Presentation on theme: "Tutorial #4 by Ma’ayan Fishelson"— Presentation transcript:
1 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 #4by Ma’ayan Fishelson
2 Hardy-Weinberg equilibrium A||aOther Possibilities: (a||A), (A||A),(a||a)What is the probability that a person chosen at random fromthe population would have each one of these genotypes ?Hardy-Weinberg equilibrium: If the allele frequencies areP(a)=p and P(A)=q, then, under some assumptions:P(Aa)= 2pq, P(AA)= p2, P(aa)= q2.Originally by Dan GeigerCorresponds to the random mating of two gametes, an egg & a sperm.
3 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.
4 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:FrequencyNature of offspringMating Typeu2A/AA/A x A/A2uv½A/A + ½A/aA/A x A/a2uwA/aA/A x a/av2¼A/A + ½A/a + ¼a/aA/a x A/a2vw½A/a + ½a/aA/a x a/aw2a/aa/a x a/aP(A/A) = u2 + uv + ¼v2 = (u + ½v)2P(A/a) = uv + 2uw + ½v2 + vw = 2(u + ½v)(½v + w)¼v2 + vw + w2 = (½v + w)2
5 Hardy-Weinberg equilibrium If we define the frequencies of the alleles as:p = P(A) = u + v/2q = P(a) = v/2 + wthen, the genotype frequencies are:P(A/A) = p2P(A/a) = 2pqP(a/a) = q2Second generation respects the same distribution:P(AA)= (p2 + ½2 pq)2 = [p(p+q)]2 = p2P(Aa)= 2(p2 + ½2pq) (½2pq +q2) =2p(p+q)q(p+q)= 2pqP(aa) = (½2pq + q2)2 = [q(p + q)]2 = q2
6 There is no association between alleles at any 2 loci. Linkage equilibriumA a B ba ab bA aThere 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 femaleand male recombination fractions, respectively.
7 Question #1You 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”.
8 Question #2A 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 = 0.45B = 0.55The population is exposed to a high temp. whichproves to be lethal for all individuals homozygousfor 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 ?
9 Hardy-Weinberg Equilibrium for X-linked Loci 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/A2A1/A2A1/A1p222p1p2p12
10 Question #3The 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 ?
11 Question #4Only 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 ?
12 Question #5 – Linkage Equilibrium The degree of Linkage Disequilibrium D can be measured using the following equation:D = PABPab – PaBPAbWhat value of D do we expect to see when the population is in Linkage Equilibrium ?
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