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Doss [256]
3 years ago
13

In the same mouse species, a third unlinked gene (gene C/c) also has an epistatic effect on fur color. The presence of the domin

ant allele C (for color), allows the A/a and B/b genes to be expressed normally. The presence of two recessive alleles (cc), on the other hand, prevents any pigment from being formed, resulting in an albino (white) mouse. An albino mouse with a solid white coloration and the cc (two lowercase letters) genotype. Match the phenotypes on the labels at left to the genotypes listed below. Labels can be used once, more than once, or not at all.
Biology
1 answer:
Cerrena [4.2K]3 years ago
8 0

Answer:

The phenotypes will be -

AaBbcc - Albino,

AaBBCC - Agouti Black

Aabbcc - Albino

AAbbCc - Agouti Brown

aaBbCc - Solid color, Black

AABBcc - Albino

Explanation:

The C/c gene is epistatic to both the A/a and B/b genes, any offspring with the cc allele combination will be albino. Otherwise, the A/a and B/b genes are expressed normally.

               

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Alleles A and a reside at a locus on the same chromosome as a locus with alleles B and b. Aa Bb is crossed with aa bb and the fo
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Answer:

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Explanation:

<em>The conclusion that can be drawn about the arrangement of the genes on the chromosome in the Aa Bb parent is that the two genes are linked together. That is, they do not assort independently.</em>

When two genes are linked together, the number of recombinant offspring would be more than the number of offspring with parental traits. The frequency of recombination would also not exceed 50%.

<em>Recombination frequency = number of recombinant offspring/total number of offspring x 100%</em>

Number of recombinant offspring = 45 + 45 = 90

Total number of offspring = 45 + 45 + 5 + 5 = 100

Recombination frequency = 90/100 x 100% = 10%

Hence, the two genes Aa and Bb are said to be linked together.

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