Presentation is loading. Please wait.

Presentation is loading. Please wait.

D4: The Hardy-Weinberg Principle 2 hours. D.4.1 Explain how the Hardy–Weinberg equation is derived. TT = p 2 = freq of homoz domTT = p 2 = freq of homoz.

Similar presentations


Presentation on theme: "D4: The Hardy-Weinberg Principle 2 hours. D.4.1 Explain how the Hardy–Weinberg equation is derived. TT = p 2 = freq of homoz domTT = p 2 = freq of homoz."— Presentation transcript:

1 D4: The Hardy-Weinberg Principle 2 hours

2 D.4.1 Explain how the Hardy–Weinberg equation is derived. TT = p 2 = freq of homoz domTT = p 2 = freq of homoz dom Tt = pq = freq of heterozygoteTt = pq = freq of heterozygote tt = q 2 = freq of homoz rectt = q 2 = freq of homoz rec Punnett Square... Tt x TtPunnett Square... Tt x Tt p 2 + 2pq + q 2 = 100% (1)p 2 + 2pq + q 2 = 100% (1) p = freq of dominant allelep = freq of dominant allele q = freq of recessive alleleq = freq of recessive allele

3 D.4.2 Calculate allele, genotype and phenotype frequencies for two alleles of a gene, using the Hardy–Weinberg equation. Examples from class!Examples from class!

4 D.4.3 State the assumptions made when the Hardy–Weinberg equation is used. It must be assumed that: --a population is large --with random mating --constant allele frequency over time This implies --no allele-specific mortality --no mutation --no emigration --no immigration

5 POPULATION GENETICS Predicting inheritance in a population © 2008 Paul Billiet ODWSODWS

6 Predictable patterns of inheritance in a population so long as… the population is large enough not to show the effects of a random loss of genes by chance events i.e. there is no genetic drift the mutation rate at the locus of the gene being studied is not significantly high mating between individuals is random (a panmictic population) new individuals are not gained by immigration or lost be emigration i.e. there is no gene flow between neighbouring populations the genes allele has no selective advantage or disadvantage © 2008 Paul Billiet ODWSODWS

7 SUMMARY Genetic drift Mutation Mating choice Migration Natural selection All can affect the transmission of genes from generation to generation Genetic Equilibrium If none of these factors is operating then the relative proportions of the alleles (the GENE FREQUENCIES) will be constant © 2008 Paul Billiet ODWSODWS

8 THE HARDY WEINBERG PRINCIPLE Step 1 Calculating the gene frequencies from the genotype frequencies Easily done for codominant alleles (each genotype has a different phenotype) © 2008 Paul Billiet ODWSODWS

9 Iceland Population (2007 est) Area km 2 Distance from mainland Europe 970 km © 2008 Paul Billiet ODWSODWS Google Earth

10 Example Icelandic population: The AB blood group 2 M n alleles per person 1 M n allele per person 1 M m allele per person 2 M m alleles per person Contribution to gene pool Numbers747 BBABAAGenotypes Type BType ABType APhenotypesSample Population © 2008 Paul Billiet ODWSODWS

11 AB blood group in Iceland Total A alleles = (2 x 233) + (1 x 385) = 851 Total B alleles = (2 x 129) + (1 x 385) = 643 Total of both alleles =1494 = 2 x 747 (humans are diploid organisms) p=Frequency of the A allele = 851/1494 = 0.57 or 57% q=Frequency of the B allele = 643/1494 = 0.43 or 43% © 2008 Paul Billiet ODWSODWS

12 In general for a diallellic gene A and a If the frequency of the A allele = p and the frequency of the a allele = q Then p+q = 1 © 2008 Paul Billiet ODWSODWS

13 Step 2 Using the calculated gene frequency to predict the EXPECTED genotypic frequencies in the NEXT generation OR to verify that the PRESENT population is in genetic equilibrium © 2008 Paul Billiet ODWSODWS

14 BB 0.18AB 0.25 AA 0.32 B 0.43 A 0.57 B 0.43A 0.57 Assuming all the individuals mate randomly SPERMS EGGS NOTE the ALLELE frequencies are the gamete frequencies too © 2008 Paul Billiet ODWSODWS p*p= p 2 p*q q*q= q 2

15 Close enough for us to assume genetic equilibrium GenotypesExpected frequencies Observed frequencies AA p 2 = = 0.31 AB 2pq = = 0.52 BB q 2 = = 0.17 © 2008 Paul Billiet ODWSODWS

16 SPERMS A pa q EGGS A pAA p 2 Aa pq a qAa pqaa q 2 In general for a diallellic gene A and a Where the gene frequencies are p and q Then p + q = 1 and © 2008 Paul Billiet ODWSODWS

17 THE HARDY WEINBERG EQUATION So the genotype frequencies are: AA=p 2 Aa=2pq aa=q 2 or p 2 + 2pq + q 2 = 1 © 2008 Paul Billiet ODWSODWS

18 DEMONSTRATING GENETIC EQUILIBRIUM Using the Hardy Weinberg Equation to determine the genotype frequencies from the gene frequencies may seem a circular argument © 2008 Paul Billiet ODWSODWS

19 Only one of the populations below is in genetic equilibrium. Which one? Population sampleGenotypesAllele frequencies AAAaaaAa © 2008 Paul Billiet ODWSODWS

20 Only one of the populations below is in genetic equilibrium. Which one? aAaaAaAA Gene frequenciesGenotypesPopulation sample © 2008 Paul Billiet ODWSODWS

21 Only one of the populations below is in genetic equilibrium. Which one? Population sampleGenotypesGene frequencies AAAaaaAa © 2008 Paul Billiet ODWSODWS

22 SICKLE CELL ANAEMIA IN WEST AFRICA, A BALANCED POLYMORPHISM haemoglobin gene Normal allele Hb N Sickle allele Hb S PhenotypesNormalSickle Cell Trait Sickle Cell Anaemia Alleles GenotypesHb N Hb N Hb S Hb S Hb N Hb S Observed frequencies Expected frequencies © 2008 Paul Billiet ODWSODWS

23 SICKLE CELL ANAEMIA IN WEST AFRICA, A BALANCED POLYMORPHISM haemoglobin gene Normal allele Hb N Sickle allele Hb S Expected frequencies Observed frequencies Hb S Hb N Hb S Hb N Hb S Hb N Genotypes AllelesSickle Cell Anaemia Sickle Cell Trait NormalPhenotypes © 2008 Paul Billiet ODWSODWS

24 SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION PhenotypesNormalSickle Cell Trait Sickle Cell Anaemia Alleles GenotypesHb N Hb N Hb S Hb S Hb N Hb S Observed frequencies Expected frequencies © 2008 Paul Billiet ODWSODWS

25 SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION Expected frequencies Observed frequencies Hb S Hb N Hb S Hb N Hb S Hb N Genotypes AllelesSickle Cell Anaemia Sickle Cell Trait NormalPhenotypes © 2008 Paul Billiet ODWSODWS

26 RECESSIVE ALLELES EXAMPLE ALBINISM IN THE BRITISH POPULATION Frequency of the albino phenotype = 1 in or © 2008 Paul Billiet ODWSODWS

27 PhenotypesGenotypesHardy Weinberg frequencies Observed frequencies NormalAAp2p2 NormalAa2pq Albinoaaq2q2 A = Normal skin pigmentation allele Frequency = p a = Albino (no pigment) allele Frequency = q © 2008 Paul Billiet ODWSODWS

28 PhenotypesGenotypesHardy Weinberg frequencies Observed frequencies NormalAAp2p NormalAa2pq Albinoaaq2q A = Normal skin pigmentation allele Frequency = p a = Albino (no pigment) allele Frequency = q © 2008 Paul Billiet ODWSODWS Use p+q=1 to determine

29 Albinism gene frequencies Normal allele = A = p = ? Albino allele = q = ? © 2008 Paul Billiet ODWSODWS

30 Albinism gene frequencies Normal allele = A = p = ? Albino allele = q = ? If q 2 = …then q = … ( ) = or 0.7% © 2008 Paul Billiet ODWSODWS

31 HOW MANY PEOPLE IN BRITAIN ARE CARRIERS FOR THE ALBINO ALLELE (Aa)? a allele = = q A allele = p But p + q = 1 Therefore p= 1- q = 1 – = or 99.3% The frequency of heterozygotes (Aa) = 2pq = 2 x x = or 1.4% © 2008 Paul Billiet ODWSODWS

32 Heterozygotes for rare recessive alleles can be quite common Genetic inbreeding leads to rare recessive mutant alleles coming together more frequently Therefore outbreeding is better Outbreeding leads to hybrid vigour © 2008 Paul Billiet ODWSODWS

33 Example: Rhesus blood group in Europe What is the probability of a woman who knows she is rhesus negative (rhrh) marrying a man who may put her child at risk (rhesus incompatibility Rh – mother and a Rh + fetus)? © 2008 Paul Billiet ODWSODWS

34 Rhesus blood group R = Rh+ r = Rh- A rhesus positive fetus is possible if the father is rhesus positive RR x rr 100% chance Rr x rr 50% chance © 2008 Paul Billiet ODWSODWS

35 Rhesus blood group Rhesus positive allele is dominant R Frequency = p Rhesus negative allele is recessive r Frequency = q Frequency of r allele = 0.4 = q Ifp + q = 1 Therefore R allele = p = 1 – q = 1 – 0.4 = 0.6 © 2008 Paul Billiet ODWSODWS

36 Rhesus blood group Frequency of the rhesus positive phenotype = RR + Rr = p 2 + 2pq = (0.6) 2 + (2 x 0.6 x 0.4) = = 0.84 or 84% © 2008 Paul Billiet ODWSODWS

37 Rhesus blood group Therefore, a rhesus negative, European woman in Europe has an 84% chance of having husband who is rhesus positive… of which 36% will only produce rhesus positive children and 48% will produce rhesus positive child one birth in two PhenotypesGenotypesHardy Weinberg frequencies Observed frequencies Rhesus positiveRRp2p (0.84 total) 0.48 Rhesus positiveRr2pq Rhesus negativeRrq2q © 2008 Paul Billiet ODWSODWS


Download ppt "D4: The Hardy-Weinberg Principle 2 hours. D.4.1 Explain how the Hardy–Weinberg equation is derived. TT = p 2 = freq of homoz domTT = p 2 = freq of homoz."

Similar presentations


Ads by Google