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