# Pedigree Analysis and Inheritance

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1. A man and woman marry who are both Rh+ and both heterozygous for the Rh factor. (Rh+ is dominant over Rh-.)

A. Give the genotypes of the parents. (2 points) Man ______________ Woman ______________

B. What is the probability that their first child will be Rh-? (2 points)

C. What is the probability of a Rh+ male or a Rh- female? (2 points)

D. What if the probability that they will have a Rh+ male, a Rh- female, and a Rh+ female? (2 points)

E. Calculate the probability of 2 Rh+ and 4 Rh- children? (4 points)

2. Red-green color blindness is X-linked recessive. MN blood type is determined by an autosomal gene with two codominant alleles, LM and LN. A woman whose father was color-blind has type MN blood. She is going to have a child with a man who has normal color vision and type MN blood.

What is the probability the child will be a boy, with normal color vision and type M blood? (5 points)

3. The accompanying pedigree is for a rare, but relatively mild, hereditary disorder of the skin.

A. What is the mode of inheritance for this rare skin disorder? Explain. (4 points)

B. Use appropriate genetic symbols and give the genotype of the following individuals (4 points)

I-1 _____ II-1 _____ III-2 _____ III-7

C. What is the probability that III-1 is heterozygous? (2 points)

D. If III-1 and III-8 marry:

(1) what is the probability that their first child will have the skin disorder? (2 points)

(2) what is the probability that their second child with be a normal male? (2 points)

4. The trait in the pedigree below is inherited as a single dominant gene.

A. What is the probability that II-1 is heterozygous? (2 points)

B. What is the probability of an affected child if the following cousins marry?

(1) III-1 and III-3 (2 points)

(2) III-2 and III-4 (2 points)

5. A student crossed lobed-eyed, tan-bodied female Drosophila with normal-eyed, black-bodied males. All the F1's were lobed-eyed and tan-bodied. She crossed the F1's and obtained the results given below.

Phenotype Number

lobed-eyed, tan-bodied 675

lobed-eyed, black-bodied 97

normal-eyed, tan-bodied 78

normal-eyed, black-bodied 150

A. Assume independent assortment and state the null hypothesis. (2 points)

B. Test your null hypothesis using chi-square analysis. (10 points)

6. In humans, right-handedness (L) is dominant to left-handedness (l). Assume a mating between 2 individuals that are heterozygous.

A. What is the probability that a child will be left-handed? (1 point)

B. If a child from this mating is right-handed, what is the probability that s/he is heterozygous? (2 points)

C. What is the probability of having a family of 3 right-handed and 1 left-handed children? (2 points)

#### Solution Preview

1. A man and woman marry who are both Rh+ and both heterozygous for the Rh factor. (Rh+ is dominant over Rh-.)

A. Give the genotypes of the parents. (2 points) Man ______________ Woman ______________

B. What is the probability that their first child will be Rh-? (2 points)

C. What is the probability of a Rh+ male or a Rh- female? (2 points)

D. What if the probability that they will have a Rh+ male, a Rh- female, and a Rh+ female? (2 points)

E. Calculate the probability of 2 Rh+ and 4 Rh- children? (4 points)

The man and the woman at both Rh+ Rh-.

To have an Rh- phenotype, you have to get two copies of the Rh- allele. The probability of that is 0.25 (0.5 that you get Rh- from the mother x 0.5 that you get Rh- from the father).

The probability of an Rh+ male is 0.75 (that he's Rh+) x 0.5 (that it's a boy) = 0.375. The probability of an Rh- female is 0.25 (that she's Rh-) x 0.5 (that it's a girl) = 0.125. The probability that either of these happen is 0.375 + 0.125 = 0.5.

The probability of an Rh+ male is 0.375, an Rh- female is 0.125, and an Rh+ female is 0.375 (same logic as an Rh+ male). The probability of having all three of these children is 0.375 x 0.125 x 0.375 = 0.0176. However, there are 6 possible birth-orders for the children, so the actually probability is 6 x 0.0176 = 0.03.

The probability of 2 Rh+ children is 0.75 x 0.75. The probability of 4 Rh- children is 0.25 x 0.25 x 0.25 x 0.25. The probability of all of those children being born is those probabilities multiplied together -- 0.002. However, there are 15 possible birth-orders for the children, so the actually probability is 15 x 0.002 = 0.1056.

2. Red-green color blindness is X-linked recessive. MN blood type is determined by an autosomal gene with two codominant alleles, LM and LN. A woman whose father was color-blind has type MN ...

#### Solution Summary

This problem set has 6 multi-part questions about genetic inheritance, gene dominance, and pedigree analysis. Complete, step-by-step explanations are included.