Question

1. a man with type o blood marries a woman with heterozygous type a blood.

o. what are the possible genotypes of the offspring?
p. what are the possible phenotypes of the offspring?
q. what is the probability of each genotype?
r. what is the probability of each phenotype?

Explanation:

Step1: Determine Genotypes of Parents

The man has Type O blood, so his genotype is \( ii \) (since Type O is recessive, requiring two \( i \) alleles). The woman is heterozygous Type A, so her genotype is \( I^A i \) (Type A can be \( I^A I^A \) or \( I^A i \); heterozygous means one \( I^A \) and one \( i \)).

Step2: Set Up Punnett Square

	\( I^A \)	\( i \)
\( i \)	\( I^A i \)	\( ii \)

Step3: Analyze Offspring Genotypes (Question o)

From the Punnett square, the possible genotypes are \( I^A i \) and \( ii \).

Step4: Analyze Offspring Phenotypes (Question p)

• Genotype \( I^A i \) corresponds to Type A blood (since \( I^A \) is dominant over \( i \)).
• Genotype \( ii \) corresponds to Type O blood. So possible phenotypes are Type A and Type O.

Step5: Calculate Genotype Probabilities (Question q)

There are 4 total squares in the Punnett square.

• \( I^A i \) appears in 2 squares: Probability = \( \frac{2}{4} = \frac{1}{2} \).
• \( ii \) appears in 2 squares: Probability = \( \frac{2}{4} = \frac{1}{2} \).

Step6: Calculate Phenotype Probabilities (Question r)

• Type A (from \( I^A i \)): Probability = \( \frac{2}{4} = \frac{1}{2} \).
• Type O (from \( ii \)): Probability = \( \frac{2}{4} = \frac{1}{2} \).

Answer:

(for each sub-question):

• o. Possible genotypes: \( I^A i \), \( ii \)
• p. Possible phenotypes: Type A blood, Type O blood
• q. Probability of \( I^A i \): \( \frac{1}{2} \); Probability of \( ii \): \( \frac{1}{2} \)
• r. Probability of Type A: \( \frac{1}{2} \); Probability of Type O: \( \frac{1}{2} \)