Gene frequencies (Homework help)

[Pieman]

I'm stuck on this one, so I thought why not see if RZ can help me?

Assignment:
A particular form of albinism is caused by a deviation of a single gene. The allele for this form of albinism is recessive. The frequency of this allele is 0,01.
Two parents with normal pigmentation conceive a child. In this assignment it's assumed no mutations occur.
Calculate the chance of the child having this form of albinism.

This is how I did it, but I'm not sure if it's correct.

I use "p" to indicate the frequency of the dominant allele and "q" for the recessive allele
I use "A" to indicate the dominant allele and "a" to indicate the recessive allele

q = 0,01 (see text)
p + q = 1 -> p = 0,99

To be a bearer of albinism the genotype needs to be Aa. So,

f(Aa) = 2pq = 2*0,99 * 0,01 = 0,0198

The chance of two bearers mating = 0,0198 * 0,0198 = 0,0198^2

the chance of said parents actually getting an albino baby,
A a
A AA Aa
a Aa aa
(1/4)

So the chance of 2 bearing parents to mate and conceive an albino child = 0,198^2 * 1/4 = 9,801 * 10^-5

Can anyone tell me if I did it right?

 

[MasterBubba]

Calculation of allele frequencies from genotype frequencies
If f(AA), f(Aa), and f(aa) are the frequencies of the three genotypes at a locus with two alleles, then the frequency p of the A-allele and the frequency q of the a-allele are obtained by counting alleles. Because each homozygote AA consists only of A-alleles, and because half of the alleles of each heterozygote Aa are A-alleles, the total frequency p of A-alleles in the population is calculated as

frequency of A
Similarly, the frequency q of the a allele is given by

frequency of a
It would be expected that p and q sum to 1, since they are the frequencies of the only two alleles present. Indeed they do:


and from this we get:

q = 1 − p and p = 1 − q
If there are more than two different allelic forms, the frequency for each allele is simply the frequency of its homozygote plus half the sum of the frequencies for all the heterozygotes in which it appears. Allele frequency can always be calculated from genotype frequency, whereas the reverse requires that the Hardy-Weinberg conditions of random mating apply. This is partly due to the three genotype frequencies and the two allele frequencies. It is easier to reduce from three to two.


[edit] An example population
Consider a population of ten individuals and a given locus with two possible alleles, A and a. Suppose that the genotypes of the individuals are as follows:

AA, Aa, AA, aa, Aa, AA, AA, Aa, Aa, and AA
Then the allele frequencies of allele A and allele a are:


so if an individual is chosen at random there is a 70% chance it will carry that allele


and there is a 30% chance that an individual chosen at random will have the a-allele


[edit] The effect of mutation
Let

 

[Pieman]

Calculation of allele frequencies from genotype frequencies
If f(AA), f(Aa), and f(aa) are the frequencies of the three genotypes at a locus with two alleles, then the frequency p of the A-allele and the frequency q of the a-allele are obtained by counting alleles. Because each homozygote AA consists only of A-alleles, and because half of the alleles of each heterozygote Aa are A-alleles, the total frequency p of A-alleles in the population is calculated as

frequency of A
Similarly, the frequency q of the a allele is given by

frequency of a
It would be expected that p and q sum to 1, since they are the frequencies of the only two alleles present. Indeed they do:


and from this we get:

q = 1 − p and p = 1 − q
If there are more than two different allelic forms, the frequency for each allele is simply the frequency of its homozygote plus half the sum of the frequencies for all the heterozygotes in which it appears. Allele frequency can always be calculated from genotype frequency, whereas the reverse requires that the Hardy-Weinberg conditions of random mating apply. This is partly due to the three genotype frequencies and the two allele frequencies. It is easier to reduce from three to two.


[edit] An example population
Consider a population of ten individuals and a given locus with two possible alleles, A and a. Suppose that the genotypes of the individuals are as follows:

AA, Aa, AA, aa, Aa, AA, AA, Aa, Aa, and AA
Then the allele frequencies of allele A and allele a are:


so if an individual is chosen at random there is a 70% chance it will carry that allele


and there is a 30% chance that an individual chosen at random will have the a-allele


[edit] The effect of mutation
Let

 

[DrMido0o89]

oooh i forgot this gene problems xD
i liked it so much btw

 

[MasterBubba]

Sorry, didnt read full post.
I do believe that you are correct.

 

[Pfa]

I believe it is correct. I could help ya out if you get into multi-gene frequency calculations, for example:

AABB X aabb.

GL HF