Evolution
Evolution
4th Edition
ISBN: 9781605356051
Author: Douglas Futuyma, Mark Kirkpatrick
Publisher: SINAUER
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Chapter 4, Problem 4PDT

a.

Summary Introduction

To describe: What is surprising about the observed number of two-locus genotypes.

Introduction: In two-locus genotype, four different types of alleles are included to form differentgenotypes. In this problem, there are two alleles A1 and A2, which are located at one locus and B1 and B2 are located on the second locus.

b.

Summary Introduction

To explain: Whether each locus is at Hardy-Weinberg equilibrium or not.

Introduction: If the value of the Hardy-Weinberg equation is equal to one, it means that the locus is atHardy-Weinberg equilibrium. If the value exceeds 1, it shows that there is an excess of homozygous genotypes. On the other hand, if the value of the Hardy-Weinberg equation is less than 1, it says that there is an excess of heterozygous genotypes.

c.

Summary Introduction

To explain: Whether the two loci are close to one another, or far apart in the genome.

Introduction: Loci are just sites on the chromosome from which some portion of gene interchange with the segment of another chromosome during crossing over and recombination.

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What is the expected genotype frequency of the heterozygous genotype under the Hardy-Weinberg equation P = 0.7?
Pretend 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…
Through PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 90 individuals and H5/H5=85 individuals. Considering the Hardy Weinberg equilibrium and calculating the Chi-square statistic, do you reject or fail to reject the null-hypothesis?  1. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I reject the null hypothesis. 2. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis.  3. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I fail to reject the null hypothesis. 4. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the…
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What is Evolution?; Author: Stated Clearly;https://www.youtube.com/watch?v=GhHOjC4oxh8;License: Standard Youtube License