Biology (Genetics)

Question

In a population, the frequency of alleles determining the ABO blood types are

estimated as 0.74 for i, 0.16 for A, 0.1 for B. Assuming random mating, what are

the expected frequencies for AB blood type?

Answer 1

The AB blood type occurs only with the genotype I^A I^B.

Under random mating, the probability of inheriting an I^A allele from one parent and an I^B allele from the other is calculated as follows:

Frequency(AB) = 2 × (Frequency of I^A) × (Frequency of I^B)

Plugging in the given values:

Frequency of I^A = 0.16

Frequency of I^B = 0.10

Calculation: Frequency(AB) = 2 × (0.16) × (0.10) = 2 × 0.016 = 0.032

So, the expected frequency of the AB blood type is 0.032 or 3.2%.

Answer 2

1. Recall what determines ABO blood type

1. Alleles: I^A, I^B, i
2. Genotypes and phenotypes:
3. I^AI^A or I^Ai ⇒ type A
4. I^BI^B or I^Bi ⇒ type B
5. I^AI^B ⇒ type AB
6. ii ⇒ type O

We are asked specifically about type AB, which comes only from the genotype I^AI^B.

2. Use allele frequencies

The problem gives:

1. p(i) = 0.74
2. p(I^A) = 0.16
3. p(I^B) = 0.10

Since the population mates randomly, genotype frequencies can be estimated with the Hardy–Weinberg principle.

3. Calculate frequency of AB genotype

f(I^AI^B) = 2 × p(I^A) × p(I^B)

f(I^AI^B) = 2 × (0.16) × (0.10) = 0.032

The expected frequency of type AB blood in the population is 0.032, or 3.2%.

Final Answer:



The expected frequency of AB blood type is 3.2% of the population.

Answer 3

To find the expected frequency of the AB blood type under random mating, we use the Hardy-Weinberg principle for multiple alleles.

The ABO blood group system is determined by three alleles:

1. IAI^AIA (A allele, frequency = 0.16)
2. IBI^BIB (B allele, frequency = 0.10)
3. iii (O allele, frequency = 0.74)

Step 1: Use the Hardy-Weinberg formula for AB genotype

AB blood type occurs when an individual has one A allele and one B allele, i.e., genotype IAIBI^A I^BIAIB.

The frequency of this genotype is calculated as:

Frequency of AB=2×(frequency of IA)×(frequency of IB)\text{Frequency of AB} = 2 \times (\text{frequency of } I^A) \times (\text{frequency of } I^B)Frequency of AB=2×(frequency of IA)×(frequency of IB) =2×0.16×0.10=0.032= 2 \times 0.16 \times 0.10 = 0.032=2×0.16×0.10=0.032

Answer: The expected frequency of the AB blood type is 0.032 or 3.2%.

Answer 4

There are three alleles, so we will use: (p + q + r)² = 1

p = frequency of allele A = 0.16

q = frequency of allele B = 0.10

r = frequency of allele i = 0.74.

For an A blood type you can have either an AA or AO, so p² + 2pr

For a B blood type you can have either a BB or BO, so q² + 2qr

For an O blood type you would have ii, so r²

For an AB blood type you would use 2pq, so 2(0.16)(0.10) = 0.032