect MC

4440_fc27

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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.
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

2,648

2,616

136

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 N and T

A. distance =

½(136) + 3×(0)

5,400

68 + 0

5,400

= 0.0126 = 1.26 cM Incorrect B. distance =

½(136) + 3×(136)

5,400

68 + 408

5,400

476

5,400

= 0.0881 = 8.81 cM Incorrect C. distance =

½(0) + 3×(136)

5,400

0 + 408

5,400

408

5,400

= 0.0756 = 7.56 cM Incorrect D. distance =

½(2,616) + 3×(0)

5,400

1,308 + 0

5,400

1,308

5,400

= 0.2422 = 24.22 cM Incorrect E. distance =

½(2,616) + 3×(136)

5,400

1,308 + 408

5,400

1,716

5,400

= 0.3178 = 31.78 cM Correct MC

d319_4725

Unordered Tetrad Two Gene Mapping

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 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.

Set #

Tetrad Genotypes

Progeny
Count

2,011

1,526

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 D and M

A. distance =

½(1,526) + 3×(63)

3,600

763 + 189

3,600

952

3,600

= 0.2644 = 26.44 cM Correct B. distance =

½(63) + 3×(63)

3,600

31.5 + 189

3,600

220.5

3,600

= 0.0612 = 6.12 cM Incorrect C. distance =

½(1,526 + 2,011) + 3×(63)

3,600

1768.5 + 189

3,600

1957.5

3,600

= 0.5437 = 54.37 cM Incorrect D. distance =

½(0) + 3×(63)

3,600

0 + 189

3,600

189

3,600

= 0.0525 = 5.25 cM Incorrect E. distance =

½(1,526) + 3×(0)

3,600

763 + 0

3,600

763

3,600

= 0.2119 = 21.19 cM Incorrect MC

c4fb_adcd

Unordered Tetrad Two Gene Mapping

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 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

1,310

1,044

TOTAL =

2,400

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 F and X

A. distance =

½(46) + 3×(46)

2,400

23 + 138

2,400

161

2,400

= 0.0671 = 6.71 cM Incorrect B. distance =

½(0) + 3×(46)

2,400

0 + 138

2,400

138

2,400

= 0.0575 = 5.75 cM Incorrect C. distance =

½(1,044) + 3×(0)

2,400

522 + 0

2,400

522

2,400

= 0.2175 = 21.75 cM Incorrect D. distance =

½(46) + 3×(0)

2,400

23 + 0

2,400

= 0.0096 = 0.96 cM Incorrect E. distance =

½(1,044) + 3×(46)

2,400

522 + 138

2,400

660

2,400

= 0.2750 = 27.50 cM Correct MC

0e98_dc06

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

988

1,373

TOTAL =

2,400

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 P and Y

A. distance =

½(39) + 3×(0)

2,400

19.5 + 0

2,400

19.5

2,400

= 0.0081 = 0.81 cM Incorrect B. distance =

½(39) + 3×(39)

2,400

19.5 + 117

2,400

136.5

2,400

= 0.0569 = 5.69 cM Incorrect C. distance =

½(1,373) + 3×(0)

2,400

686.5 + 0

2,400

686.5

2,400

= 0.2860 = 28.60 cM Incorrect D. distance =

½(988) + 3×(39)

2,400

494 + 117

2,400

611

2,400

= 0.2546 = 25.46 cM Correct E. distance =

½(0) + 3×(39)

2,400

0 + 117

2,400

117

2,400

= 0.0488 = 4.88 cM Incorrect MC

7293_0b2c

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

150

3,082

3,368

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 H and N

A. distance =

½(150) + 3×(0)

6,600

75 + 0

6,600

= 0.0114 = 1.14 cM Incorrect B. distance =

½(3,082) + 3×(150)

6,600

1,541 + 450

6,600

1,991

6,600

= 0.3017 = 30.17 cM Correct C. distance =

½(3,368) + 3×(0)

6,600

1,684 + 0

6,600

1,684

6,600

= 0.2552 = 25.52 cM Incorrect D. distance =

½(0) + 3×(150)

6,600

0 + 450

6,600

450

6,600

= 0.0682 = 6.82 cM Incorrect E. distance =

½(3,082 + 3,368) + 3×(150)

6,600

3,225 + 450

6,600

3,675

6,600

= 0.5568 = 55.68 cM Incorrect MC

ba4f_6788

Unordered Tetrad Two Gene Mapping

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 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

5,120

3,358

122

TOTAL =

8,600

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(3,358 + 5,120) + 3×(122)

8,600

4,239 + 366

8,600

4,605

8,600

= 0.5355 = 53.55 cM Incorrect B. distance =

½(122) + 3×(122)

8,600

61 + 366

8,600

427

8,600

= 0.0497 = 4.97 cM Incorrect C. distance =

½(5,120) + 3×(122)

8,600

2,560 + 366

8,600

2,926

8,600

= 0.3402 = 34.02 cM Incorrect D. distance =

½(3,358) + 3×(122)

8,600

1,679 + 366

8,600

2,045

8,600

= 0.2378 = 23.78 cM Correct E. distance =

½(122) + 3×(0)

8,600

61 + 0

8,600

= 0.0071 = 0.71 cM Incorrect MC

69a4_38a0

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

Gene N is associated 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 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

5,112

3,366

122

TOTAL =

8,600

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 N and X

A. distance =

½(3,366) + 3×(122)

8,600

1,683 + 366

8,600

2,049

8,600

= 0.2383 = 23.83 cM Correct B. distance =

½(5,112) + 3×(0)

8,600

2,556 + 0

8,600

2,556

8,600

= 0.2972 = 29.72 cM Incorrect C. distance =

½(122) + 3×(122)

8,600

61 + 366

8,600

427

8,600

= 0.0497 = 4.97 cM Incorrect D. distance =

½(3,366) + 3×(0)

8,600

1,683 + 0

8,600

1,683

8,600

= 0.1957 = 19.57 cM Incorrect E. distance =

½(0) + 3×(122)

8,600

0 + 366

8,600

366

8,600

= 0.0426 = 4.26 cM Incorrect MC

15c3_ec99

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

3,123

2,748

129

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 H and W

A. distance =

½(3,123) + 3×(0)

6,000

1561.5 + 0

6,000

1561.5

6,000

= 0.2602 = 26.02 cM Incorrect B. distance =

½(3,123) + 3×(129)

6,000

1561.5 + 387

6,000

1948.5

6,000

= 0.3247 = 32.48 cM Incorrect C. distance =

½(0) + 3×(129)

6,000

0 + 387

6,000

387

6,000

= 0.0645 = 6.45 cM Incorrect D. distance =

½(2,748) + 3×(129)

6,000

1,374 + 387

6,000

1,761

6,000

= 0.2935 = 29.35 cM Correct E. distance =

½(2,748 + 3,123) + 3×(129)

6,000

2935.5 + 387

6,000

3322.5

6,000

= 0.5537 = 55.38 cM Incorrect MC

c1b7_3ff5

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

2,982

2,688

130

TOTAL =

5,800

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 M and W

A. distance =

½(2,982) + 3×(0)

5,800

1,491 + 0

5,800

1,491

5,800

= 0.2571 = 25.71 cM Incorrect B. distance =

½(130) + 3×(130)

5,800

65 + 390

5,800

455

5,800

= 0.0784 = 7.84 cM Incorrect C. distance =

½(0) + 3×(130)

5,800

0 + 390

5,800

390

5,800

= 0.0672 = 6.72 cM Incorrect D. distance =

½(2,688) + 3×(130)

5,800

1,344 + 390

5,800

1,734

5,800

= 0.2990 = 29.90 cM Correct E. distance =

½(2,982) + 3×(130)

5,800

1,491 + 390

5,800

1,881

5,800

= 0.3243 = 32.43 cM Incorrect MC

50fa_f065

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

4,755

2,944

101

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 H and Y

A. distance =

½(101) + 3×(101)

7,800

50.5 + 303

7,800

353.5

7,800

= 0.0453 = 4.53 cM Incorrect B. distance =

½(4,755) + 3×(0)

7,800

2377.5 + 0

7,800

2377.5

7,800

= 0.3048 = 30.48 cM Incorrect C. distance =

½(2,944) + 3×(101)

7,800

1,472 + 303

7,800

1,775

7,800

= 0.2276 = 22.76 cM Correct D. distance =

½(2,944 + 4,755) + 3×(101)

7,800

3849.5 + 303

7,800

4152.5

7,800

= 0.5324 = 53.24 cM Incorrect E. distance =

½(0) + 3×(101)

7,800

0 + 303

7,800

303

7,800

= 0.0388 = 3.88 cM Incorrect MC

2c94_6ed7

Unordered Tetrad Two Gene Mapping

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 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

112

2,554

3,134

TOTAL =

5,800

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 M

A. distance =

½(112) + 3×(112)

5,800

56 + 336

5,800

392

5,800

= 0.0676 = 6.76 cM Incorrect B. distance =

½(2,554) + 3×(0)

5,800

1,277 + 0

5,800

1,277

5,800

= 0.2202 = 22.02 cM Incorrect C. distance =

½(3,134) + 3×(112)

5,800

1,567 + 336

5,800

1,903

5,800

= 0.3281 = 32.81 cM Incorrect D. distance =

½(2,554) + 3×(112)

5,800

1,277 + 336

5,800

1,613

5,800

= 0.2781 = 27.81 cM Correct E. distance =

½(2,554 + 3,134) + 3×(112)

5,800

2,844 + 336

5,800

3,180

5,800

= 0.5483 = 54.83 cM Incorrect MC

2eae_bfdd

Unordered Tetrad Two Gene Mapping

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 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

2,528

3,779

TOTAL =

6,400

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 F and W

A. distance =

½(3,779) + 3×(0)

6,400

1889.5 + 0

6,400

1889.5

6,400

= 0.2952 = 29.52 cM Incorrect B. distance =

½(0) + 3×(93)

6,400

0 + 279

6,400

279

6,400

= 0.0436 = 4.36 cM Incorrect C. distance =

½(2,528 + 3,779) + 3×(93)

6,400

3153.5 + 279

6,400

3432.5

6,400

= 0.5363 = 53.63 cM Incorrect D. distance =

½(93) + 3×(93)

6,400

46.5 + 279

6,400

325.5

6,400

= 0.0509 = 5.09 cM Incorrect E. distance =

½(2,528) + 3×(93)

6,400

1,264 + 279

6,400

1,543

6,400

= 0.2411 = 24.11 cM Correct MC

4ca6_fb05

Unordered Tetrad Two Gene Mapping

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 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

5,118

2,986

TOTAL =

8,200

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 E and N

A. distance =

½(2,986 + 5,118) + 3×(96)

8,200

4,052 + 288

8,200

4,340

8,200

= 0.5293 = 52.93 cM Incorrect B. distance =

½(96) + 3×(96)

8,200

48 + 288

8,200

336

8,200

= 0.0410 = 4.10 cM Incorrect C. distance =

½(96) + 3×(0)

8,200

48 + 0

8,200

= 0.0059 = 0.59 cM Incorrect D. distance =

½(0) + 3×(96)

8,200

0 + 288

8,200

288

8,200

= 0.0351 = 3.51 cM Incorrect E. distance =

½(2,986) + 3×(96)

8,200

1,493 + 288

8,200

1,781

8,200

= 0.2172 = 21.72 cM Correct MC

58d2_fe02

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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.
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

1,318

1,830

TOTAL =

3,200

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 H and W

A. distance =

½(1,318) + 3×(52)

3,200

659 + 156

3,200

815

3,200

= 0.2547 = 25.47 cM Correct B. distance =

½(1,318) + 3×(0)

3,200

659 + 0

3,200

659

3,200

= 0.2059 = 20.59 cM Incorrect C. distance =

½(1,830) + 3×(0)

3,200

915 + 0

3,200

915

3,200

= 0.2859 = 28.59 cM Incorrect D. distance =

½(52) + 3×(52)

3,200

26 + 156

3,200

182

3,200

= 0.0569 = 5.69 cM Incorrect E. distance =

½(0) + 3×(52)

3,200

0 + 156

3,200

156

3,200

= 0.0488 = 4.88 cM Incorrect MC

d65e_e33a

Unordered Tetrad Two Gene Mapping

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 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

2,903

2,206

TOTAL =

5,200

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 R

A. distance =

½(2,206 + 2,903) + 3×(91)

5,200

2554.5 + 273

5,200

2827.5

5,200

= 0.5437 = 54.37 cM Incorrect B. distance =

½(91) + 3×(0)

5,200

45.5 + 0

5,200

45.5

5,200

= 0.0088 = 0.88 cM Incorrect C. distance =

½(2,903) + 3×(0)

5,200

1451.5 + 0

5,200

1451.5

5,200

= 0.2791 = 27.91 cM Incorrect D. distance =

½(2,206) + 3×(91)

5,200

1,103 + 273

5,200

1,376

5,200

= 0.2646 = 26.46 cM Correct E. distance =

½(0) + 3×(91)

5,200

0 + 273

5,200

273

5,200

= 0.0525 = 5.25 cM Incorrect MC

14e1_ca08

Unordered Tetrad Two Gene Mapping

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 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

4,318

2,596

TOTAL =

7,000

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 F

A. distance =

½(4,318) + 3×(86)

7,000

2,159 + 258

7,000

2,417

7,000

= 0.3453 = 34.53 cM Incorrect B. distance =

½(0) + 3×(86)

7,000

0 + 258

7,000

258

7,000

= 0.0369 = 3.69 cM Incorrect C. distance =

½(86) + 3×(0)

7,000

43 + 0

7,000

= 0.0061 = 0.61 cM Incorrect D. distance =

½(2,596) + 3×(0)

7,000

1,298 + 0

7,000

1,298

7,000

= 0.1854 = 18.54 cM Incorrect E. distance =

½(2,596) + 3×(86)

7,000

1,298 + 258

7,000

1,556

7,000

= 0.2223 = 22.23 cM Correct MC

e719_90f4

Unordered Tetrad Two Gene Mapping

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 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.

Set #

Tetrad Genotypes

Progeny
Count

2,206

3,517

TOTAL =

5,800

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 T

A. distance =

½(77) + 3×(77)

5,800

38.5 + 231

5,800

269.5

5,800

= 0.0465 = 4.65 cM Incorrect B. distance =

½(0) + 3×(77)

5,800

0 + 231

5,800

231

5,800

= 0.0398 = 3.98 cM Incorrect C. distance =

½(77) + 3×(0)

5,800

38.5 + 0

5,800

38.5

5,800

= 0.0066 = 0.66 cM Incorrect D. distance =

½(2,206) + 3×(77)

5,800

1,103 + 231

5,800

1,334

5,800

= 0.2300 = 23 cM Correct E. distance =

½(3,517) + 3×(0)

5,800

1758.5 + 0

5,800

1758.5

5,800

= 0.3032 = 30.32 cM Incorrect MC

abe3_6535

Unordered Tetrad Two Gene Mapping

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 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

3,764

1,978

TOTAL =

5,800

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 N and P

A. distance =

½(3,764) + 3×(58)

5,800

1,882 + 174

5,800

2,056

5,800

= 0.3545 = 35.45 cM Incorrect B. distance =

½(58) + 3×(0)

5,800

29 + 0

5,800

= 0.0050 = 0.50 cM Incorrect C. distance =

½(1,978 + 3,764) + 3×(58)

5,800

2,871 + 174

5,800

3,045

5,800

= 0.5250 = 52.50 cM Incorrect D. distance =

½(1,978) + 3×(0)

5,800

989 + 0

5,800

989

5,800

= 0.1705 = 17.05 cM Incorrect E. distance =

½(1,978) + 3×(58)

5,800

989 + 174

5,800

1,163

5,800

= 0.2005 = 20.05 cM Correct MC

7160_b993

Unordered Tetrad Two Gene Mapping

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 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

108

3,342

5,750

TOTAL =

9,200

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 H

A. distance =

½(5,750) + 3×(108)

9,200

2,875 + 324

9,200

3,199

9,200

= 0.3477 = 34.77 cM Incorrect B. distance =

½(3,342) + 3×(108)

9,200

1,671 + 324

9,200

1,995

9,200

= 0.2168 = 21.68 cM Correct C. distance =

½(3,342) + 3×(0)

9,200

1,671 + 0

9,200

1,671

9,200

= 0.1816 = 18.16 cM Incorrect D. distance =

½(0) + 3×(108)

9,200

0 + 324

9,200

324

9,200

= 0.0352 = 3.52 cM Incorrect E. distance =

½(108) + 3×(0)

9,200

54 + 0

9,200

= 0.0059 = 0.59 cM Incorrect MC

37e6_e7f6

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

166

3,254

3,380

TOTAL =

6,800

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 M and X

A. distance =

½(3,380) + 3×(166)

6,800

1,690 + 498

6,800

2,188

6,800

= 0.3218 = 32.18 cM Incorrect B. distance =

½(0) + 3×(166)

6,800

0 + 498

6,800

498

6,800

= 0.0732 = 7.32 cM Incorrect C. distance =

½(3,380) + 3×(0)

6,800

1,690 + 0

6,800

1,690

6,800

= 0.2485 = 24.85 cM Incorrect D. distance =

½(3,254 + 3,380) + 3×(166)

6,800

3,317 + 498

6,800

3,815

6,800

= 0.5610 = 56.10 cM Incorrect E. distance =

½(3,254) + 3×(166)

6,800

1,627 + 498

6,800

2,125

6,800

= 0.3125 = 31.25 cM Correct MC

e488_a739

Unordered Tetrad Two Gene Mapping

Characteristics of Recessive Phenotypes

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 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

1,066

1,900

TOTAL =

3,000

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 W

A. distance =

½(34) + 3×(34)

3,000

17 + 102

3,000

119

3,000

= 0.0397 = 3.97 cM Incorrect B. distance =

½(34) + 3×(0)

3,000

17 + 0

3,000

= 0.0057 = 0.57 cM Incorrect C. distance =

½(1,066) + 3×(34)

3,000

533 + 102

3,000

635

3,000

= 0.2117 = 21.17 cM Correct D. distance =

½(0) + 3×(34)

3,000

0 + 102

3,000

102

3,000

= 0.0340 = 3.40 cM Incorrect E. distance =

½(1,066 + 1,900) + 3×(34)

3,000

1,483 + 102

3,000

1,585

3,000

= 0.5283 = 52.83 cM Incorrect MC

5709_93fa

Unordered Tetrad Two Gene Mapping

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 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

2,182

1,182

TOTAL =

3,400

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 Y

A. distance =

½(1,182) + 3×(36)

3,400

591 + 108

3,400

699

3,400

= 0.2056 = 20.56 cM Correct B. distance =

½(2,182) + 3×(36)

3,400

1,091 + 108

3,400

1,199

3,400

= 0.3526 = 35.26 cM Incorrect C. distance =

½(36) + 3×(0)

3,400

18 + 0

3,400

= 0.0053 = 0.53 cM Incorrect D. distance =

½(36) + 3×(36)

3,400

18 + 108

3,400

126

3,400

= 0.0371 = 3.71 cM Incorrect E. distance =

½(0) + 3×(36)

3,400

0 + 108

3,400

108

3,400

= 0.0318 = 3.18 cM Incorrect