# Tutorial #4 by Ma’ayan Fishelson

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Tutorial #4 by Ma’ayan Fishelson
Hardy Weinberg Equilibrium & Linkage Equilibrium. References: Kenneth Lange “Mathematical and Statistical Methods for Genetic Analysis” Jurg Ott “Analysis of Human Genetic Linkage” Tutorial #4 by Ma’ayan Fishelson

Hardy-Weinberg equilibrium
A||a Other Possibilities: (a||A), (A||A),(a||a) What is the probability that a person chosen at random from the population would have each one of these genotypes ? Hardy-Weinberg equilibrium: If the allele frequencies are P(a)=p and P(A)=q, then, under some assumptions: P(Aa)= 2pq, P(AA)= p2, P(aa)= q2. Originally by Dan Geiger Corresponds to the random mating of two gametes, an egg & a sperm.

Hardy-Weinberg equilibrium
This model relies on the following assumptions: Infinite population size. Discrete generations. Random mating. No selection. No migration. No mutation. Equal initial genotype frequencies in the two sexes.

Hardy-Weinberg equilibrium
Let the initial genotype frequencies be: P(A/A) = u, P(A/a) = v, P(a/a) = w. The next generation satisfies: Frequency Nature of offspring Mating Type u2 A/A A/A x A/A 2uv ½A/A + ½A/a A/A x A/a 2uw A/a A/A x a/a v2 ¼A/A + ½A/a + ¼a/a A/a x A/a 2vw ½A/a + ½a/a A/a x a/a w2 a/a a/a x a/a P(A/A) = u2 + uv + ¼v2 = (u + ½v)2 P(A/a) = uv + 2uw + ½v2 + vw = 2(u + ½v)(½v + w) ¼v2 + vw + w2 = (½v + w)2

Hardy-Weinberg equilibrium
If we define the frequencies of the alleles as: p = P(A) = u + v/2 q = P(a) = v/2 + w then, the genotype frequencies are: P(A/A) = p2 P(A/a) = 2pq P(a/a) = q2 Second generation respects the same distribution: P(AA)= (p2 + ½2 pq)2 = [p(p+q)]2 = p2 P(Aa)= 2(p2 + ½2pq) (½2pq +q2) =2p(p+q)q(p+q)= 2pq P(aa) = (½2pq + q2)2 = [q(p + q)]2 = q2

There is no association between alleles at any 2 loci.
Linkage equilibrium A a B b a a b b A a There is no association between alleles at any 2 loci. Linkage equilibrium: P(Aa,Bb) = P(Aa)P(Bb) Hardy-Weinberg equilibrium: P(Aa) = P(A)P(a), P(Bb)=P(B)P(b) Convergence to equilibrium is at the rate of (1 – θ)n. θ = (θf + θm) / 2, where θf and θm are the female and male recombination fractions, respectively.

Question #1 You have sampled a population in which you know that the percentage of the homozygos recessive genotype (aa) is 36%. Calculate the following: The frequency of the “aa” genotype. The frequency of the “a” allele. The frequency of the “A” allele. The frequencies of the genotype “AA” and “Aa”. The frequencies of the two possible phenotypes if “A” is completely dominant over “a”.

Question #2 A temp. dependant enzyme that affects the survival of organisms has 2 alleles (A,B). In a population of 1000 individuals, these alleles have the following frequencies: A = 0.45 B = 0.55 The population is exposed to a high temp. which proves to be lethal for all individuals homozygous for the A allele. Given that the population is maintained at this higher temp., what will be the genotypic and allelic frequencies of the next generation ?

Suppose we have two alleles, A1 and A2 with equilibrium frequencies p1 and p2. The female genotypes have the following frequencies: In males the genotypes A1 and A2 have the frequencies p1 and p2. A2/A2 A1/A2 A1/A1 p22 2p1p2 p12

Question #3 The red cell antigen Xg(a) is an X-linked dominant allele with a frequency of approximately p=0.65 in Caucasians. How many males and how many females carry the antigen ?

Question #4 Only about 70% of all white North Americans can taste the chemical phenylthiocabamide. The ability to taste is determined by the dominant allele T. The inability to taste is determined by the recessive allele t. Assume that the population is in Hardy-Weinberg equilibrium. What are the genotypic and allelic frequencies in this population ?