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may5_no_background - A? B? C? Calculating allele frequency...

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Unformatted text preview: A? B? C? Calculating allele frequency Proportion of all copies of a gene in a population that are of a given allelic type Diploid population, each individual has two alleles E.g., population of 16 individuals have 32 copies of cystic fibrosis gene 8 RR 16 copies of R 2 Rr 2 copies of R 6 rr 0 copies of R 16 + 2 + 0 = 18 copies of R allele Proportion (frequency) of R alleles in population is 18/32 = 9/16 = 0.56 Try same exercise to calculate frequency of r allele in population Applying allele frequencies to Punnett Square illustrates HardyWeinberg equilibrium A (p) A (p) a (q) AA (p2) Aa (pq) a (q) Aa (pq) aa (q2) Refers to gene pool, not one set of parents. Allele frequencies do not change from generation to generation in a population at Hardy-Weinberg equilibrium. A Hardy-Weinberg population achieves the genotype frequencies of p2, 2pq, and q2 in just one generation and maintains them in subsequent generations. If we know (assume) a population is in equilibrium, we can determine the frequency of heterozygotes Cystic fibrosis is determined by a recessive allele and occurs in about 1/2000 whites of northern and central European origin. What is the heterozygote frequency in the population? q2 = 1/2000 q = .022 2pq = .043 p=1-q=.978 2pq p2 + 2pq 2/3 1/4 What is the chance the indicated couple's child will have cystic fibrosis? p(aa) = (2/3)(1/4)2pq/(p2 + 2pq) p(aa) = (1/6)(.043)/(.956+.043) = .0072/.999 =.0072 p2 pq pq q2 X-linked alleles are counted in males Presence of the enzyme glucose-6-phosphate dehydrosgenase (G-6-PD) is determined by a sex-linked dominant gene, its absence by the recessive allele. In one Hardy-Weinberg population at equilibrium 15% of the males are G-6-PD negative. What is the expected frequency of G-6-PD negative females in the population? Males: q=.15 p=.85 Females: q2=.0225 or 2.25% Review materials from text Two steps in translating the genotype frequencies from one generation to the next Step 1 Calculate allele frequency of gametes same as adults Fig. 21.2 Step 2 Use gamete allele frequency to calculate genotype frequencies in the zygotes of next generation Fig. 21.3 Albanism example Population of 100,000 people 100 aa albinos = 200 a alleles 1,800 Aa carriers = 1,800 A alleles and 1,800 a alleles 98,100 AA individuals = 196,200 A alleles A allele frequency is 198,000/200,000 = 0.99 p = 0.99 q = 0.01 a allele frequency is 2,000/200,000 = 0.01 This is also the allele frequencies of the gametes. Hardy-Weinberg equation for population p2 + 2pq + q2 = (0.99)2 + 2(0.99 x 0.01) + (0.01)2 = 0.9801 + 0.0198 + 0.0001 = 1 Next Generation of 100,000 people 100,000 X 0.9801 AA individuals 100,000 X 0.0198 Aa individuals 100,000 X 0.0001 aa individuals Frequency of p and q allele in next generation p+q=1 Thus the frequency of the p allele in the next generation is p2 + [2p(1-p)] = p2 + p(1-p) = p2 + p p2 q2 + [2q (1-q)] = q2 + q (1-q) = q2 + q q2 and the frequency of the q allele in the next generation is For albinism where p = 0.99 and q = 0.01 The frequency of A allele in second generation is 0.98 + 0.99 0.98 = 0.99 0.0001 + 0.01 0.0001 = 0.01 The frequency of the a allele is Genotype frequencies changed, but allele frequencies stay the same for both dominant and recessive alleles. The Hardy-Weinberg Law Phenotype frequency proportion of individuals in a population that are a particular phenotype E.g., population of 16, 6 of which have the recessive disease cystic fibrosis 6/16 = 3/8 diseased (rr genotypes) 10/16 = 5/8 normal (RR, Rr genotypes) Genotype frequency proportion of individuals in a population that are a particular genotype E.g., molecular analysis shows 8/16 individuals are type RR = , 2/16 are Rr = 1/8, and 6/16 are rr = 3/8. The Hardy-Weinberg law clarifies the relations between genotype and allele frequency within a generation and from one generation to the next. Five assumptions Infinitely large population Individuals mate at random. No new mutations appear in gene pool. No migration into or out of population No genotype-dependent differences in ability to survive and reproduce If all assumptions hold, population is in Hardy-Weinberg equilibrium. All natural populations violate one or more assumptions of Hardy-Weinberg law. However, equations derived based on assumptions are remarkably robust. Hardy-Weinberg law can be used as a null model. Calculating the frequency of heterozygous carriers when you only know the frequency of diseased individuals for a recessive trait PKU phenylketonuria Autosomal recessive mutation 1 in 3600 Caucasians in USA have PKU q2 = 1/3600 q = q2 = 1/3600 = 0.0167 p = 1-q = 0.9833 2pq = 2 X 0.0167 X 0.9833 = 0.0328 Frequency of carriers is thus about 3.3% II. Is a given population in HardyWeinberg Equilibrium? Measure allele frequencies using genes with codominant (or incomplete dominant) alleles and compare the genotype frequencies predicted by HWE to the measured frequencies (p2 + 2pq + q2) = 1 4/100 = (.2)2 64/100 = (.8)2 32/100 = 2(.2)(.8) M/N Blood types are frequently used to determine if populations are in equilibrium A population shows 744 type M, 1149 type MN and 429 type N people. Allele frequencies? Predicted genotype frequencies? p(M) = 2(744) +1149 = .57 2(2322) q(N) = 1-p = 1-.57 = .43 p(M) = .57, q(N) =.43 Genotype frequencies assuming HWE gives us our expected frequencies f(MM) = p2 = .32 # = .325(2322) = 754 f(MN) = 2pq= 2(.57)(.43) = .49 # = .49(2322) = 1138 f(NN) = q2 = .185 # = .185(2322)= 430 Null hypothesis: observed and expected values are not significantly different from one another Genotypes: Observed: Expected: MM 744 754 MN 1149 1138 NN 429 430 Degrees of freedom? Must be 1 because there are only two classes: p and q, df = 2-1 X2 = (744-754)2 754 + (1149- 1138)2 + (429- 430)2 1138 430 X2 = .13 + .11 + .0023 = .2423 p >.5 Fail to reject hypothesis: Therefore, population is in equilibrium III. What determines allele frequencies? Measuring how mutation and selection cause changes in allele frequency Evolution sometimes defined as change in allele frequency over multiple generations Microevolution changes that occur from generation to generation within a species Macroevolution changes that occur through geologic time among species Can use Hardy-Weinberg law to examine microevolution Violations of assumptions to Hardy-Weinberg can be used to analyze evolutionary forces causing deviations in allele frequencies. 1. Evolutionary equilibrium: A balance between mutation to a new allele and selection against the allele Mutations are ultimate source of new variation Rate in mammals is 10-4 10-6 per gene per generation ~50,000 genes in human genome or 0.05 5 new mutations in each human gamete 2. Natural selection acts on differences in fitness to alter allele frequencies. Fitness individual's relative ability to survive and transmit genes to next generation Viability and reproductive success Natural selection individuals with higher fitness survive and reproduce more than individuals with lower fitness How a recessive genetic condition influences allele frequency of a population rr genotype has decreased fitness Fitness or RR and Rr same WRR = 1, WRr = 1, Wrr = 1-s s = selection coefficient against rr which varies from 0 (no affect) to 1 (lethal) Fig. 21.7 Calculating the relative values of selection coefficients for sickle cell Sickle cell heterozygotes resistant to malaria, homozygous recessives high probability of early death qe lies is about 0.17 0.17 = s1/(s1 + s2) S1 = 0.2S2 If S2 = 0 (those with sickle cell never reproduce), S1 = 0.2. Relative fitness of wild-type genotype is 0.8. Fitness of heterozygote is 1.0. 1.0/0.8 = 1.25 which is the relative fitness advantage of heterozygotes for sickle cell allele over noncarriers. Many common disorders (diabetes, heart disease, etc) are not selected against! Time of onset of disease can also influence frequency. If onset is in middle or late age, after reproduction No affect on fitness Sustain little or no negative selection 3. Genetic drift has unpredictable effects on evolutionary equilibrium. Chance fluctuations in allele frequency that have neutral effect on fitness Smaller populations are more affected than larger populations. Sampling biases are more pronounced in small populations, just as coin flips are more biased towards more heads or tails with fewer tosses. Out of Africa migration The effects of genetic drift Fig. 21.8 Summary of evolutionary equilibrium between mutation and selection New alleles arise in populations by mutation. When allele has affect on fitness, selection will drive frequency towards an equilibrium with wild-type allele. Equilibrium value is determined by relative selection coefficients for heterozygous and homozygous individuals for new allele. If new allele has no effect on fitness, genetic drift will determine its frequency. Opener IV. What happens when you have more ABO blood types than two alleles? let p = the frequency of the A allele let q = the frequency of the B allele, and let r = the frequency of the i allele. such that p + q + r = 1 Then, p2 + 2pq + q2 + 2pr + r2 +2qr = 1 The frequencies of the i, A and B alleles for the US white population are as follows: Type Frequency i A B 0.67 .26 .07 What is the genotype frequency of individuals with type B? f(BB) + f(Bi) = (.07)2 + 2(.67)(.07) = .0049 + .0938 = .0987 ...
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This note was uploaded on 07/08/2009 for the course LIFESCI ls 4 taught by Professor Merriam during the Winter '08 term at UCLA.

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