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snow_tiger [21]
3 years ago
6

How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE?

Biology
1 answer:
dexar [7]3 years ago
7 0
The answer is B) 8.

There are 5 genes. There are two possibilities for each of the heterozygous genes and only one possibility for each of homozygous genes. So, the possibilities for each gene are:
Aa - 2: A or a;
Bb - 2: B or b;
CC - 1: D;
Dd - 2: D or d;
EE - 1: E.
Genotype AaBbCCDdEE will have
                2 x 2 x 1 x 2 x 1 gametes

Thus, there will be 8 unique gametes produced by an <span>individual with the genotype AaBbCCDdEE</span>
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Answer:

The question lacks options, the options are:

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A.100% red/white-feather chickens.

B.50% white-feather chickens and 50% red/white-feather chickens.

C.33% white-feather chickens, 33% red-feather chickens, and 33% red/white-feather chickens.

D.25% white-feather chickens, 25% red-feather chickens, and 50% red/white-feather chickens

The Correct answer is D

Explanation:

According to the question, the hybrid chicken has a mixture of red and white feather phenotypes expressed at the same time. This tells us that the red and white alleles are CODOMINANT in the chicken i.e. none is dominant or recessive over the other.

Let's assume that (R) allele is the red feather colour while (r) allele is for white feather colour. Therefore, the red/white feathered chicken will have a Rr genotype. In a cross between two red/white feathered chicken (Rr), the following offsprings will be produced RR, Rr, Rr and rr

RR- Homozygous red feather d chicken (1/4 × 100=25%)

Rr- Hybrid Red/White feathered chicken (2/4 × 100=50%)

rr- Homozygous white feathered chicken (1/4 × 100=25%)

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