1. In the doormouse, an autosomal allele causing banded fur ( b ) is recessive t
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Question
1. In the doormouse, an autosomal allele causing banded fur (b) is recessive to the allele for unbanded fur (B). Alleles of a gene at a different locus determine the background color of the fur; yellow (c) is recessive to brown (C). A banded, yellow mouse is crossed with a homozygous brown, unbanded mouse. The F1 are crossed with a banded, yellow mouse (a testcross).
a. What will be the results of the tescross (offspring phenotype and ratios) if the loci assort independently? 0.5 pts
b. What will be the results of the testcross if the loci are linked with no crossing over? Give offspring classes and percent expected of each type. 0.5 pts
c. What will be the results of the testcross if the loci are linked and 30 map units apart, allowing for crossing over? Give offspring phenotypic classes and percent expected of each type.1 pt
2. In these mice, the dominant allele, G of the X-linked gene Greasy produces shiny fur, while the recessive wild type allele G+ determines normal fur. The dominant allele B of the X-linked Broadhead gene causes skeletal abnormalities including broad heads and snouts, while the recessive wild type B+ allele yields normal skeletons. Female mice heterozygous for the two alleles of both genes were mated with wild type males. Among 500 male progeny of this cross, 240 had shiny fur, 245 had skeletal abnormalities, 7 had shiny fur and skeletal abnormalities, and 8 was wild type.
a. Diagram the cross described and calculate the distance between the two genes. 2 pts
b. What would have been the results if you had counted 100 female progeny of the cross? 2 pts
Explanation / Answer
1. Parents for the cross will be bbcc and BBCC
The gametes from each parent will be bc and BC.
When the parents are crossed the F1 will have genotype BbCc. This F1 is now testcrossed with bbcc.
The punnett square for the cross will be:
From this punnett square we can answer the questions given
a. The results of the tescross if the loci assort independently will be:
Unbanded, brown - 1, Unbanded, yellow- 1, banded, brown- 1, banded, yellow- 1
Thus, the phenotypic ratio will be 1:1:1:1.
b. If loci are linked with no crossing over then the classes and percent expected of each type will be:
banded, yellow - 1/2 or 50%
unbanded, brown - 1/2 or 50%
c. If loci are linked and 30 map units apart then percent expected of each type will be:
30 mu
=30% /2 = 15%
Banded, brown 15%
Unbanded, yellow 15%
2. The diagram of cross is as follows:
G B+/G+ B X G+ B+/Y
The distance between these two genes will be
=(7+8)/500=0.03
= 3 map units
b. If we counted 100 female progeny of the cross then the result will be same as 100 progeny brothers would be counted.
BC Bc bC bc bc BbCc Bbcc bbCc bbccRelated Questions
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