7 Multiple AllelesAll chicken have combs on their heads, but it does not always look the same. The comb is a fleshy growth on the top of the chicken's head. Both male and female chickens have combs, but the ones on the male are larger. Combs of different breeds may look different in shape and even in color.Chicken comb features:rrpp = singlerrP_ = peaR_pp = roseR_P_ = walnut1. What must be the genotypes of the two parents for the outcome to always be a walnut offspring? (neither can be walnut to begin with) ____________ x _____________ = walnut (R_P_)2. Show a Punnett square for the following cross and describe the phenotypic ratios. RrPp x RrPp3. Show a punnett square for the following cross and describe the phenotypic ratios. rrpp x RrPp4. Show a punnett square for the following cross and describe the phenotypic ratios. rrpp x rrPp5. A rose crossed with a pea produces six walnut and five rose offspring. What must be the genotypes of the parents? Show the cross.
8 Hardy-Weinberg Principle Natural selection acts on individuals, but only populations evolveExplain this statement
9 Hardy-Weinberg Principle Describes a non-evolving populationThe gene pool does not changeGene pool - all alleles for all loci in the populationAllele frequencies remain constant in non-evolving populationsWhy do this?Allelic variation within a population can be modeled by the Hardy-Weinberg equations.
10 A Population in Hardy-Weinberg Equilibrium http://zoology. okstate The seven assumptions underlying Hardy–Weinberg equilibrium are as follows:organisms are diploidonly sexual reproduction occursgenerations are non overlappingmating is randompopulation size is infinitely largeallele frequencies are equal in the sexesthere is no migration, mutation or selectionFig. 23-6
11 Interpretation:A population with 80% dominant and 20% recessive allelesthat meets the conditions for Hardy-Weinberg equilibriumwill pass 80% dominant and 20% recessive alleles to the next generationAlleles in the populationFrequencies of allelesGametes producedp = frequency ofEach egg:Each sperm:CR allele = 0.8q = frequency of80%chance20%chance80%chance20%chanceCW allele = 0.2
12 Hardy-Weinberg equations: p + q = 1p2 + 2pq + q2 = 1Describes alleles in a gene poolThis is the equation for a trait with 2 allelesCan be used to predict genotypesp – represents dominant alleleq – represents the recessive alleleWorksheet 3, 4 and 5
14 Gametes of this generation: 64% CRCR, 32% CRCW, and 4% CWCWGametes of this generation:64% CR + 16% CR = 80% CR = 0.8 = p4% CW + 16% CW = 20% CW = 0.2 = qGenotypes in the next generation:Figure 23.7 The Hardy-Weinberg principle64% CRCR, 32% CRCW, and 4% CWCW plants
15 Reality Hardy-Weinberg - hypothetical population In real populations, allele and genotype frequencies change over time
16 You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following: A. What is the frequency of the "aa" genotype in the population? B. What is the allele frequency of the "a" allele? C. What is the frequency of the "A" allele in the population? D. What is the frequencies of the genotype "Aa” in the population? E. What is the frequencies of the two possible phenotypes if "A" is completely dominant over "a."The winged trait isdominant.