Lecture 31: Population Genetics 2 Flashcards
(14 cards)
What is the gene pool?
The gene pool is the number of all of the alleles present in a population.
We can describe the population of size N in terms of the number of genotypes and the number of alleles (2N)
e.g. the Cystic Fibrosis Transmembrane Receptor (CFTR)
What does it mean if a gene pool is broad or narrow?
A broad gene pool has lots of different alleles.
A narrow gene pool does not have many different alleles.
What does it mean if a gene pool is deep or shallow?
A deep gene pool has alleles with very high frequency.
A shallow gene pool does not have any alleles with a very high frequency.
What are allele frequencies?
Genotypes AA, Aa and aa occur at frequencies fA/A, fA/a and fa/a respectively.
Define frequency of A allele = p
p = fA/A + ½fA/a
Define frequency of a allele = q
q = fa/a + ½fA/a
N.B. p+q = fA/A + fA/a + fa/a = 1
so q = 1 - p
What is the probability that a sperm or egg contains the allele a) A and b) a?
a) p
b) q
What type of probability is used when calculating the probably of a certain fertilisation event?
The egg having A or a is an independent event to the sperm having A or a.
Independent - product rule
P(A + A) fertilisation = pxp = p2
P(A + a) fertilisation = pxq + pxq = 2pq
P(a + a) fertilisation = qxq = q2
What is the Hardy-Weinberg equilibrium?
p2 + 2pq + q2 = 1
What is the probability of two unaffected people having a child with cystic fibrosis, a recessive disease with a frequency of 1 in 2400 births in the UK?
Recessive disease so aa required to have disease, so fa/a = 1/2400 = 0.00042, P(a/a) = q^2 so q = 0.02 so p = 1 - q = 1 - 0.02 p = 0.98
Recessive disease, so both unaffected parents have to be heterozygous, Aa
P(A/a) = 2pq
= 2(0.98)(0.02)
= 0.039
The probability that a normal person is a heterozygote:
Normal people: AA and Aa
P(AA + Aa) = p^2 + 2pq = (0.98)^2 + 0.039
So the proportion of normal people who are heterozygotes:
0.039 / (0.98)^2 + 0.039 = 0.039
Probability that they have affected child ?
=probability each heterozygotes times probability of affected child being born from heterozygotes (independent events):
= (0.039)(0.039)(1/4) ≈ 3.8 x 10-4 or 1 in 2600
What are the genotype frequencies for X-linked genes?
- For X linked genes, females have normal genotype frequencies (2 X chromosomes):
faa = q2, fAa = 2pq, fAA = p2 - Males one X chromosome and so fa = q and fA = p
So if f(affected males) is 0.01, then q=0.01 and f(affected females) is 0.01^2
e.g. red/green colour blindness affects 8% of males, 0.64% of females
What are the allele frequencies when a single autosomal gene has multiple alleles?
Consider three alleles A1, A2, A3 - frequencies p1, p2, p3.
Frequency of any homozygote = square of allele frequency
Frequency of any heterozygote = 2 x product of allele frequencies
What are the limitations to Hardy-Weinberg?
No assortative mating:
(Assortative mating is a mating pattern in which individuals with similar genotypes and/ or phenotypes mate with one another more frequently than would be expected under a random mating pattern)
e.g. height
e.g. Attraction to different MHC alleles
(Women prefer the scent of men whose MHC haplotypes were different from their own)
- Large population size required
- Genetically static
no migration in or out
no mutation
no selection
What a the genetics of blue eyes?
A single SNP (A to G) defines all blue eyed people - this occurs within a 50kbp conserved haplotype.
The SNP rs12913832 that controls blue eye colour is located upstream from OCA2 in an intron of HERC2 on chromosome 15.
Blue eyes first appeared about 6-10,000 years ago.
Blue eyes predominate in northern Europe around the Baltic - GEOGRAPHICAL ISOLATION prevents equilibration according to Hardy-Weinberg equation.
How can the SNP mutation rate of a species’ genome be calculated?
Sequence large parts of the genome from different generations and look for new mutations.
What can the SNP mutation rate of a species’ genome be used to estimate?
From this figure, can estimate the number of generations / time between related genomes.
