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
What is the expected genotype frequency of the heterozygous genotype under the Hardy-Weinberg equation P = 0.7?
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.
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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- 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_forwardA 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_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_forward
- Orange 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_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_forwardWhat is the mathematical expression for the genetic equilibrium for genes with two alleles? Is this statistical distribution the same as the statistical distribution of the respective phenotypes?arrow_forward
- One particularly useful feature of the Hardy-Weinberg equation is that it allows us to estimate the frequency of heterozygotes for recessive genetic diseases, assuming that Hardy-Weinberg equilibrium exists. As an example, let’s consider cystic fibrosis, which is a human genetic disease involving a gene that encodes a chloride transporter. Persons with this disorder have an irregularity in salt and water balance. One of the symptoms is thick mucus in the lungs that can contribute to repeated lung infections. In populations of Northern European descent, the frequency of affected individuals is approximately 1 in 2500. Because this is a recessive disorder, affected individuals are homozygotes. Assuming that the population is in Hardy-Weinberg equilibrium, what is the frequency of individuals who are heterozygous carriers?arrow_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_forward8% of XY individuals are color blind in a population. Assume Hardy-Weinberg conditions. Submit your answer as it is. a) What is the percentage of color-blind XX individuals? b) What is the percentage of XX individuals who are carriers? c) If this population has 1000 individuals with 50% of male and 50% of female, how many carriers are present in this population? Submit your answer as it is. Do not round up.arrow_forward
- "he equation p + 2pq + q° = 1 representing the lardy-Weinberg proportions examines genes with nly two alleles in a population. . Derive a similar equation describing the equilib- rium proportions of genotypes for a gene with three alleles. [Hint: Remember that the Hardy- Weinberg equation can be written as the binomial expansion (p + q)*.] A single gene with three alleles (r^, 1", and i) is responsible for the ABO blood groups. Individuals with blood type A can be either rr or ^ i; those with blood type B can be either 1" 1 or 1" i; people with AB blood are A ", and type O individuals are ii. Among Armenians, the frequency of is 0.360, the frequency of " is 0.104, and the frequency of iarrow_forwardIn a randomly mating laboratory population of Drosophila, 4 percent of the flies have black bodies (encodedby the autosomal recessive b), and 96 percent havebrown bodies (the wild type, encoded by B). If this population is assumed to be in Hardy–Weinberg equilibrium, what are the allele frequencies of B and b and thegenotypic frequencies of B/B and B/b?arrow_forwardWhat is the first variable that can be calculated given these data, and what is the final variable we are requesting you to calculate? "In Finland, 256 people out of 10,000 are homozygous for the CCR5 allele mutation resulting in HIV resistance. Assuming the locus is in Hardy-Weinberg equilibrium, what is the expected genotype frequency of heterozygous carriers for the CCR5 mutation?"arrow_forward
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