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Chapter 5: Problem 27

A homozygous \(A A B B\) mouse was crossed to a homozygous recessive aabb mouse. The \(\mathrm{F}_{1}\) were backcrossed to a homozygous recessive aabb mouse, and the offspring were analyzed for \(A\) and \(B\). The genetic distance between \(A\) and \(B\) was found to be \(10 \mathrm{cM}(\mathrm{mu})\). If 200 mice were analyzed, deduce the phenotypes observed in the offspring.

Short Answer

Expert verified
Answer: The phenotypes observed in the offspring will include both dominant traits (A-B + a-b) in approximately 180 mice and recombinant traits (Ab + aB) in approximately 20 mice.

Step by step solution

01

Determine the genotype of F1 offspring

In the first cross (AA BB) x (aabb), the offspring will inherit one A allele and one a allele, and one B allele and one b allele. Therefore, the genotype of F1 offspring will be AB Ab aB ab.
02

Calculate recombination frequency

Given that the genetic distance between A and B is 10 cM, this indicates that there is a 10% recombination frequency between these genes. This means that in 10% of the offspring, the alleles A and B will be separated during gamete formation due to crossing over.
03

Identify possible genotypes and calculate probabilities from recombination frequency

The possible genotypes of the offspring are: 1. A-B gametes: 45% (non-recombination) of 50% F1 genotype 2. Ab or aB gametes: 5% recombination frequency × 50% F1 genotype 3. a-b gametes: 45% (non-recombination) of 50% F1 genotype
04

Calculate expected genotype distribution in the offspring from the backcross

In the backcross with a homozygous recessive aabb mouse, the genotype probabilities are: 1. A-B + a-b: (45% × 50%) + (45% × 50%) = 90% 2. Ab + aB: (5% × 50%) + (5% × 50%) = 10%
05

Determine the phenotypes observed in the offspring

From expected genotype distribution, we can estimate the number of offspring with each genotype, out of the 200 mice analyzed: 1. A-B + a-b offspring: 90% × 200 = 180 2. Ab + aB offspring: 10% × 200 = 20 As a result, the phenotypes observed in the offspring will include both dominant traits (A-B + a-b) and recombinant traits (Ab + aB).

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Most popular questions from this chapter

In a certain plant, fruit is either red or yellow, and fruit shape is either oval or long. Red and oval are the dominant traits. Two plants, both heterozygous for these traits, were testcrossed, with the following results. Determine the location of the genes relative to one another and the genotypes of the two parental plants.

The genes dumpy\((d p),\) clot \((c l),\) and apterous \((a p)\) are linked on chromosome II of Drosophila. In a series of two-point mapping crosses, the following genetic distances were determined. What is the sequence of the three genes?

What possible conclusions can be drawn from the observations that in male Drosophila, no crossing over occurs, and that during meiosis, synaptonemal complexes are not seen in males but are observed in females where crossing over occurs?

In Drosophila, a female fly is heterozygous for three mutations, Bar eyes \((B),\) miniature wings \((m),\) and ebony body \((e)\) Note that Bar is a dominant mutation. The fly is crossed to a male with normal eyes, miniature wings, and ebony body. The results of the cross are as follows. Interpret the results of this cross. If you conclude that linkage is involved between any of the genes, determine the map distance(s) between them.

Why does more crossing over occur between two distantly linked genes than between two genes that are very close together on the same chromosome?
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