The reason that the HD allele acts as a dominant allele is A. The mutant HD alle

Question

The reason that the HD allele acts as a dominant allele is A. The mutant HD allele suppress protein production from the normal HD allele. B. The HD mutations results in a protein that can damage nerve cells even in the presence of the normal protein C. The normal HD allele does not normally prod protein but the mutant HD allele does D. The protein produce from the mutant HD allele is non-functional. Suppose that bin plants, smooth seeds (S) are dominant to wrinkle-seeds (s) and tall plants (T) are dominant to short plants (t). A tall plant with smooth seeds was backcrossed to a parent that was short and wrinkled. What proportion of the progeny is expected to be homozygous for short and wrinkled? A. 1/2 B. 1/4 C. 1/8 D. 1/16 E. 0 Sickle cell anemia is a recessive trait in humans, In a cross between a father who has sickle cell anemia and a mother who is heterozygous for the gene, what is the probability that their first three children will have the normal phenotype? A. 1/4 B. 1/2 C. 0 D. 1/8 E. 1/16 Using the product rule, one would calculate the probability of parents having six children where are all boys as (1/2)^r A. True B. False

Explanation / Answer

1. (B) The HD allele acts as a dominant allele as the mutant HTT (gene product of mutant HD allele) can damage the nerve cells even in the presence of a normal functioning HTT. The mutant HTT is an altered form of the protein which causes neuronal degeneration.

2. Since, one parent of the tall plant with smooth seeds was a short plant with wrinkled seeds, this plan must be heterozygous for tall and smooth. The genotype of the plant is (TtSs) and that of its parent is (ttss). The expected progeny will be
1/4 TtSs - tall and smooth
1/4 Ttss - tall and wrinkled
1/4 ttSs - short and smooth
1/4 ttss - short and wrinkled.
Therefore the proportion of progeny that would be homozygous for short and wrinkled is 1/4.

Unless, there is perfect linkage (present on the same chromosome) between the genes coding for tall and smooth, this cross would yield
1/2 TtSs - tall and smooth
1/2 ttss - short and wrinkled, thereby in this case the proportion of progeny that would be homozygous for short and wrinkles is 1/2.

3. (D) Sickle cell anemia is reccessive, and not sex linked (A - normal, a - anemic). Father has sickle cell anemia (aa) and mother is heterozygous for the gene (Aa).

The probability that one child will have normal phenotype is 1/2, therefore, the probability that all three children will show normal phenotype is (1/2)3 = 1/8.

4. (A) True, as all the six events are independent of each other. Probability of one child being a boy is 1/2, and hence, the probability of all six children being boys is (1/2)6.

a a A Aa Aa a aa aa

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