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**D4: The Hardy-Weinberg Principle**

2 hours

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**D.4.1 Explain how the Hardy–Weinberg equation is derived.**

TT = p2 = freq of homoz dom Tt = pq = freq of heterozygote tt = q2 = freq of homoz rec Punnett Square... Tt x Tt p2 + 2pq + q2 = 100% (1) p = freq of dominant allele q = freq of recessive allele

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

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**--a population is large --with random mating **

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

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**Predicting inheritance in a population**

POPULATION GENETICS Predicting inheritance in a population © 2008 Paul Billiet ODWS

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**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 gene’s allele has no selective advantage or disadvantage © 2008 Paul Billiet ODWS

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**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 ODWS

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**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 ODWS

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**Iceland Population 313 337 (2007 est) Area 103 000 km2**

Distance from mainland Europe 970 km Google Earth © 2008 Paul Billiet ODWS

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**Example Icelandic population: The AB blood group**

129 385 233 Numbers 747 BB AB AA Genotypes Type B Type AB Type A Phenotypes Sample Population 2 Mn alleles per person 1 Mn allele per person 1 Mm allele per person 2 Mm alleles per person Contribution to gene pool © 2008 Paul Billiet ODWS

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**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 = or 57% q=Frequency of the B allele = 643/1494 = or 43% © 2008 Paul Billiet ODWS

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**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 ODWS

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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 ODWS

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**Assuming all the individuals mate randomly**

NOTE the ALLELE frequencies are the gamete frequencies too SPERMS B A B 0.43 A 0.57 EGGS AA AB p*p= p2 p*q AB BB q*q= q2 p*q © 2008 Paul Billiet ODWS

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**Close enough for us to assume genetic equilibrium**

Genotypes Expected frequencies Observed frequencies AA p2 = 0.32 233 747 = 0.31 AB 2pq =0.50 385 747 = 0.52 BB q2 =0.18 129 747 = 0.17 © 2008 Paul Billiet ODWS

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**In general for a diallellic gene A and a**

Where the gene frequencies are p and q Then p + q = 1 and SPERMS A p a q EGGS AA p2 Aa pq aa q2 © 2008 Paul Billiet ODWS

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**THE HARDY WEINBERG EQUATION**

So the genotype frequencies are: AA = p2 Aa = 2pq aa = q2 or p2 + 2pq + q2 = 1 © 2008 Paul Billiet ODWS

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**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 ODWS

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**Only one of the populations below is in genetic equilibrium. Which one?**

Population sample Genotypes Allele frequencies AA Aa aa A a 100 20 80 36 48 16 50 30 60 40 © 2008 Paul Billiet ODWS

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**Only one of the populations below is in genetic equilibrium. Which one?**

40 60 100 30 20 50 16 48 36 0.4 0.6 80 a A aa Aa AA Gene frequencies Genotypes Population sample 0.4 0.6 0.4 0.6 0.4 0.6 © 2008 Paul Billiet ODWS

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**Only one of the populations below is in genetic equilibrium. Which one?**

Population sample Genotypes Gene frequencies AA Aa aa A a 100 20 80 0.6 0.4 36 48 16 50 30 60 40 © 2008 Paul Billiet ODWS

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**SICKLE CELL ANAEMIA IN WEST AFRICA, A BALANCED POLYMORPHISM**

haemoglobin gene Normal allele HbN Sickle allele HbS Phenotypes Normal Sickle Cell Trait Sickle Cell Anaemia Alleles Genotypes HbNHbN HbN HbS HbS HbS HbN HbS Observed frequencies 0.56 0.4 0.04 Expected frequencies © 2008 Paul Billiet ODWS

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**SICKLE CELL ANAEMIA IN WEST AFRICA, A BALANCED POLYMORPHISM**

haemoglobin gene Normal allele HbN Sickle allele HbS Expected frequencies 0.04 0.4 0.56 Observed frequencies HbS HbN HbS HbS HbN HbS HbNHbN Genotypes Alleles Sickle Cell Anaemia Sickle Cell Trait Normal Phenotypes 0.24 0.76 0.06 0.36 0.58 © 2008 Paul Billiet ODWS

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**SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION**

Phenotypes Normal Sickle Cell Trait Sickle Cell Anaemia Alleles Genotypes HbNHbN HbN HbS HbS HbS HbN HbS Observed frequencies 0.9075 0.09 0.0025 Expected frequencies © 2008 Paul Billiet ODWS

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**SICKLE CELL ANAEMIA IN THE AMERICAN BLACK POPULATION**

Expected frequencies 0.0025 0.09 0.9075 Observed frequencies HbS HbN HbS HbS HbN HbS HbNHbN Genotypes Alleles Sickle Cell Anaemia Sickle Cell Trait Normal Phenotypes 0.09 0.91 0.0081 0.16 0.8281 © 2008 Paul Billiet ODWS

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**RECESSIVE ALLELES EXAMPLE ALBINISM IN THE BRITISH POPULATION**

Frequency of the albino phenotype = 1 in or © 2008 Paul Billiet ODWS

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**Hardy Weinberg frequencies**

A = Normal skin pigmentation allele Frequency = p a = Albino (no pigment) allele Frequency = q Phenotypes Genotypes Hardy Weinberg frequencies Observed frequencies Normal AA p2 Aa 2pq Albino aa q2 © 2008 Paul Billiet ODWS

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**Hardy Weinberg frequencies**

A = Normal skin pigmentation allele Frequency = p a = Albino (no pigment) allele Frequency = q Phenotypes Genotypes Hardy Weinberg frequencies Observed frequencies Normal AA p2 Aa 2pq Albino aa q2 Use p+q=1 to determine © 2008 Paul Billiet ODWS

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**Albinism gene frequencies**

Normal allele = A = p = ? Albino allele = q = ? © 2008 Paul Billiet ODWS

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**Albinism gene frequencies**

Normal allele = A = p = ? Albino allele = q = ? If q2 = …then q = … ( ) = or 0.7% © 2008 Paul Billiet ODWS

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**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 – 0.007 = or 99.3% The frequency of heterozygotes (Aa) = 2pq = 2 x x 0.007 = or 1.4% © 2008 Paul Billiet ODWS

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**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 ODWS

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**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 ODWS

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**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 ODWS

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**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 If p + q = 1 Therefore R allele = p = 1 – q = 1 – 0.4 = 0.6 © 2008 Paul Billiet ODWS

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Rhesus blood group Frequency of the rhesus positive phenotype = RR + Rr = p2 + 2pq = (0.6)2 + (2 x 0.6 x 0.4) = = 0.84 or 84% © 2008 Paul Billiet ODWS

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**Hardy Weinberg frequencies**

Rhesus blood group Phenotypes Genotypes Hardy Weinberg frequencies Observed frequencies Rhesus positive RR p2 0.36 (0.84 total) 0.48 Rr 2pq Rhesus negative q2 0.16 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 © 2008 Paul Billiet ODWS

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