Skip to main content
Pearson+ LogoPearson+ Logo
Ch. 8 - Chromosome Mutations: Variation in Number and Arrangement
Klug - Concepts of Genetics  12th Edition
Klug12th EditionConcepts of Genetics ISBN: 9780135564776Not the one you use?Change textbook
Chapter 8, Problem 23

In a cross in Drosophila, a female heterozygous for the autosomally linked genes a, b, c, d, and e (abcde/ +++++) was testcrossed with a male homozygous for all recessive alleles. Even though the distance between each of the loci was at least 3 map units, only four phenotypes were recovered, yielding the following data:
Table showing four Drosophila phenotypes with corresponding fly counts totaling 1000: 440, 460, 48, and 52 flies.
Why are many expected crossover phenotypes missing? Can any of these loci be mapped from the data given here? If so, determine map distances.

Verified step by step guidance
1
Step 1: Identify the parental and recombinant phenotypes from the data. The two most frequent phenotypes (+++++ with 440 flies and abcde with 460 flies) represent the parental types, while the less frequent phenotypes (++ ++e with 48 flies and abcd+ with 52 flies) represent single crossover recombinants. The absence of other recombinant phenotypes suggests double or multiple crossovers are not observed.
Step 2: Understand why many expected crossover phenotypes are missing. Since the genes are autosomally linked and the distance between each locus is at least 3 map units, double or multiple crossovers are expected but not observed here. This is likely due to crossover interference, where one crossover event reduces the probability of another nearby crossover, leading to fewer double crossover phenotypes than expected.
Step 3: Determine which loci can be mapped from the data. Because only single crossover classes involving the last gene 'e' are observed, it suggests that only the distance between gene 'd' and gene 'e' can be reliably estimated from this data. Other distances cannot be mapped accurately without additional recombinant classes.
Step 4: Calculate the recombination frequency between loci 'd' and 'e'. Use the formula for recombination frequency (RF): \[\text{RF} = \frac{\text{Number of recombinant offspring}}{\text{Total number of offspring}} \times 100\] Here, the recombinant offspring are those with phenotypes ++++e and abcd+, totaling 48 + 52 flies.
Step 5: Convert the recombination frequency to map units (centiMorgans, cM). Since 1% recombination frequency corresponds to 1 map unit, the distance between loci 'd' and 'e' is numerically equal to the recombination frequency calculated in Step 4.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
4m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Genetic Linkage and Recombination

Genetic linkage occurs when genes are located close together on the same chromosome and tend to be inherited together. Recombination during meiosis can separate linked genes, producing new allele combinations. The frequency of recombination between two loci is proportional to their physical distance, measured in map units or centimorgans.
Recommended video:
Guided course
07:33
Chi Square and Linkage

Testcross and Phenotypic Ratios

A testcross involves crossing an individual heterozygous for certain genes with a homozygous recessive individual to reveal the genotype of the heterozygote. The phenotypic ratios of offspring reflect the parental and recombinant types, allowing inference of gene order and recombination frequencies.
Recommended video:
Guided course
10:48
Mutations and Phenotypes

Interference and Double Crossovers

Interference is the phenomenon where one crossover event reduces the likelihood of another nearby crossover, leading to fewer double crossover phenotypes than expected. This can cause missing or underrepresented recombinant classes in genetic data, complicating mapping of loci.
Recommended video:
Guided course
06:59
RNA Interference
Related Practice
Textbook Question

A couple planning their family are aware that through the past three generations on the husband's side a substantial number of stillbirths have occurred and several malformed babies were born who died early in childhood. The wife has studied genetics and urges her husband to visit a genetic counseling clinic, where a complete karyotype-banding analysis is performed. Although the tests show that he has a normal complement of 46 chromosomes, banding analysis reveals that one member of the chromosome 1 pair (in group A) contains an inversion covering 70 percent of its length. The homolog of chromosome 1 and all other chromosomes show the normal banding sequence.

How would you explain the high incidence of past stillbirths?

653
views
Textbook Question

A couple planning their family are aware that through the past three generations on the husband's side a substantial number of stillbirths have occurred and several malformed babies were born who died early in childhood. The wife has studied genetics and urges her husband to visit a genetic counseling clinic, where a complete karyotype-banding analysis is performed. Although the tests show that he has a normal complement of 46 chromosomes, banding analysis reveals that one member of the chromosome 1 pair (in group A) contains an inversion covering 70 percent of its length. The homolog of chromosome 1 and all other chromosomes show the normal banding sequence.

What can you predict about the probability of abnormality/normality of their future children?

430
views
Textbook Question

A couple planning their family are aware that through the past three generations on the husband's side a substantial number of stillbirths have occurred and several malformed babies were born who died early in childhood. The wife has studied genetics and urges her husband to visit a genetic counseling clinic, where a complete karyotype-banding analysis is performed. Although the tests show that he has a normal complement of 46 chromosomes, banding analysis reveals that one member of the chromosome 1 pair (in group A) contains an inversion covering 70 percent of its length. The homolog of chromosome 1 and all other chromosomes show the normal banding sequence.

Would you advise the woman that she will have to bring each pregnancy to term to determine whether the fetus is normal? If not, what else can you suggest?

449
views
Textbook Question

A woman who sought genetic counseling is found to be heterozygous for a chromosomal rearrangement between the second and third chromosomes. Her chromosomes, compared to those in a normal karyotype, are diagrammed to the right.

What kind of chromosomal aberration is shown?

632
views
Textbook Question

A woman who sought genetic counseling is found to be heterozygous for a chromosomal rearrangement between the second and third chromosomes. Her chromosomes, compared to those in a normal karyotype, are diagrammed to the right.

Using a drawing, demonstrate how these chromosomes would pair during meiosis. Be sure to label the different segments of the chromosomes.

478
views
Textbook Question

A woman who sought genetic counseling is found to be heterozygous for a chromosomal rearrangement between the second and third chromosomes. Her chromosomes, compared to those in a normal karyotype, are diagrammed to the right.

This woman is phenotypically normal. Does this surprise you? Why or why not? Under what circumstances might you expect a phenotypic effect of such a rearrangement?

627
views