The ear tuft allele in chicken is autosomal and produces feathered skin projections near the ear on each side of the head. This gene is dominant embryos do not hatch from the egg. Assume that in a population of 6000 chickens, 2000 have no tufts, and 4000 have ear tufts. what are the frequencies of the normal vs. ear tuft alleles in population?

Question

The ear tuft allele in chicken is autosomal and produces feathered skin projections near the ear on each side of the head. This gene is dominant embryos do not hatch from the egg. Assume that in a population of 6000 chickens, 2000 have no tufts, and 4000 have ear tufts. what are the frequencies of the normal vs. ear tuft alleles in population?

Answer 1

wait i am missing a part to it,

this is the final question in complete XD i skipped a sentence

The ear tuft allele in chicken is autosomal and produces feathered skin projections near the ear on each side of the head. This gene is dominant and is lethal in the homozygous state. IN other words, homozygous dominant embryos do not hatch from the egg. Assume that in a population of 6000 chickens, 2000 have no tufts, and 4000 have ear tufts. what are the frequencies of the normal vs. ear tuft alleles in population?

Answer 2

To determine the frequencies of the normal and ear tuft alleles in the population, we will need to calculate the allele frequencies using the given information.

Let's assign variables as follows:
- N: Represents the normal allele
- T: Represents the ear tuft allele

Given that the gene for ear tufts is dominant, we can assume that the 4000 chickens with ear tufts must be homozygous dominant TT or heterozygous NT. However, since we don't know the genotypes of these chickens, we will have to infer it from the observed phenotypic ratio.

We know that the phenotype ratio is 2000 (no tufts) : 4000 (ear tufts). If we assume that the observed ratio of phenotypes corresponds to the expected ratio of genotypes in a population at Hardy-Weinberg equilibrium, we can proceed with this assumption.

Let's set up the following equations:
- 2000 = 2nN + nT (Equation 1)
- 4000 = nT + 2nTT (Equation 2)

Here, nN represents the frequency of the normal allele, and nT represents the frequency of the ear tuft allele.

From Equation 1, we can solve for 2nN:
2nN = 2000 - nT

Now, substitute this value into Equation 2:
4000 = nT + 2nTT
4000 = nT + 2(2000 - nT) [substituting 2nN from above]
4000 = nT + 4000 - 2nT
0 = -nT + 4000

Simplifying, we find:
nT = 4000

Now, substituting this value back into Equation 1:
2nN = 2000 - 4000
2nN = -2000
nN = -1000

Obviously, negative frequencies are not meaningful, so we made a mistake somewhere in our calculations or assumptions.

It seems that the observed phenotypic ratio provided is not consistent with the expected genotype ratios at Hardy-Weinberg equilibrium. Further information is needed to accurately calculate the allele frequencies.

wait i am missing a part to it,

hbhg

To determine the frequencies of the normal and ear tuft alleles in the population, we will need to calculate the allele frequencies using the given information.