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The following questions are based on information in the Ritland & Marshall (2001

ID: 93483 • Letter: T

Question

The following questions are based on information in the Ritland & Marshall (2001) abstract (see OA2 folder). Choose the best answer for each question.

What is the frequency of heterozygotes if the population is under Hardy Weinberg equilibrium?

How many bears are expected to be heterozygotes given the population is at Hardy Weinberg equilibrium?

What is the frequency of the Cys/ Cys genotype?

What is the frequency of the black bear genotype Tyr / Tyr if the population is at Hardy Weinberg equilibrium?

What is the frequency of the Cys allele?

https://www.ncbi.nlm.nih.gov/pubmed/11566108?dopt=Abstract this is the link for the abstract

0.438

96

0.1

0.46

0.32

      -       A.       B.       C.       D.       E.   

What is the frequency of heterozygotes if the population is under Hardy Weinberg equilibrium?

      -       A.       B.       C.       D.       E.   

How many bears are expected to be heterozygotes given the population is at Hardy Weinberg equilibrium?

      -       A.       B.       C.       D.       E.   

What is the frequency of the Cys/ Cys genotype?

      -       A.       B.       C.       D.       E.   

What is the frequency of the black bear genotype Tyr / Tyr if the population is at Hardy Weinberg equilibrium?

      -       A.       B.       C.       D.       E.   

What is the frequency of the Cys allele?

https://www.ncbi.nlm.nih.gov/pubmed/11566108?dopt=Abstract this is the link for the abstract

A.

0.438

B.

96

C.

0.1

D.

0.46

E.

0.32

Explanation / Answer

When a population is in Hardy Weinberg equilibrium then, p+q=1, where p and q are the gene frequency of two allele.

If P is the frequency of wild type gene and q is the Mutant gene then,

q2=22/220=0.1, then q=0.316

Then p=1-q=0.684, then 2pq=0.432.

Thus the heterozygous Frequency is 0.432 or 43.2%.

2. Thus the number of heterozygous bare in the population is approx. 95.

3.The mutant Cyc/cys genotype frequency is q2=0.1or 10%.

4.The Tyr / Tyr genotype frequency is p2= 0.468 or 46.8%.

5.Frequency of the Cys allele is q=0.316