1. The cross AB/ab x ab/ab is performed. The progeny that result ar...

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1. The cross AB/ab x ab/ab is performed. The progeny that result are:
852 AB/ab
825 ab/ab
146 Ab/ab
177 aB/ab
How far apart are genes A and B?
2. In Drosophila, the genes dumpy (d, small wings) and apterous (ap, wingless) are 40 cM apart. If a d(+) ap(+)/d(+) ap(+) female is mated to a d ap/d ap male,
a. What type of gametes are produced by the F1 males? In what proportion?
b. What type of gametes are produced by the F1 females? In what proportion?
3. You perform a three-point testcross in Drosophila, crossing triply heterozygous females to males that are homozygous for three linked recessive traits: glassy eye, coal-colored bodies and striped thoraxes.
a. List the 8 possible phenotypic classes you expect to observe in the offspring from this cross.
b. You only observe 7 of the 8 possible phenotypes. Assuming that all 8 of the possible phenotypes are equally viable, propose a logical explanation for why you fail to observe offspring from one of the phenotypic classes (think about which class is mostly likely missing, and why it might be missing).

4. Singed bristles (sn), crossveinless wings (cv) and vermilion eyes (v) are three recessive X-linked mutations in Drosophila. When a female heterozygous for each of the three genes was testcrossed to a sn cv v male, the following progeny were obtained:

3       singed, crossveinless, vermillion
392    crossveinless, vermilion
34      vermilion
61      crossveinless
32      singed, crossveinless
65      singed, vermilion
410    singed
3       wild type

a. What are the non-recombinant classes of offspring?
b. Which classes represent single crossover events?
c. Which classes represent double crossover events?
d. What was the genotype of the heterozygous female? Be sure to indicate which alleles are on each of the two homologous chromosomes.
e. What is the order of these genes on the X chromosome?
f. What is the distance between each pair of genes?
5. In Drosophila, the recessive mutations curved (c), plexus (px) and cinnabar (cn) generate curved wings, extra wing veins and bright red eyes, respectively. The three genes are linked, and are located on the second chromosome. When a triple heterozygous female was crossed to a homozygous mutant male, the following offspring were observed:

329    plexus
296    curved, cinnabar
110    plexus, cinnabar
95      curved
76      wild type
69      curved, plexus, cinnabar
15      curved, plexus
10      cinnabar

a. What are the genotypes of the parental generation (the parents of the F1 female)? You can assume that the parents have homozygous genotypes.
b. What is the order of the genes on chromosome 2?
c. What is the distance between the genes?
d. Is there interference? Explain your answer.
e. Calculate values for the coefficient of coincidence (C) and interference (I).
6. In rabbits, the dominant C allele is required for colored fur, while rabbits that are homozygous for the genotype cc are albino. A second gene determines whether the fur is black (B, dominant) or brown (b, recessive). A brown rabbit with the genotype bb CC was crossed to an albino rabbit with the genotype BB cc to generate F1 heterozygotes. The F1 rabbits were crossed to an albino rabbit with the genotype bb cc, and 200 offspring were generated (they breed like rabbits, after all!).
a. If these genes are unlinked, what type of phenotypes will be present among the 200 offspring, and in what proportions? How many rabbits with each phenotype will you expect to observe?
b. Among the 200 offspring, you observe 34 black rabbits, 66 brown rabbits and 100 albino rabbits. Do these numbers suggest that the B and C genes are linked, or unlinked? Use chisquare analysis to support your answer.

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852 AB/ab
825 ab/ab
146 Ab/ab
177 aB/ab

In the above data, the parental types are 852 AB/ab, and 825 ab/ab. And the cross-over types are 146 Ab/ab and 177 aB/ab. The total number of progenies is 852+825+146+177 = 1970. The recombinants are 146+177 = 323. Hence, the % cross-over types are16.39%.
Thus the genes A and B are 16.39 centiMorgan apart....

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