Question

question 9 (1 point)
use the following information to answer the next question.
inheritance of abo blood types is controlled by three alleles. the alleles for type a blood ((i^a)) and type b blood ((i^b)) are co - dominant, and the allele for type o blood (i) is recessive.
some descriptions of blood type in a family
generation i: the father has type a blood and the mother has type ab blood.
generation ii: the oldest child is a son with type a blood; the second - oldest child is a daughter with type b blood; and the youngest child is a son whose blood type is unknown.
if the two parents in generation i have another child, what is the probability that this child will be a daughter with type b blood?
answer:
(record your answer as a frequency between 0 and 1 rounded to two decimal places in the space below)

Explanation:

Step1: Determine parental genotypes

Father has type A blood. Since one child has type B blood (genotype \(I^B i\) or \(I^B I^B\), but mother is \(I^A I^B\), so father must be \(I^A i\) (because to have a child with B blood, father must contribute \(i\) as mother contributes \(I^B\)). Mother is \(I^A I^B\) (type AB).

Step2: Find possible gametes

Father (\(I^A i\)) produces gametes: \(I^A\), \(i\) (each with probability \(0.5\)).
Mother (\(I^A I^B\)) produces gametes: \(I^A\), \(I^B\) (each with probability \(0.5\)).

Step3: Determine possible offspring genotypes for blood type B

Blood type B requires genotype with \(I^B\). So:
- Father’s \(i\) + Mother’s \(I^B\): \(I^B i\) (type B).
- Father’s \(I^A\) + Mother’s \(I^B\): \(I^A I^B\) (type AB, not B).
So only when father gives \(i\) and mother gives \(I^B\), offspring has type B. Probability of this: \(P(\text{father } i) \times P(\text{mother } I^B) = 0.5 \times 0.5 = 0.25\).

Step4: Probability of daughter

Probability of female offspring is \(0.5\) (since sex ratio is 50% male, 50% female).

Step5: Combine probabilities

Probability of daughter with type B blood = Probability (type B) × Probability (daughter) = \(0.25 \times 0.5 = 0.125\), rounded to two decimal places is \(0.13\)? Wait, wait, let's recheck. Wait, father's genotype: let's confirm. Father is type A, mother is AB. Children: oldest son A, second daughter B. So for daughter to be B, her genotype is \(I^B i\) (since mother is AB, she can give \(I^B\), father must give \(i\)). So father's genotype: if he were \(I^A I^A\), he can't give \(i\), so he must be \(I^A i\). So gametes: father: \(I^A\) (0.5), \(i\) (0.5); mother: \(I^A\) (0.5), \(I^B\) (0.5).

Offspring with type B: only when father gives \(i\) and mother gives \(I^B\): \(i \times I^B\) → probability \(0.5 \times 0.5 = 0.25\) (for blood type B). Then probability of daughter: 0.5. So total probability: \(0.25 \times 0.5 = 0.125\), which is \(0.13\) when rounded to two decimals? Wait, 0.125 rounded to two decimals is 0.13? Wait, no: 0.125 is exactly halfway, but typically rounded to 0.13? Wait, no, 0.125 rounded to two decimal places: the third digit is 5, so round up the second digit: 0.13. Wait, but let's recalculate the blood type probability again. Wait, possible offspring genotypes:

Father (\(I^A i\)) × Mother (\(I^A I^B\)):

| Father \ Mother | \(I^A\) (0.5) | \(I^B\) (0.5) |
|-----------------|---------------|---------------|
| \(I^A\) (0.5) | \(I^A I^A\) (A) | \(I^A I^B\) (AB) |
| \(i\) (0.5) | \(I^A i\) (A) | \(I^B i\) (B) |

So the probability of type B blood: number of type B genotypes over total. Total possible genotypes: 4 (each cell is 0.5×0.5=0.25 probability). Type B is \(I^B i\), which is 1 cell (probability 0.25). So probability of type B is \(0.25\). Then probability of daughter: 0.5. So \(0.25 \times 0.5 = 0.125\), which is \(0.13\) when rounded to two decimal places? Wait, 0.125 rounded to two decimals: the second decimal is 2, third is 5, so we round up the second decimal: 0.13. Wait, but maybe I made a mistake in father's genotype. Wait, is father \(I^A I^A\) possible? If father is \(I^A I^A\), then his gametes are only \(I^A\). Then offspring with mother \(I^A I^B\) would be \(I^A I^A\) (A), \(I^A I^B\) (AB). No type B offspring, but there is a daughter with type B, so father must be \(I^A i\). So that part is correct. So probability of type B: 0.25, probability of daughter: 0.5, so 0.25*0.5=0.125, rounded to two decimals is 0.13? Wait, 0.125 is exactly 1/8, which is 0.125, so when rounded to two decimal places, it's 0.13 (since the third digit is 5, we round the second digit up: 2 becomes 3). Wait, but maybe the correct calculation is:

Wait, let's list all possible offspring:

From father \(I^A i\) (gametes \(I^A\), \(i\)) and mother \(I^A I^B\) (gametes \(I^A\), \(I^B\)):

1. \(I^A\) (father) + \(I^A\) (mother): \(I^A I^A\) (A) – male or female? Probability 0.5 (father) * 0.5 (mother) = 0.25, 50% chance female: 0.125.
2. \(I^A\) (father) + \(I^B\) (mother): \(I^A I^B\) (AB) – 0.25, 50% female: 0.125.
3. \(i\) (father) + \(I^A\) (mother): \(I^A i\) (A) – 0.25, 50% female: 0.125.
4. \(i\) (father) + \(I^B\) (mother): \(I^B i\) (B) – 0.25, 50% female: 0.125.

Wait, so the probability of a daughter with type B is the probability of the fourth case: 0.25 (for the genotype) * 0.5 (for female) = 0.125, which is 0.13 when rounded to two decimal places? Wait, no, 0.125 rounded to two decimal places is 0.13? Wait, 0.125 is 1/8, which is 0.125. The second decimal is 2, the third is 5. In rounding, 0.125 to two decimals: look at the third digit (5), so we round the second digit up: 2 becomes 3, so 0.13. Wait, but maybe the correct answer is 0.13? Wait, no, wait: 0.125 is exactly halfway between 0.12 and 0.13, but in standard rounding, 0.125 rounds to 0.13. Alternatively, maybe I made a mistake in parental genotypes. Wait, mother is AB (\(I^A I^B\)), father is A. If father were \(I^A I^A\), then all offspring would be A or AB, but there is a child with B blood, so father must be \(I^A i\). So that's correct. So the probability of type B is 0.25 (only \(I^B i\) from \(i\) and \(I^B\)), and probability of daughter is 0.5, so 0.25*0.5=0.125, which is 0.13 when rounded to two decimal places. Wait, but let's check again. Wait, the possible genotypes for type B: \(I^B I^B\) or \(I^B i\). But mother is AB, so she can't give \(I^B I^B\) (she only has one \(I^B\)). So offspring with type B must be \(I^B i\), which requires father to give \(i\) and mother to give \(I^B\). So that's correct. So the probability is 0.25 (for type B) * 0.5 (for daughter) = 0.125, rounded to two decimals is 0.13? Wait, no, 0.125 rounded to two decimal places is 0.13? Wait, 0.125 is 1/8, which is 0.125. The first decimal is 1, second is 2, third is 5. So when rounding to two decimals, we look at the third digit (5) and round the second digit up: 2 becomes 3, so 0.13. Yes.

Answer:

0.13