Question

important information: use this information for the next few questions. handedness r = right r = left earlobes a = attached a = free 30. a male with attached earlobes (heterozygous) has children with a female with attached earlobes (heterozygous). what proportion of their children will have attached earlobes? a. 1 of 4 b. 2 of 4 c. 3 of 4 d. 1 of 8 e. none of these are correct 31. using the cross in the problem above, what proportion of their children would be expected to be homozygous? a. 0 of 4 b. 1 of 4 c. 2 of 4 d. 3 of 4 e. 4 of 4 32. for another set of parents, rraa and rraa, what proportion of their children will be left - handed and attached lobed? a. 1 of 4 b. 2 of 4 c. 3 of 4 d. 4 of 4 e. none of these are correct

Explanation:

Step1: Set up Punnett - square for question 30

The male and female both have genotype Aa for earlobes. The possible gametes from each parent are A and a. The Punnett - square has 4 cells: AA, Aa, aA, aa. Attached earlobes are dominant (A), so AA, Aa, and aA all result in attached earlobes. So the proportion of children with attached earlobes is $\frac{3}{4}$.

Step2: Analyze homozygous proportion for question 31

From the Punnett - square of Aa x Aa (question 30), the homozygous genotypes are AA and aa. There are 2 homozygous genotypes out of 4 possible genotypes, so the proportion of homozygous children is $\frac{1}{2}$ or 2 of 4.

Step3: Set up Punnett - square for question 32

For handedness, parents are Rr x Rr, and for earlobes, they are AA x aa. For handedness, the Punnett - square for Rr x Rr gives genotypes RR, Rr, rR, rr. For earlobes, all offspring will be Aa. The probability of being left - handed (rr) is $\frac{1}{4}$, and the probability of having attached earlobes (Aa) is 1. Using the multiplication rule of independent events, the probability of being left - handed and having attached earlobes is $\frac{1}{4}\times1=\frac{1}{4}$.

Answer:

1. c. 3 of 4
2. c. 2 of 4
3. a. 1 of 4