8: Gene Mapping

Students determine gene order on chromosomes using recombination data from test crosses, calculate map distances between linked genes, and use deletion mutants to establish gene arrangement.

LibreTexts reference: Chapter 8: Gene Mapping and Recombination

Gene Order Determination

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Using Deletion Mutants to Determine Gene Order

Deletion mutants are an essential tool in genetics for uncovering the order of five (5) genes on a chromosome. Deletions remove specific regions of the chromosome, allowing researchers to observe the effects of the missing genes on the phenotype of the organism. This approach is particularly useful for identifying the locations of recessive genes, which are only revealed when the corresponding wildtype copies are absent.
In a test cross involving deletion mutants, one parent carries a full-length wildtype chromosome and a second chromosome with a deletion, while the other parent is homozygous recessive for all five genes. Offspring inheriting the full-length wildtype chromosome display the dominant phenotype for all five genes in the test cross. However, offspring inheriting the chromosome with the deletion will display some recessive traits. These recessive traits uncover the missing genes in the deleted region. By analyzing which genes are uncovered in a series of different deletion mutants, the linear order of the genes can be determined.
In organisms such as Drosophila melanogaster, polytene chromosomes from the salivary glands provide a physical map for studying deletions. Polytene chromosomes are giant chromosomes with distinct banding patterns, allowing researchers to directly visualize which regions of the chromosome are deleted. This visual representation complements the genetic data obtained from test crosses.
For this problem, deletion mutants have been generated for a chromosome containing five genes. Your goal is to analyze the phenotypic data resulting from these deletions and determine the correct linear order of the genes.

Step-by-Step Instructions for Solving Deletion Mutant Problems

• Step 1: Simplify the information.
• List the genes and deletions provided in the question.
• Organize the deletions in a clear table or list format for easier analysis.
• Step 2: Create a template for the gene order.
• Start with placeholders for each gene (e.g., _ _ _ _).
• Insert known genes based on hints (e.g., the first or last gene).
• Step 3: Identify deletions containing the first gene.
• Analyze deletions that include the first gene to determine its neighbors.
• Use deletions that overlap to narrow down adjacent genes.
• Step 4: Analyze deletions containing the next genes.
• Look for deletions that include specific pairs of genes.
• Identify deletions that exclude certain genes to resolve ambiguities.
• Step 5: Verify the answer using all of the listed deletions.
• Deletion questions can be hard to solve, but once you have an answer, it is easy to check if it is correct!
• Go through each deletion and confirm that the proposed gene order matches the genes included in that deletion.
• If any deletion is inconsistent with the proposed order, your answer is wrong.

	Gene 1	Gene 2	Gene 3	Gene 4	Gene 5
Del #1
Del #2
Del #3
Del #4

There are five (5) genes, A, K, P, R, and Z, closely linked in a single chromosome. However, their order is unknown. In the region, four (4) deletions have been identified. These deletions uncover recessive alleles of the genes as follows:

• Deletion #1: K, P, and Z
• Deletion #2: A, K, P, and Z
• Deletion #3: A, R, and Z
• Deletion #4: A, P, R, and Z

Requirement: Enter your answer in the blank using only five (5) letters, or one comma every three (3) letters. Do not include extra commas or spaces in your answer.
Hint: The correct answer is a random sequence of five (5) letters.
What is the correct order of the five (5) genes?

• A. KPRAZ:   gene order of K, P, R, A, and Z
• B. KPRZA:   gene order of K, P, R, Z, and A
• C. KPZAR:   gene order of K, P, Z, A, and R
• D. KPZRA:   gene order of K, P, Z, R, and A
• E. KZPRA:   gene order of K, Z, P, R, and A

Check Answer

 

Gene Order Determination

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Using Deletion Mutants to Determine Gene Order

Deletion mutants are an essential tool in genetics for uncovering the order of five (5) genes on a chromosome. Deletions remove specific regions of the chromosome, allowing researchers to observe the effects of the missing genes on the phenotype of the organism. This approach is particularly useful for identifying the locations of recessive genes, which are only revealed when the corresponding wildtype copies are absent.
In a test cross involving deletion mutants, one parent carries a full-length wildtype chromosome and a second chromosome with a deletion, while the other parent is homozygous recessive for all five genes. Offspring inheriting the full-length wildtype chromosome display the dominant phenotype for all five genes in the test cross. However, offspring inheriting the chromosome with the deletion will display some recessive traits. These recessive traits uncover the missing genes in the deleted region. By analyzing which genes are uncovered in a series of different deletion mutants, the linear order of the genes can be determined.
In organisms such as Drosophila melanogaster, polytene chromosomes from the salivary glands provide a physical map for studying deletions. Polytene chromosomes are giant chromosomes with distinct banding patterns, allowing researchers to directly visualize which regions of the chromosome are deleted. This visual representation complements the genetic data obtained from test crosses.
For this problem, deletion mutants have been generated for a chromosome containing five genes. Your goal is to analyze the phenotypic data resulting from these deletions and determine the correct linear order of the genes.

Step-by-Step Instructions for Solving Deletion Mutant Problems

• Step 1: Simplify the information.
• List the genes and deletions provided in the question.
• Organize the deletions in a clear table or list format for easier analysis.
• Step 2: Create a template for the gene order.
• Start with placeholders for each gene (e.g., _ _ _ _).
• Insert known genes based on hints (e.g., the first or last gene).
• Step 3: Identify deletions containing the first gene.
• Analyze deletions that include the first gene to determine its neighbors.
• Use deletions that overlap to narrow down adjacent genes.
• Step 4: Analyze deletions containing the next genes.
• Look for deletions that include specific pairs of genes.
• Identify deletions that exclude certain genes to resolve ambiguities.
• Step 5: Verify the answer using all of the listed deletions.
• Deletion questions can be hard to solve, but once you have an answer, it is easy to check if it is correct!
• Go through each deletion and confirm that the proposed gene order matches the genes included in that deletion.
• If any deletion is inconsistent with the proposed order, your answer is wrong.

	Gene 1	Gene 2	Gene 3	Gene 4	Gene 5
Del #1
Del #2
Del #3
Del #4

There are five (5) genes, A, E, P, R, and S, closely linked in a single chromosome. However, their order is unknown. In the region, four (4) deletions have been identified. These deletions uncover recessive alleles of the genes as follows:

• Deletion #1: A, E, P, and S
• Deletion #2: A, E, P, and R
• Deletion #3: E, P, and S
• Deletion #4: A and E

Requirement: Enter your answer in the blank using only five (5) letters, or one comma every three (3) letters. Do not include extra commas or spaces in your answer.
Hint: The correct answer is an English dictionary word of length five (5).
What is the correct order of the five (5) genes?

• A. APERS:   gene order of A, P, E, R, and S
• B. APRES:   gene order of A, P, R, E, and S
• C. ASPER:   gene order of A, S, P, E, and R
• D. SPARE:   gene order of S, P, A, R, and E
• E. SPEAR:   gene order of S, P, E, A, and R

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Genetic Distance Calculation

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Ordered Tetrads in Neurospora crassa

Background Information

Neurospora crassa is an organism that has significantly contributed to the understanding of genetics. This fungus exhibits a distinctive genetic feature: the formation of ordered tetrads. These ordered tetrads result from the typical two rounds of meiotic divisions followed by a single round of mitotic division within an ascus, resulting in eight ascospores arranged in a predictable sequence. The position of each ascospore reflects the series of genetic events during cell division, providing a snapshot of the meiotic process.
The analysis of these ordered tetrads in Neurospora crassa allows for the classification of ascospores based on their allele arrangements into different segregation patterns.
The central principle in this analysis is the distinction between first-division and second-division segregation, which is based on the behavior of alleles in the presence or absence of crossover between a gene and its centromere. When alleles separate during the first meiotic division, it indicates first-division segregation. Conversely, if alleles separate during the second division, this suggests that a crossover event has occurred, leading to second-division segregation.
Counting the frequency of second-division segregation events within these ordered tetrads can provide an estimate of the genetic distance between a gene and its centromere. This frequency, reflective of the crossover events during meiosis, is used to calculate the recombination frequency. Such estimates are crucial for constructing genetic maps, which serve as a guide to the genetic landscape of Neurospora crassa, enhancing our understanding of genetic linkage and the location of genes relative to centromeres.

Experimental Data

In the table below, the six different patterns of ordered asci in Neurospora crassa are listed along with the counts found in an experiment.

Octad	Asci Count
+ + + + m m m m	1,047
+ + m m + + m m	440
+ + m m m m + +	480
m m + + + + m m	487
m m + + m m + +	549
m m m m + + + +	997
TOTAL	4,000

Distance Formula

distance between a gene and its centromere	=	½ × (asci with second-division segregation patterns)
		total number of asci

Question

Using the numbers of asci for each pattern shown in the table above, determine the genetic distance between gene M and its centromere.

• A. ^{½×(440 + 549 + 997 + 1,047)}/_4,000 = ^½×3,033/_4,000 = 0.3791 = 37.91 cM
  
• B. ^{½×(440 + 487 + 549 + 1,047)}/_4,000 = ^½×2,523/_4,000 = 0.3154 = 31.54 cM
  
• C. ^{½×(440 + 487 + 1,047)}/_4,000 = ^½×1,974/_4,000 = 0.2467 = 24.68 cM
  
• D. ^{½×(440 + 487 + 997 + 1,047)}/_4,000 = ^½×2,971/_4,000 = 0.3714 = 37.14 cM
  
• E. ^{½×(440 + 480 + 487 + 549)}/_4,000 = ^½×1,956/_4,000 = 0.2445 = 24.45 cM
  
• F. ^{½×(997 + 1,047)}/_4,000 = ^½×2,044/_4,000 = 0.2555 = 25.55 cM
  

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Gene Order and Map Distances

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Unordered Tetrad Three Gene Mapping

In this problem, you will use unordered tetrads to determine the order of three genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.

Characteristics of Recessive Phenotypes

• Gene E is related 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 K is analogous to the 'knotted' phenotype. A budding yeast that is homozygous recessive for Gene K cells grow in twisted, coiled shapes, resulting in a knotted or gnarled appearance.
• Gene T is linked with the 'toxic' phenotype. A budding yeast that is homozygous recessive for Gene T secretes a toxic compound that inhibits or kills other microbial colonies nearby.

Set #	Tetrad Genotypes				Progeny Count
1	+ + +	+ + +	e k t	e k t	53
2	+ + t	+ + t	e k +	e k +	7,314
3	+ + t	+ k +	e + t	e k +	2,076
4	+ + t	+ k t	e + +	e k +	6,192
5	+ k +	+ k +	e + t	e + t	25
6	+ k t	+ k t	e + +	e + +	240
TOTAL =					15,900

The resulting phenotypes are summarized in the table above.

Question

Using the table above, determine the order of the genes and the distances between them. Once calculated, fill in the following four blanks:

• The distance between genes E and K is cM (EK)
• The distance between genes E and T is cM (ET)
• The distance between genes K and T is cM (KT)
• From this, the correct order of the genes is (gene order).

Step-by-Step Instructions

• Step 1: Find the row for the Parental Type for all three genes.
• Step 2: Pick any two genes and 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
• Step 5: Go back to Step 2 and pick a new pair of genes until all pairs are complete.

Important Answer Guidelines

• Important Tip 1: Your calculated distances between each pair of genes should be a whole number. Finding a decimal in your answer, such as 5.5, indicates a mistake was made. Please provide your answer as a complete number without fractions or decimals.
• Important Tip 2: Your answer should be written as a numerical value only, with no spaces, commas, or units such as "cM" or "map units". For example, if the distance is fifty one centimorgans, simply write "51".
• Important Tip 3: Your gene order answer should be written as three letters only, with no spaces, commas, hyphens, or other characters allowed. For example, if the gene order is B - A - C, simply write "bac" or "cab".

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Determining Map Distance Between Genes

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Unordered Tetrad Three Gene Mapping

In this problem, you will use unordered tetrads to determine the between a single pair of genes and calculate the distances between them. The yeast Saccharomyces cerevisiae is used in this study. A cross has been performed to study the linkage relationships among three genes, and the resulting genotypes are summarized in the table below.

Characteristics of Recessive Phenotypes

• Gene B is connected with the 'bubbly' phenotype. A budding yeast that is homozygous recessive for Gene B produces excessive gas bubbles during growth, causing foamy appearance of the media.
• Gene F is affiliated with the 'fuzzy' phenotype. A budding yeast that is homozygous recessive for Gene F colonies are covered in soft, fine filaments, giving them a fuzzy, cotton-like texture.
• Gene R is connected with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.

Set #	Tetrad Genotypes				Progeny Count
1	+ + +	+ + +	b f r	b f r	150
2	+ + r	+ + r	b f +	b f +	234
3	+ + r	+ f r	b + +	b f +	3,696
4	+ f +	+ f +	b + r	b + r	366
5	+ f +	+ f r	b + +	b + r	4,404
6	+ f r	+ f r	b + +	b + +	6,150
TOTAL =					15,000

The resulting phenotypes are summarized in the table above.

Step-by-Step Instructions

• Step 1: Find the row for the Parental Type for all three genes.
• Step 2: Looking at only your two genes, assign PD, NPD, TT.
• Step 3: Determine if the two genes are linked.
• PD >> NPD → linked; PD ≈ NPD → unlinked
• Step 4: Determine the map distance between the two genes.
• D = ½ (TT + 6 NPD) / total = (3 NPD + ½ TT) / total

Determine the distance between the two genes B and R

• A. distance =

  ½(4,404) + 3×(150 + 366)

  15,000

  =

  2,202 + 1,548

  15,000

  =

  3,750

  15,000

  = 0.2500 = 25 cM
• B. distance =

  ½(3,696 + 4,404) + 3×(234 + 366)

  15,000

  =

  4,050 + 1,800

  15,000

  =

  5,850

  15,000

  = 0.3900 = 39 cM
• C. distance =

  ½(3,696) + 3×(150 + 234)

  15,000

  =

  1,848 + 1,152

  15,000

  =

  3,000

  15,000

  = 0.2000 = 20 cM
• D. distance =

  ½(3,696 + 4,404) + 3×(150)

  15,000

  =

  4,050 + 450

  15,000

  =

  4,500

  15,000

  = 0.3000 = 30 cM
• E. distance =

  ½(3,696 + 4,404) + 3×(150 + 234 + 366)

  15,000

  =

  4,050 + 2,250

  15,000

  =

  6,300

  15,000

  = 0.4200 = 42 cM

Check Answer

 

Gene Linkage and Map Distance Calculation

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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 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
1	+ +	+ +	e n	e n	5,118
2	+ +	+ n	e +	e n	2,986
3	+ n	+ n	e +	e +	96
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
• B. distance =

  ½(96) + 3×(96)

  8,200

  =

  48 + 288

  8,200

  =

  336

  8,200

  = 0.0410 = 4.10 cM
• C. distance =

  ½(96) + 3×(0)

  8,200

  =

  48 + 0

  8,200

  =

  48

  8,200

  = 0.0059 = 0.59 cM
• D. distance =

  ½(0) + 3×(96)

  8,200

  =

  0 + 288

  8,200

  =

  288

  8,200

  = 0.0351 = 3.51 cM
• E. distance =

  ½(2,986) + 3×(96)

  8,200

  =

  1,493 + 288

  8,200

  =

  1,781

  8,200

  = 0.2172 = 21.72 cM

Check Answer

 

Linkage Between Genes

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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 R is connected with the 'rusty' phenotype. A budding yeast that is homozygous recessive for Gene R colonies develop a reddish-brown pigmentation, reminiscent of rusted metal.

Set #	Tetrad Genotypes				Progeny Count
1	+ +	+ +	h r	h r	113
2	+ +	+ r	h +	h r	3,278
3	+ r	+ r	h +	h +	5,209
TOTAL =					8,600

The resulting phenotypes are summarized in the table above.

Step-by-Step Instructions

• Step 1: Find the row with the Parental Type for both genes.
• Step 2: Assign PD, NPD, TT for the other rows
• Step 3: Determine if the two genes are linked.
• PD >> NPD → linked; PD ≈ NPD → unlinked

Unordered Tetrad Two Gene Determine Linkage

The yeast Saccharomyces cerevisiae has unordered tetrads. A cross is made to study the linkage relationships among two genes.
Using the table above, determine the linkage between the two genes.

• A. The two genes are BOTH linked AND unlinked.
• B. The two genes are LINKED and on the SAME chromosome.
• C. The two genes are LINKED but are likely on DIFFERENT chromosomes.
• D. The two genes are UNLINKED but may be on either the SAME or DIFFERENT chromosomes.
• E. The two genes are NEITHER linked NOR unlinked.

Check Answer

 

Gene Order and Distance Calculation

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Three-Point Test Cross Problem

A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a three-point test cross examines three (3) genes at the same time to learn about their assortment in gamete formation.
A standard three-point test cross involves crossing a heterozygous organism for all three genes with an organism that is homozygous recessive for all three genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for three genes was conducted to understand their genetic interactions.

Characteristics of Recessive Phenotypes

• Gene H is related to the 'horsey' phenotype. A fruit fly that is homozygous recessive for Gene H is quite big and strong-looking, much larger than your typical fruit fly.
• Gene N is affiliated with the 'nerdy' phenotype. A fruit fly that is homozygous recessive for Gene N has large, prominent eyes that stand out, much like thick-rimmed glasses.
• Gene R is correlated with the 'rusty' phenotype. A fruit fly that is homozygous recessive for Gene R has a reddish-brown color, much like rusted iron metal.

Phenotype	Genotypes			Progeny Count
horsey, nerdy, rusty	h	n	r	376
horsey, nerdy	h	n	+	842
horsey, rusty	h	+	r	17
horsey	h	+	+	149
nerdy, rusty	+	n	r	145
nerdy	+	n	+	25
rusty	+	+	r	824
wildtype	+	+	+	422
TOTAL =				2,800

Question

Using the table above, determine the order of the genes and the distances between them. Once calculated, fill in the following four blanks:

• The distance between genes H and N is cM (HN)
• The distance between genes H and R is cM (HR)
• The distance between genes N and R is cM (NR)
• From this the correct order of the genes is (gene order).

Hints

• Important Tip 1: Your calculated distances between each pair of genes should be a whole number. Finding a decimal in your answer, such as 5.5, indicates a mistake was made. Please provide your answer as a complete number without fractions or decimals.
• Important Tip 2: Your answer should be written as a numerical value only, with no spaces, commas, or units such as "cM" or "map units". For example, if the distance is fifty one centimorgans, simply write "51".
• Important Tip 3: Your gene order answer should be written as three letters only, with no spaces, commas, hyphens, or other characters allowed. For example, if the gene order is B - A - C, simply write "bac" or "cab".

Check Answer

 

Three-Point Test Cross Interference

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Three-Point Test Cross Problem

A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a three-point test cross examines three (3) genes at the same time to learn about their assortment in gamete formation.
A standard three-point test cross involves crossing a heterozygous organism for all three genes with an organism that is homozygous recessive for all three genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for three genes was conducted to understand their genetic interactions.

Characteristics of Recessive Phenotypes

• Gene H is analogous to the 'horsey' phenotype. A fruit fly that is homozygous recessive for Gene H is quite big and strong-looking, much larger than your typical fruit fly.
• Gene K is correlated with the 'kidney' phenotype. A fruit fly that is homozygous recessive for Gene K has a body shape that is curved, similar to a kidney bean.
• Gene T is related to the 'tipsy' phenotype. A fruit fly that is homozygous recessive for Gene T moves in an erratic path, suggesting a lack of coordination, as if intoxicated.

Phenotype	Genotypes			Progeny Count
horsey, kidney, tipsy	h	k	t	1,109
horsey, kidney	h	k	+	167
horsey, tipsy	h	+	t	422
horsey	h	+	+	9
kidney, tipsy	+	k	t	8
kidney	+	k	+	411
tipsy	+	+	t	190
wildtype	+	+	+	1,084
TOTAL =				3,400

The resulting phenotypes are summarized in the table above.

• The distance between genes K and H is 25 cM
• The distance between genes K and T is 35 cM
• The distance between genes H and T is 11 cM
• The correct gene order determined from these distances is KHT

Step-by-Step Instructions for Calculating Interference

• Step 1: Count the observed number of double crossovers from the data table.
• Step 2: Calculate the probability of independent crossovers between distant genes.
• Multiply the two individual crossover probabilities (based on their distance) for both adjacent gene pairs.
• Step 3: Determine the expected number of double crossovers.
• Multiply the combined probability (from Step 2) by the total progeny count.
• Step 4: Calculate the Coefficient of Coincidence (CoC).
• Divide the observed number of double crossovers (from Step 1) by the expected number (from Step 3).
• Step 5: Calculate Interference.
• Interference is given by the formula: Interference = 1 - CoC.

In genetic studies, interference refers to the phenomenon where the occurrence of a crossover in one region of a chromosome reduces the likelihood of another crossover occurring nearby, thereby affecting the expected genetic ratios.

Question

Based on the traits expressed in the offspring, select the correct fraction that represents the interference level between genes K and T.

• A. no interference, 0%
• B. ¹/₁₁ = 0.0909 = 9%
• C. ²/₁₁ = 0.1818 = 18%
• D. ³/₁₁ = 0.2727 = 27%
• E. ⁵/₁₁ = 0.4545 = 45%
• F. ¹/₂ = 0.5000 = 50%
• G. ⁷/₁₁ = 0.6364 = 64%
• H. ⁹/₁₁ = 0.8182 = 82%
• I. complete interference, 100%

Check Answer

 

Genetic Distance Calculation

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Three-Point Test Cross Problem

A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a three-point test cross examines three (3) genes at the same time to learn about their assortment in gamete formation.
A standard three-point test cross involves crossing a heterozygous organism for all three genes with an organism that is homozygous recessive for all three genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for three genes was conducted to understand their genetic interactions.

Characteristics of Recessive Phenotypes

• Gene K is affiliated with the 'kidney' phenotype. A fruit fly that is homozygous recessive for Gene K has a body shape that is curved, similar to a kidney bean.
• Gene N is linked with the 'nerdy' phenotype. A fruit fly that is homozygous recessive for Gene N has large, prominent eyes that stand out, much like thick-rimmed glasses.
• Gene P is connected with the 'prickly' phenotype. A fruit fly that is homozygous recessive for Gene P is covered with sharp bristles, giving it a spiky texture.

Phenotype	Genotypes			Progeny Count
kidney, nerdy, prickly	k	n	p	81,728
kidney, nerdy	k	n	+	29,972
kidney, prickly	k	+	p	17,503
kidney	k	+	+	5,610
nerdy, prickly	+	n	p	5,494
nerdy	+	n	+	17,698
prickly	+	+	p	29,556
wildtype	+	+	+	82,439
TOTAL =				270,000

The resulting phenotypes are summarized in the table above.
The resulting phenotypes are summarized in the table above.

Question

With the progeny data from the table, and using only the genotypes that result from crossover events between the two genes K and N during meiosis.
calculate the genetic distance between the two genes K and N, expressing your answer in centimorgans (cM)

• A. ^{(5,610 + 17,503 + 29,972)}/_270,000 = ^53,085/_270,000 = 0.1966 = 19.66 cM
• B. ^5,610/_270,000 = ^5,610/_270,000 = 0.0208 = 2.08 cM
• C. ^81,728/_270,000 = ^81,728/_270,000 = 0.3027 = 30.27 cM
• D. ^{(81,728 + 82,439)}/_270,000 = ^164,167/_270,000 = 0.6080 = 60.80 cM
• E. ^{(5,494 + 5,610 + 17,503 + 17,698)}/_270,000 = ^46,305/_270,000 = 0.1715 = 17.15 cM
• F. ^{(5,494 + 5,610 + 17,698 + 29,556)}/_270,000 = ^58,358/_270,000 = 0.2161 = 21.61 cM

Check Answer

 

Double Crossover Genotypes

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Three-Point Test Cross Problem

A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a three-point test cross examines three (3) genes at the same time to learn about their assortment in gamete formation.
A standard three-point test cross involves crossing a heterozygous organism for all three genes with an organism that is homozygous recessive for all three genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for three genes was conducted to understand their genetic interactions.

Characteristics of Recessive Phenotypes

• Gene B is affiliated with the 'bumpy' phenotype. A fruit fly that is homozygous recessive for Gene B has a skin texture that is not smooth, but rough with small bumps all over.
• Gene E is correlated with the 'eery' phenotype. A fruit fly that is homozygous recessive for Gene E appears to have something off, crooked limbs and other twisted appendages.
• Gene T is analogous to the 'tipsy' phenotype. A fruit fly that is homozygous recessive for Gene T moves in an erratic path, suggesting a lack of coordination, as if intoxicated.

Phenotype	Genotypes			Progeny Count
bumpy, eery, tipsy	b	e	t	34
bumpy, eery	b	e	+	1,448
bumpy, tipsy	b	+	t	198
bumpy	b	+	+	507
eery, tipsy	+	e	t	527
eery	+	e	+	220
tipsy	+	+	t	1,434
wildtype	+	+	+	32
TOTAL =				4,400

The resulting phenotypes are summarized in the table above.
The resulting phenotypes are summarized in the table above.

Question

Based on the traits expressed in the offspring, identify the double crossover genotype combinations. These allele combinations are a result of two genetic crossover events.
More than one genotype will be correct. Select all that apply.

• A. bet
• B. be+
• C. b+t
• D. b++
• E. +et
• F. +e+
• G. ++t
• H. +++

Check Answer Clear Selection

 

Recombinant Genotypes for Genes H and F

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Three-Point Test Cross Problem

A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a three-point test cross examines three (3) genes at the same time to learn about their assortment in gamete formation.
A standard three-point test cross involves crossing a heterozygous organism for all three genes with an organism that is homozygous recessive for all three genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for three genes was conducted to understand their genetic interactions.

Characteristics of Recessive Phenotypes

• Gene D is correlated with the 'dewy' phenotype. A fruit fly that is homozygous recessive for Gene D appears moist, with its body covered in tiny droplets of water.
• Gene E is connected with the 'eery' phenotype. A fruit fly that is homozygous recessive for Gene E appears to have something off, crooked limbs and other twisted appendages.
• Gene F is associated with the 'fuzzy' phenotype. A fruit fly that is homozygous recessive for Gene F is covered in a dense layer of hairs, giving it a soft appearance.

Phenotype	Genotypes			Progeny Count
dewy, eery, fuzzy	d	e	f	169
dewy, eery	d	e	+	1,677
dewy, fuzzy	d	+	f	435
dewy	d	+	+	29
eery, fuzzy	+	e	f	40
eery	+	e	+	416
fuzzy	+	+	f	1,658
wildtype	+	+	+	176
TOTAL =				4,600

The resulting phenotypes are summarized in the table above.
The resulting phenotypes are summarized in the table above.

Question

Based on the traits expressed in the offspring, identify the all recombinant genotypes for genes E and F. These genotypes result from crossover events that occur between the two genes E and F during meiosis.
More than one genotype will be correct. Select all that apply.

• A. def
• B. de+
• C. d+f
• D. d++
• E. +ef
• F. +e+
• G. ++f
• H. +++

Check Answer Clear Selection

 

Parental Genotype Combinations in a Three-Point Test Cross

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Three-Point Test Cross Problem

A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a three-point test cross examines three (3) genes at the same time to learn about their assortment in gamete formation.
A standard three-point test cross involves crossing a heterozygous organism for all three genes with an organism that is homozygous recessive for all three genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for three genes was conducted to understand their genetic interactions.

Characteristics of Recessive Phenotypes

• Gene D is associated with the 'dewy' phenotype. A fruit fly that is homozygous recessive for Gene D appears moist, with its body covered in tiny droplets of water.
• Gene F is affiliated with the 'fuzzy' phenotype. A fruit fly that is homozygous recessive for Gene F is covered in a dense layer of hairs, giving it a soft appearance.
• Gene W is affiliated with the 'waxy' phenotype. A fruit fly that is homozygous recessive for Gene W has a thick protective layer that is water resistant and opague.

Phenotype	Genotypes			Progeny Count
dewy, fuzzy, waxy	d	f	w	661
dewy, fuzzy	d	f	+	71
dewy, waxy	d	+	w	218
dewy	d	+	+	1,552
fuzzy, waxy	+	f	w	1,559
fuzzy	+	f	+	241
waxy	+	+	w	82
wildtype	+	+	+	716
TOTAL =				5,100

The resulting phenotypes are summarized in the table above.
The resulting phenotypes are summarized in the table above.

Question

Based on the traits expressed in the offspring, identify the parental genotype combinations. These are the allele combinations that the parent fruit flies originally carried.
More than one genotype will be correct. Select all that apply.

• A. dfw
• B. df+
• C. d+w
• D. d++
• E. +fw
• F. +f+
• G. ++w
• H. +++

Check Answer Clear Selection

 

Two-Point Test Cross Configuration

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Two-Point Test Cross Problem

A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.

Characteristics of Recessive Phenotypes

• Gene C is affiliated with the 'chummy' phenotype. A fruit fly that is homozygous recessive for Gene C shows behavior where it always maintains a close distance to other flies.
• Gene H is affiliated with the 'horsey' phenotype. A fruit fly that is homozygous recessive for Gene H is quite big and strong-looking, much larger than your typical fruit fly.

Phenotype	Genotypes		Progeny Count
chummy, horsey	c	h	1,192
chummy	c	+	1,894
horsey	+	h	1,950
wildtype	+	+	1,164
TOTAL =			6,200

The phenotype counts resulting from the cross are summarized in the table above.

Question

Using the data presented in the table to determine the configuration of the alleles on the parental chromosomes. Determine whether the alleles for the two genes are in a cis (on the same chromosome) or trans (on different chromosomes) configuration.

• A. cis
• B. trans
• C. both
• D. neither
• E. cannot be determined

Check Answer

 

Two-Point Test Cross Configuration

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Two-Point Test Cross Problem

A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.

Characteristics of Recessive Phenotypes

• Gene E is correlated with the 'eery' phenotype. A fruit fly that is homozygous recessive for Gene E appears to have something off, crooked limbs and other twisted appendages.
• Gene J is linked with the 'jerky' phenotype. A fruit fly that is homozygous recessive for Gene J moves in rapid and sudden movements, displaying an unpredictable flight pattern.

Phenotype	Genotypes		Progeny Count
eery, jerky	e	j	467
eery	e	+	39
jerky	+	j	41
wildtype	+	+	453
TOTAL =			1,000

The resulting phenotypes are summarized in the table above.
The phenotype counts resulting from the cross are summarized in the table above.

Question

Using the data presented in the table to determine the configuration of the alleles on the parental chromosomes. Determine whether the alleles for the two genes are in a cis (on the same chromosome) or trans (on different chromosomes) configuration.

• A. cis
• B. trans
• C. both
• D. neither
• E. cannot be determined

Check Answer

 

Genetic Distance Calculation

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Two-Point Test Cross Problem

A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.

Characteristics of Recessive Phenotypes

• Gene T is correlated with the 'tipsy' phenotype. A fruit fly that is homozygous recessive for Gene T moves in an erratic path, suggesting a lack of coordination, as if intoxicated.
• Gene Y is related to the 'yucky' phenotype. A fruit fly that is homozygous recessive for Gene Y gives off an unpleasant odor and has a generally unappealing look.

Phenotype	Genotypes		Progeny Count
tipsy, yucky	t	y	2,370
tipsy	t	+	7,678
yucky	+	y	7,496
wildtype	+	+	2,456
TOTAL =			20,000

The resulting phenotypes are summarized in the table above.
The resulting phenotypes are summarized in the table above.

Question

With the progeny data from the table, calculate the genetic distance between the two genes, expressing your answer in centimorgans (cM)

• A. ^{(2,370 + 2,456)}/_20,000 = ^4,826/_20,000 = 0.2413 = 24.13 cM
• B. ^7,496/_20,000 = ^7,496/_20,000 = 0.3748 = 37.48 cM
• C. ^2,456/_20,000 = ^2,456/_20,000 = 0.1228 = 12.28 cM
• D. ^{(7,496 + 7,678)}/_20,000 = ^15,174/_20,000 = 0.7587 = 75.87 cM
• E. ^7,678/_20,000 = ^7,678/_20,000 = 0.3839 = 38.39 cM
• F. ^2,370/_20,000 = ^2,370/_20,000 = 0.1185 = 11.85 cM

Check Answer

 

Two-Point Test Cross Problem

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Two-Point Test Cross Problem

A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.

Characteristics of Recessive Phenotypes

• Gene M is related to the 'mushy' phenotype. A fruit fly that is homozygous recessive for Gene M feels soft to the touch and unusually squishy, unlike the usual firmness.
• Gene X is related to the 'xanthic' phenotype. A fruit fly that is homozygous recessive for Gene X has a fluorescent bright yellow coloring.

Phenotype	Genotypes		Progeny Count
mushy, xanthic	m	x	177
mushy	m	+	2,679
xanthic	+	x	2,773
wildtype	+	+	171
TOTAL =			5,800

The resulting phenotypes are summarized in the table above.
The resulting phenotypes are summarized in the table above.

Question

Review the phenotype counts shown in the table. Based on the traits expressed in the offspring, identify the possible parental genotype combinations. These are the allele combinations that the parent fruit flies originally carried. More than one combination will be correct. Select all that apply.

• A. mx
• B. m+
• C. +x
• D. ++

Check Answer Clear Selection

 

Two-Point Test Cross

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Two-Point Test Cross Problem

A test cross is a way to explore the relationship between genes and their respective alleles. It is a useful tool for genetic mapping and deciphering the inheritance of traits. Specifically, a two-point test cross examines two (2) genes at the same time to learn about their assortment in gamete formation.
A standard two-point test cross involves crossing a heterozygous organism for both genes with an organism that is homozygous recessive for both genes
For this problem, a test cross using a fruit fly (Drosophila melanogaster) heterozygous for two genes was conducted to understand their genetic interactions.

Characteristics of Recessive Phenotypes

• Gene A is connected with the 'artsy' phenotype. A fruit fly that is homozygous recessive for Gene A has wings that are colorful and distinctive patterns.
• Gene J is linked with the 'jerky' phenotype. A fruit fly that is homozygous recessive for Gene J moves in rapid and sudden movements, displaying an unpredictable flight pattern.

Phenotype	Genotypes		Progeny Count
artsy, jerky	a	j	1,128
artsy	a	+	1,615
jerky	+	j	1,745
wildtype	+	+	1,112
TOTAL =			5,600

The resulting phenotypes are summarized in the table above.
The resulting phenotypes are summarized in the table above.

Question

Review the phenotype counts shown in the table. Based on the traits expressed in the offspring, identify the possible recombinant genotype combinations. These allele combinations have occurred due to genetic crossover. More than one combination will be correct. Select all that apply.

• A. aj
• B. a+
• C. +j
• D. ++

Check Answer Clear Selection