Study Guide for Campbell Biology
11th Edition
ISBN: 9780134443775
Author: Lisa A. Urry, Michael L. Cain, Steven A. Wasserman, Peter V. Minorsky, Jane B. Reece, Martha R. Taylor, Michael A. Pollock
Publisher: PEARSON
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Textbook Question
Chapter 23, Problem 1SYK
- a. What is Hardy-Weinberg equilibrium?
- b. Define the variables of the equation for Hardy-Weinberg equilibrium. Make sure you can use this equation to determine allele frequencies and predict genotype frequencies.
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What are some general tips in calculating allele and genotype frequencies?
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What 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?
The frequency of the A allele is represented by p, while the a allele is represented by q. At hardy-weinburg equilibrium, what is the frequency of the Aa genotype?
Chapter 23 Solutions
Study Guide for Campbell Biology
Ch. 23 - a. What is a major source of genetic variation for...Ch. 23 - In a population of 200 mice, 98 are homozygous...Ch. 23 - Use the allele frequencies you determined in...Ch. 23 - Practice using the Hardy-Weinberg equation so that...Ch. 23 - Prob. 5IQCh. 23 - Why hasnt the highly deleterious sickle-cell...Ch. 23 - a. What is Hardy-Weinberg equilibrium? b. Define...Ch. 23 - Prob. 2SYKCh. 23 - Prob. 1TYKCh. 23 - Prob. 2TYK
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- Describe in your own words what variance measures. Then explain why the variance of both samples must be used in the calculation of a t-test (In other words, explain why we have to factor in the variance when comparing the means).arrow_forwardWhich of the following statements does NOT apply to the Hardy-Weinberg expression: p2 + 2pq + q2? Group of answer choices p2 is the frequency of individuals with the homozygous recessive genotype. 2pq is the frequency of individuals with the heterozygous genotype. It can be used to determine the genotype and allele frequencies of the previous and the next generations. Knowing either p2 or q2, you can calculate all the other frequenciesarrow_forwardA population of 40 asters segregating at two loci with two alleles each (A/T and G/C) has 8 individuals with the AAGGgenotype. The allele frequencies of A and G are 0.5 and 0.5 in the population. What is the coefficient of gametic disequilibrium, D? What is the value of Lewontin’s D’? (that is “D prime”)arrow_forward
- How Can We Measure Allele Frequencies in Populations? What are four assumptions of the HardyWeinberg law?arrow_forwardUsing the HardyWeinberg Law in Human Genetics Suppose you are monitoring the allelic and genotypic frequencies of the MN blood group locus (see Question 2 for a description of the MN blood group) in a small human population. You find that for 1-year-old children, the genotypic frequencies are MM = 0.25, MN = 0.5, and NN = 0.25, whereas the genotypic frequencies for adults are MM = 0.3, MN = 0.4, and NN = 0.3. a. Compute the M and N allele frequencies for 1-year-olds and adults. b. Are the allele frequencies in equilibrium in this population? c. Are the genotypic frequencies in equilibrium?arrow_forwardConsider 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_forward
- 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…arrow_forwardUsing the two Hardy-Weinberg equations, calculate the allelic and genotypic frequencies. Tongue rollers [R_-]= 840, Non rollers [rr] = 160, Total = 1000 1. What are the phenotypic frequencies for tongue rollers? 0.84 2. What are the phenotypic frequencies for non-tongue rollers? 0.16 3. What is the allelic frequency for r? 0.4 4. What is the allelic frequency for R? 0.6 5. What is the homozygous dominant frequency? 6. What is the heterozygous frequency? 7. What is the homozygous recessive frequency?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
- . Full color (D) is dominant over dilute color (d). Of 325 cats observed, 194 have full color and 131 have dilute color. a. If the cats are in Hardy-Weinberg equilibrium, what is the frequency of the dilute allele? b. How many of the 194 cats with full color are likely heterozygous?arrow_forward(1 point) Humans with the genotypes DD and Dd show the Rh+ blood phenotype, whereas those with the genotype dd show the Rh- blood phenotype. In a sample of 400 Basques from Spain, 230 people were Rh+ and 170 people were Rh-. Assuming that this population is in Hardy-Weinberg proportions, what is the allele frequency of the allele D? (a) (a) 0.348 (answer) (b) (b) 0.652 (c) (c) 0.425 (d) (d) 0.575 (e) (e) 0.288 2. (2 points) In the Basque population mentioned above, what proportion of the Rh+ individuals would be expected to be heterozygote? (a) (a) 0.454 (b) (b) 0.789 (answer) (c) (c) 0.516 (d) (d) 0.250 (e) (e) 0.500 How is the answer for #2, b? please explainarrow_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 asa 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 allele frequencies (p and q )arrow_forward
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