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

Q: 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?

A. The 'assuming random mating' phrase hints that the teacher wants you to use the expectations of the Hardy Weinberg Equilibrium (HWE) equation. There are five generally recognized assumptions that have to be met for HWE to work. One of those is random mating. If your teacher is telling you they are randomly mating, its likely they are either simplifying the assumptions or have stated the rest of them elsewhere. However, without HWE being true, there is no easy way to solve this, so let's assume that is what they want here.

If I were your tutor, I would go over how and why HWE works here. Once you understand it, this part is easy.

But basically, if we accept that HWE is true for this problem then each expression within the equation tells us what our expected values for each genotype frequency are. Remember that HWE for two alleles in a 'normal' Mendelian system is p^2 + 2pq + q^2 =1 It will always equal one - it has to (an important point that people get confused on). Each part of that equation can be used as an estimate of a different genotype if you know the allele frequencies. Each squared component is equal to the genotype frequency of one of the homozygous conditions, and the 2pq is for the heterozygous.

Now, for this problem, you have a slightly more complicated equation since there are three alleles, but the pattern is still the same - we don't have to actually know the bigger equation. The frequency for any heterozygote is:

[frequency of allele 1] * [frequency of allele 2] * 2

It turns out that this is a basic statistical truism that works for any system where we know the frequency of the two things occurring. Here, those two things are 'drawing the A allele (I^A , technically)', and 'drawing the B allele'. How often you draw them out of the gene pool just depends on how common they are in the gene pool. So, if 16% of all alleles (frequency of 0.16) are allele A then the chance you would draw that allele out of the gene pool is 0.16. The chance you would draw allele B is .10 because allele B occurs at a frequency of 0.10. The chance you draw out both allele A and then B is simply .16*.1. BUT you can also draw out B and then A (.1 * .16). That is why we multiply the whole thing by 2.

The short version is that you can determine the expected value of any heterozygote genotype from a population in HWE by multiplying the frequencies of the two alleles and then multiplying that times 2.

Good luck! HWE is actually fun and handy once you get the hang of it.

Answer 2

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 3

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 4

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 5

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