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In a population of rabbits,
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Genetic Analysis: An Integrated Approach (2nd Edition)
- If the chi-square value is 0 under the degree of freedom 1. What could be the interpretation and does it fit in the population of Hardy-Weinberg equilibrium?arrow_forwardIn the F2 generation, 306 rabbits with red eyes and 71 with a white eye phenotype suppose the calculated x2 value is 0.35. Find the x2 range using the distribution chart. What is the p-value range? using these information do you accept or reject the null hypothesis? The distribution chart is attached below.arrow_forwardPretend that you are comparing the actual genotype distribution for a population with the distribution of genotypes predicted by the Hardy-Weinberg theorem. So your hypothesis is that the population is in Hardy-Weinberg equilibrium (i.e. that actual population data fit the Hardy-Weinberg expectations). If you carry out a chisquare goodness of fit test and calculate a total chisquare value of 0.03 with 1 degree of freedom (see table), what does this mean? (select all true statements)a) The data do NOT fit the hypothesized distribution.b) The data do fit the hypothesized distribution well enough, so we accept the hypothesis at this time (i.e. we cannot reject the hypothesis). c) The probability that the data came from a population in Hardy-Weinberg equilibrium is too small, so we reject the hypothesis.d) The probability that the data came from a population in Hardy-Weinberg equilibrium is too big, so we reject the hypothesis.e) The data support Hardy-Weinberg expectations – there is no…arrow_forward
- Suppose there is an autosomal locus of 2 alleles, A1 and A2, with probabilities (frequencies) p1 and p2, and the genotype probabilities (frequencies) are P(A1A1) = p1*p1, P(A1A2) = 2*p1*p2, and P(A2A2) = p2*p2, respectively. Prove the Hardy-Weinberg Law, i.e., after one generation of random mating, the genotype probabilities (frequencies) in the offspring are also P(A1A1) = p1*p1, P(A1A2) = 2*p1*p2, and P(A2A2) = p2*p2. Hint: List all possible combinations of random mating. Then list the probabilities of the resulting genotype probabilities (frequencies) in the offspring. Combine the probabilities of random mating and resulting genotype probabilities (frequencies) in the offspring.arrow_forwardIn a population of locusts, the mean wing length is 47 mm, the selection gradient on wing length is β = 0.12/mm, the phenotypic variance for wing length is P = 3.6 mm2, and the heritability of wing length is h2 = 0.27. In addition, we know that the additive genetic covariance between wing and leg length is 0.60 mm2. What is the expected evolutionary change in mean leg length due to selection on both wings and legs? Repeat these calculations to predict what will happen to wing length as a result of the selection on both wings and legs. What do you predict the average wing and leg lengths will be in the next generation?arrow_forwardPretend that you are comparing the actual genotype distribution for a population with the distribution of genotypes predicted by the Hardy-Weinberg theorem. So you hypothesize that the population is in Hardy-Weinberg equilibrium (i.e. that actual population data fit the Hardy-Weinberg expectations). If you carry out chi-square goodness of fit test and calculate a total chi-square value of 0.03 with 1 degree of freedom (see table), what does this mean?arrow_forward
- Assume that the frequency of gene B in a hypothetical population Is 0.63, that there are only two alleles (B and b) of the gee in the population, that allele B is dominant over allele b, that neither allele has a selective advantage over the other, and that the population is at equilibrium with regard to this particular gene. And how many individuals in this population are expected to be of genotype BB according to the Hardy-Weinberg formula? (Assume that the total population size is 150) 71 52 118 60 131arrow_forwardAn autosomal locus has alleles A and a. The allele frequencies in a population at Hardy Weinberg equilibrium are p = Freq(A) = 0.5 q = Freq(a) = 0.5 What is the frequency of homozygous wild-type (AA) in this population? Enter a single number between 0 and 1, for example, 0.33arrow_forwardThis is a classic data set on wing coloration in the scarlet tiger moth (Panaxia dominula). Coloration in this species had been previously shown to behave as a single-locus, two-allele system with incomplete dominance. Data for 1612 individuals are given below: White-spotted (AA) =1469 Intermediate (Aa) = 138 Little spotting (aa) =5 Calculate the following frequencies: A=, a=, AA=, Aa=, aa=arrow_forward
- Consider a Hardy-Weinburg Equilibrium population with an autosomal locus of 2 alleles, A1 and A2. If P(A1A2) = 8 * P(A1A1), what are the allele frequencies at the locus?arrow_forwardWhat is the expected frequency of Gg while in Hardy-Weinberg Equilibrium if you have an allele frequency of G = 0.1 and g = 0.9 in a two allele system?arrow_forwardplease tell us the expected average time to fixation of an allele at frequency p = 0.5 in a population of 100 individuals, we can safely assume that the allele does become fixed?arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning