Human Heredity: Principles and Issues (MindTap Course List)
11th Edition
ISBN: 9781305251052
Author: Michael Cummings
Publisher: Cengage Learning
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Chapter 19, Problem 11QP
Using the Hardy–Weinberg Law in Human Genetics
In a given population, the frequencies of the four
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A total of 1000 members of a Central American population are typed for the ABO blood group. In the sample, 421 have blood type A, 168 have blood type B, 336 have blood type
O, and 75 have blood type AB.
Part A
Use this information to determine the frequency of ABO blood group alleles in the sample.
Recall that when considering genes with three alleles whose frequencies are represented by the variables p, q, and r, the sum of genotype frequencies resulting from trinomial
expansion is:
(p+q+r)² =p² + 2pq+q2+2pr+r²+2gr = 1
in the Hardy-Weinberg equation, what do the terms p2, q2, and 2pq represent, in terms of the genetic structure of a population?
What is the expected genotype frequency of the heterozygous genotype under the Hardy-Weinberg equation P = 0.7?
Chapter 19 Solutions
Human Heredity: Principles and Issues (MindTap Course List)
Ch. 19.8 - Why dont genetic markers on the Y chromosome...Ch. 19.8 - Prob. 2GRCh. 19 - If you suspected that heterozygous carriers of a...Ch. 19 - If allele frequencies in the hemoglobin gene are...Ch. 19 - Prob. 1QPCh. 19 - How Can We Measure Allele Frequencies in...Ch. 19 - How Can We Measure Allele Frequencies in...Ch. 19 - Prob. 4QPCh. 19 - Prob. 5QPCh. 19 - How Can We Measure Allele Frequencies in...
Ch. 19 - How Can We Measure Allele Frequencies in...Ch. 19 - How Can We Measure Allele Frequencies in...Ch. 19 - Using the HardyWeinberg Law in Human Genetics...Ch. 19 - Prob. 10QPCh. 19 - Using the HardyWeinberg Law in Human Genetics In a...Ch. 19 - Prob. 12QPCh. 19 - Measuring Genetic Diversity in Human Populations...Ch. 19 - Measuring Genetic Diversity in Human Populations...Ch. 19 - Prob. 15QPCh. 19 - Measuring Genetic Diversity in Human Populations...Ch. 19 - Prob. 17QPCh. 19 - Prob. 18QPCh. 19 - Measuring Genetic Diversity in Human Populations...Ch. 19 - Natural Selection Affects the Frequency of Genetic...Ch. 19 - Prob. 21QPCh. 19 - Prob. 22QPCh. 19 - The Evolutionary History and Spread of Our Species...Ch. 19 - Prob. 24QPCh. 19 - Genomics and Human Evolution The Denisovan genome...
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- Consider the MN blood system in humans. This blood system consists of only two alleles, M and N, and they are codominant to each other. Three genotypes are possible, and these genotypes can be easily identified through simple blood typing procedures, including the heterozygous MN due to the codominant mode of inheritance. Consider the following data: Phenotype Genotype Number of individuals M MM 89 MN MN 162 N NN 79 1. How many alleles exist in this population? 2. What is the genotypic frequency of the M blood type in the population? 3. What is the allelic frequency of M? 4. What is the allelic frequency of N? 5. What is the genotypic frequency of MN? Write your answers directly. You do not need to show your solution.arrow_forwardIn a population of 200 people, an allele F has a frequency of 84%. What is the frequency of allele f? Using the Hardy-Weinberg equation, estimate the numbers of homozygous dominant, heterozygous, and homozygous recessive genotypes. (Remember that the formula is: p2 + 2pq + q2 = 1, where p represents the dominant allele and q represents the recessive allele.) *Be sure to account for all 200 people in the population.arrow_forwardSuppose 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_forward
- A worldwide survey of genetic variation in human populations reported the autosomal codominant MN blood group types in a sample of 1029 Chinese individuals from Hong Kong. The sample contained 342 people with blood type M, 500 with blood type MN, and 187 with blood type N. Part A The number of people expected in each blood-type category is: Enter the values expected for MM, MN and NN blood-type categories, respectively, separated by commas and using two decimal places (example 1.02, 2.30, 3.45).arrow_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_forwardWhat is the frequency of the heterozygous genotype if the homozygous genotype frequencies are 0.1385 (BB) and 0.5897 (bb)?arrow_forward
- Nieman-Pick Syndrome involves a defective enzyme, sphyngomylinase. It is usually fatal before the age of 3. The defective allele frequency is 0.01 in Ashkenazi populations. Let’s call the healthy allele A, and the lethal allele a. a) What is the frequency of allele A? Assuming Hardy-Weinberg equilibrium, how many people do you expect to have the three genotypes in a population of 10,000? b) AA:_______ c) Aa:________ d) aa:_________arrow_forwardAn autosomal locus has alleles A and a. We are given the frequency of individuals with the autosomal recessive phenotype. Which of the following statements is TRUE? Choose all that are true. Note: HWE = Hardy-Weinberg equilibrium a) If we assume HWE, we can calculate both allele frequencies b) If we assume HWE, we can calculate the genotype frequencies that we weren't given c) We can calculate both allele frequencies even if we don't assume HWE d) We can calculate q = Freq(a) even if we don't assume HWE e) Even if we don't assume HWE, we can calculate the genotype frequencies that we weren't givenarrow_forwardOrange coat color in cats is due to an X-linked allele (XO) that iscodominant with the allele for black (X+). When genotypes at the orangelocus were determined for a sample of cats in Minneapolis and St. Paul,Minnesota, the following data were obtained:XOXO females 11XOX+ females 70X+X+ females 94XOY males 36X+Y males 112Calculate the frequencies of the XO and X+ alleles for this population.arrow_forward
- While studying the frequency of sickle-cell disease ("sickle cell anemia") in a population living in sub-Saharan Africa, you obtain the following data from a sample of n= 100 people (note that I chose a simpler system for identifying the alleles rather than using "Hb S" for sickle cell allele), which is the actual name and what you used in lab). What is the frequency of the sickle cell allele (b) in the sample below from a human population? Sample Data BB-60 individuals (No sickle cell disease) Bb-30 individuals (No sickle cell disease) bb-10 individuals (Sickle cell disease) 1. 0.25 2. 0.10 3. 0.35 4. 0.60 5. 0.20arrow_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 total of 6129 North American Caucasians were blood typed for the MN locus, which is determined by two codominant alleles, LM and LN. The following data were obtained: Blood type Number M 1787 MN 3039 N 1303 Carry out a chi-square test to determine whether this population is in Hardy–Weinberg equilibrium at the MN locus.arrow_forward
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