Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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Chapter 21, Problem 27P
In Europe, the frequency of the CF allele causing the recessive autosomal disease cystic fibrosis is about 0.04. Cystic fibrosis causes death before reproduction in virtually all cases.
| a. | Determine values of relative fitness (W) for the unaffected, carrier, and affected genotypes. Assume that no selective advantage is associated with heterozygosity for the disease allele. |
| b. | Given your answer to part (a), determine the average (mean) fitness at birth of the population as a whole with respect to the cystic fibrosis trait () and the expected change in allele frequency over one generation (Δq) when measured at the birth of the next generation. |
| c. | Suppose the European population is at equilibrium for the frequency of the CF allele because some heterozygote advantage exists. Recalculate the relative fitness values for the three genotypes under this assumption |
| d. | The CFTR protein encoded by the CF gene is a chloride ion channel. People suffering from cholera have diarrhea that pumps water and chloride ions out of the small intestine. Use these facts to explain why a heterozygote advantage might in fact exist for the CF gene |
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. In Europe, the frequency of the CF− allele causing therecessive autosomal disease cystic fibrosis is about0.04. Cystic fibrosis causes death before reproductionin virtually all cases.a. Determine values of relative fitness (W) for the unaffected, carrier, and affected genotypes. Assumethat no selective advantage is associated with heterozygosity for the disease allele.b. Given your answer to part (a), determine the average (mean) fitness at birth of the population as awhole with respect to the cystic fibrosis trait (W)and the expected change in allele frequency overone generation (Δq) when measured at the birth ofthe next generation.c. Suppose the European population is at equilibriumfor the frequency of the CF− allele because someheterozygote advantage exists. Recalculate the relative fitness values for the three genotypes underthis assumption. d. The CFTR protein encoded by the CF gene is achloride ion channel. People suffering from cholerahave diarrhea that pumps water and…
Identify each of the following as an example of allele, genotype, and/or phenotype frequency:
A. Approximately 1 in 2500 people of Northern European descent is born with cystic fibrosis.
B. The percentage of carriers of the sickle cell allele in West Africa is approximately 13%.
C. The number of new mutations for achondroplasia, a genetic disorder, is approximately 5 × 10–5.
Excessive secretion of male sex hormones results in premature sexual maturation in
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Chapter 21 Solutions
Genetics: From Genes to Genomes
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- Identify each of the following as an example of allele, genotype,and/or phenotype frequency:A. Approximately 1 in 2500 individuals of Northern Europeandescent is born with cystic fibrosis.B. The percentage of carriers of the sickle cell allele in WestAfrica is approximately 13%.C. The number of new mutations per generation resultingin achondroplasia, a genetic disorder, is approximately5 × 10−5.arrow_forwardCystic fibrosis (CF) is a recessive autosomal disorder. In certain populations of Northern European descent, the number of people born with this disorder is about 1 in 2500. Assuming Hardy- Weinberg equilibrium for this trait: A. What are the frequencies for the normal and CF alleles? B. What are the genotype frequencies of homozygous normal, heterozygous, and homozygous affected individuals? C. Assuming random mating, what is the probability that two phenotypically unaffected heterozygous carriers will choose each other as mates?arrow_forwardAllele B is a deleterious autosomal dominant. The frequency of affected individuals is 4.0 × 10−6. The reproductive capacity of these individuals is about 30 percentthat of normal individuals. Estimate μ, the rate at whichb mutates to its deleterious allele B. Assume that the frequencies of the alleles are at their equilibrium values.arrow_forward
- a. Alleles of genes on the X chromosome can also be atequilibrium, but the equilibrium frequencies underthe Hardy-Weinberg assumptions must be calculatedseparately for the two sexes. For a gene with two alleles A and a at frequencies of p and q, respectively,write expressions that describe the equilibrium frequencies for all the genotypes in men and women.b. Approximately 1 in 10,000 males in the UnitedStates is afflicted with hemophilia, an X-linkedrecessive condition. If you assume that the population is at Hardy-Weinberg equilibrium, what proportion of American females would be hemophiliacs?About how many female hemophiliacs would youexpect to find among the 170 million women livingin the United States? (Assume that all females withhemophilia are homozygous for the disease allele.)arrow_forwardA sex-linked recessive allele rg causes a red-green colorblindness in humans. A normal woman whose father was colorblind marries a colorblind man. A. What genotypes are possible for the mother of the colorblind man? B. What are the chances that the first child from this marriage will be a colorblind boy? C. Of the girls born from this union, what proportion can be expected to be colorblind? D. Of all the children of these parentsarrow_forwardIn humans, the genetic disease cystic fibrosis is caused by a recessive allele (a). The normal (healthy) allele is dominant (A). What is the genotype of someone who has cystic fibrosis? What are the two different genotypes that a healthy person could have? If two people were both heterozygous for the cystic fibrosis gene, what fraction of their children would be likely to have this disease? Hint: Draw a Punnett square to figure it out.arrow_forward
- Consider the case of a hypothetical genetic disease called WhySciEleven (WSE), an inherited disease that results to the inability to complete production of an amino acid “moduleactivity” that results in brain damage if untreated. WSE is due to a recessive allele. Given one WSE occurrence per 10,000 births. 1. What is the frequency of the dominant allele? 2. What is the frequency of the recessive allele for WSE in population? 3. What is the frequency of the heterozygous people?arrow_forwardThe chart below is showing 4 generations of a family that is affected by a hereditary disease. a. Is the disorder being tracked dominant or recessive? How do you know? b. There is only one possible genotype for person C. True or False? c. What are the possible genotypes for person A? d. What are the possible genotypes for person B?, e. If two people with the same genotypes as person C's spouse and person A's spouse had a child, what is the probability that the child will be affected by this genetic disorder? (draw a Punnett square using the correct genotypes to help you). % chance offspring will be affected % chance offspring will not be affectedarrow_forwardOne 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_forward
- Consider the following scenario: A man without freckles (freckles are a dominant trait, determined by the dominant allele “F”) is a carrier of cystic fibrosis (recall that CF is a recessive trait, determined by the recessive allele "a"), mates with a woman whose genotype is heterozygous for freckles and is also a carrier of CF. Assume the two genes in question are in different chromosomes and, therefore, assort independently. Complete the following Punnett square to generate the offspring probabilities from this couple, by entering the genotypes of the parents, the gametes, and the offspring for the two traits described above. (4) Father’s genotype ffAa Mother’s genotypeFfAa ● Sperm: fA ● Sperm: fa ● Sperm: fA ● Sperm: fa ● Egg: FA ● Egg: fA ● Egg: Fa ● Egg: fa What is the probability for this couple to have a child with freckles? What is the…arrow_forwardFor each of the following characteristics, indicate whether it would be considered a discontinuous characteristic or a quantitative characteristic. Briefly justify your answer. a. Kernel color in a strain of wheat, in which two codominant alleles segregating at a single locus determine the color. Thus, there are three phenotypes present in this strain: white, light red, and medium red. b. Body weight in a family of Labrador retrievers. An autosomal recessive allele that causes dwarfism is present in this family. Two phenotypes are recognized: dwarf (less than 13 kg) and normal (greater than 23 kg). c. Presence or absence of leprosy. Susceptibility to leprosy is determined by multiple genes and numerous environmental factors. d. Number of toes in guinea pigs, which is influenced by genes at many loci. e. Number of fingers in humans. Extra (more than five) fingers are caused by the presence of an autosomal dominant allele.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
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