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If anyone can help me with these two questions, I would greatly appreciate it!!

ID: 185989 • Letter: I

Question

If anyone can help me with these two questions, I would greatly appreciate it!! My textbook is vague, so coming here for much needed help. PLEASE PLEASE

In a population, 1/900 has an autosomal recessive genetic condition, 64/900 are heterozygotes and 835/900 are homozygous with the "normal" allele. What is the actual allele frequency of the mutation that causes the recessive genetic condition?

1/30

8/30

29/30

D. 33/900

A gene (with the common allele A) has multiple recessive mutant alleles, a1, a2 and a3. Homozygous for any of the mutant alleles will cause the development of a congenital disease. Which of the following is FALSE regarding someone with the genotype a1/a3?

This individual is a compound heterozygote.

The parents of this individual are probably related.

This individual will show the disease phenotype from birth.

This is an example of allelic heterogeneity.

A.

1/30

B.

8/30

C.

29/30

D. 33/900

A gene (with the common allele A) has multiple recessive mutant alleles, a1, a2 and a3. Homozygous for any of the mutant alleles will cause the development of a congenital disease. Which of the following is FALSE regarding someone with the genotype a1/a3?

A.

This individual is a compound heterozygote.

B.

The parents of this individual are probably related.

C.

This individual will show the disease phenotype from birth.

D.

This is an example of allelic heterogeneity.

Explanation / Answer

Ans 1:

Frequency of recessive condition,q2 =1/900

Thus, q=1/900

Which is 1/30.

A. 1/30

Ans 2:

It is given that only homozygous individuals for a1,a2,a3 will have the disease.

This means a1/a1 , a2/a2, a3/a3 will phenotypically show the disease.

But a1/a3 will not show diseased phenotype (as it is a heterozygous individual).

option C is false.

This individual will show the disease phenotype from birth

Ans 1:

Frequency of recessive condition,q2 =1/900

Thus, q=1/900

Which is 1/30.

A. 1/30

Ans 2:

It is given that only homozygous individuals for a1,a2,a3 will have the disease.

This means a1/a1 , a2/a2, a3/a3 will phenotypically show the disease.

But a1/a3 will not show diseased phenotype (as it is a heterozygous individual).

option C is false.

This individual will show the disease phenotype from birth