Genetics Question

[darkz]

(Not in the 3/4 course, but pulls from key ideas in the 1/2 course, so if anyone knows how to do the question, any help would be appreciated!)
In a particular population, it was found that the frequency of the I^A, I^B and i alleles was 0.2, 0.1 and 0.7 respectively. Given this information, what proportion of the population has group AB blood?

So the possible genotypes are IA IA, IA IB, IA i, IB IB, IB i, i i => ratio of A:B:AB:O is 2:2:1:1 when considering a population with an equal allele frequency? Not really sure how to approach the question haha, again, any help would be appreciated
 

[Sine]

(Not in the 3/4 course, but pulls from key ideas in the 1/2 course, so if anyone knows how to do the question, any help would be appreciated!)
In a particular population, it was found that the frequency of the I^A, I^B and i alleles was 0.2, 0.1 and 0.7 respectively. Given this information, what proportion of the population has group AB blood?

So the possible genotypes are IA IA, IA IB, IA i, IB IB, IB i, i i => ratio of A:B:AB:O is 2:2:1:1 when considering a population with an equal allele frequency? Not really sure how to approach the question haha, again, any help would be appreciated
 

very rusty with genetics so feel free to correct me anyone

Hardy weinberg equation for 3 alleles
(A + B + i) = 1
(A + B + i)^2 = 1
(A)^2 + (B)^2 + (i)^2 + 2AB + 2Ai + 2Bi = 1
(0.2)^2 + (0.1)^2 + (0.7)^2 + 2AB +2(0.2)(0.7) + 2(0.1)(0.7) = 1
0.04 + 0.01 + 0.49 + 2AB + 0.28 + 0.14 = 1
0.96 + 2AB = 1
2AB = 0.04

[TheBigC]

very rusty with genetics so feel free to correct me anyone

Hardy weinberg equation for 3 alleles
(A + B + i) = 1
(A + B + i)^2 = 1
(A)^2 + (B)^2 + (i)^2 + 2AB + 2Ai + 2Bi = 1
(0.2)^2 + (0.1)^2 + (0.7)^2 + 2AB +2(0.2)(0.7) + 2(0.1)(0.7) = 1
0.04 + 0.01 + 0.49 + 2AB + 0.28 + 0.14 = 1
0.96 + 2AB = 1
2AB = 0.04

Great stuff Sine. I hadn't encountered a three-allele HW eqn before...

One point though: you could have skipped a few steps after your cubic expansion.


Therefore, for proportion of population with I^A I^B genotype:
hehe.

[Sine]

Great stuff Sine. I hadn't encountered a three-allele HW eqn before...

One point though: you could have skipped a few steps after your cubic expansion.


Therefore, for proportion of population with I^A I^B genotype:
hehe.

yeah definitely I just use this method when teaching this concept for the first time to give a more intuitive solution and also to exhibit that all the genotypes = 1 especially to any skeptical student.

[darkz]

Thanks guys!

But what's the logic behind the equation, i.e. would it be possible to derive or work out the question if you didn't know the equation?

[TheBigC]

Thanks guys!

But what's the logic behind the equation, i.e. would it be possible to derive or work out the question if you didn't know the equation?

It is possible to derive it:

Let and let (two-allele population)

Then, p + q = 1 (this is a given, as if the population only contains two alleles, then the frequencies of both alleles MUST add to give 1).

[Squaring both sides]




[Expanding]



Therefore, we can derive the above equation.

The same is true for a three allele population (except with three alleles instead of two)

NOTE: The Hardy-Weinberg equation makes specific assumptions about a population:

- All phenotypes exhibit equal fitness (no natural selection) (i.e. if a heterozygous phenotype is strongly selected against then the frequency of the heterozygous allelic combination may well be 0.0 despite the equation demonstrating otherwise).
- No immigration or emigration (again, will skew assumed relative proportions of alleles)
- No genetic drift (assuming extremely large population)

(I may not have included all of the factors... a little bit coarse on the edges after all of this time... lol)