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Concept explainers
a.
To determine:
The genotype frequencies in a population.
Introduction:
The branch of genetics that studies the transmission of genetic material in a population is termed as population genetics. Genotype frequency is defined as the proportion of individuals in a given population that have a common genotype. The genotype frequency is important for understanding population genetics.
b.
To determine:
The allele frequency of GB and GG.
Introduction:
The group of individuals that belong to a single species and live in the same geographical area is termed as population. The proportion of gene copies that are of a common allele type in a population is termed as allele frequency. Allele frequency is used to study population genetics.
(c)
To determine:
The expected frequencies of the genotypes if the population is at Hardy-Weinberg equilibrium.
Introduction:
Geoffrey H. Hardy was a scientist who proposed the concept of Hardy-Weinberg equilibrium. This concept is used to associate the allele frequency with the genotype frequency. The populations that have allele frequency and the genotypic frequency at equilibrium follow the concept of Hardy-Weinberg equilibrium.
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Chapter 20 Solutions
Genetics: From Genes to Genomes, 5th edition
- Whether or not a person has cheek dimples is an autosomally inherited character due to variation in a single gene. Having cheek dimples (D) is dominant to not having cheek dimple s (d).In a population of 28,000 Greek children, 3,500 had cheek dimples and the rest did not. Assuming that the population is in Hardy-Weinberg equilibrium, A. what are the frequencies of the two alleles? B. how many of the children do you expect to be heterozygous?arrow_forwardCystic fibrosis (CF) is a recessive autosomal disorder. In certainpopulations of Northern European descent, the number of peopleborn with this disorder is about 1 in 2500. Assuming Hardy-Weinberg equilibrium for this trait:A. What are the frequencies for the common (non-disease-causing)allele and the mutant (disease-causing) allele.B. What are the genotype frequencies of homozygous unaffected,heterozygous, and homozygous affected individuals?C. Assuming random mating, what is the probability that twophenotypically unaffected heterozygous carriers will chooseeach other as mates?arrow_forwardAbout 8% of the men in a population are red-green color blind (because of a sex-linked recessive allele). Answer the following questions, assuming random mating in the population, with respect to color blindness. a. What percentage of women would be expected to be color blind? b. What percentage of women would be expected to be heterozygous? c. What percentage of men would be expected to have normal vision two generations later?arrow_forward
- In a population of 200 mice, 98 are the homozygous dominant for brown fur (BB), 84 are heterozygous (Bb), and 18 are homozygous recessive for white fur (bb). a. The genotypes frequencies of this population arearrow_forwardAlbinism is due to a recessive allele of an autosomal gene. Let a represent the albino allele and A represent the normal non-albino allele. If one out of every 10,000 people were albino and assuming that the population is in Hardy-Weinberg equilibrium, A. what would be the phenotype frequencies of albino and non-albino ? B. what would be the allele frequencies of a and of A? C. what fraction of people can be expected to be of the genotype Aa? D. what fraction of people can be expected to be of the genotype AA?arrow_forwardA dominant mutation in Drosophila called Deltacauses changes in wing morphology in Delta/+ heterozygotes. Homozygosity for this mutation (Delta /Delta) is lethal prior to the adult stage. In a population of 150 flies, it was determined that 60 hadnormal wings and 90 had abnormal wings.a. What are the allele frequencies in this population?b. Using the allele frequencies calculated in part (a),how many total zygotes must be produced by thispopulation in order for you to count 160 viableadults in the next generation?c. Given that there is random mating, no migration,and no mutation, and ignoring the effects ofgenetic drift, what are the expected numbers ofthe different genotypes in the next generation if160 viable offspring of the population in part(a) are counted?d. Is this next generation at Hardy-Weinberg equilibrium? Why or why not?740 Chapter 21 Variation and Selection in Populationsarrow_forward
- In a population with 1000 individuals and at Hardy-Weinberg equilibrium, the frequency of individuals with galactosemia is 4%.a. What are the frequencies of the dominant and recessive alleles?b. What is the expected frequency of the homozygous dominant in the population?c. What would be the expected frequency of the heterozygous dominant after five generations?arrow_forwardExcessive secretion of male sex hormones results in premature sexual maturation in males and masculization of the sex characters in females. This disorder is called the adrenogenital syndrome, and in Switzerland there is an autosomal recessive form of the disease that affects about one in 5000 newborns. a. Assuming random mating, what is the allele frequency of the recessive? b. What is the frequency of heterozygous carriers?arrow_forwardImagine you are studying a population of clownfish and have genotyped two biallelic loci. The first locus has alleles with frequencies 0.5 and 0.5 in the population. The second locus has alleles with frequencies 0.2 and 0.8. You have genotyped one young fish at these two loci. a. What is the probability that we can exclude a father as the true parent (assuming that he is not the true parent)? b.If you could genotype a third locus to increase your probability of paternal exclusion, would you prefer a locus with allele frequencies of 0.55/0.45 or 0.2/0.8? Why? c.Is the power from part a) good enough to use for a paternity exclusion study without adding other loci? Why or why not?arrow_forward
- A mouse mutation with incomplete dominance (t =tailless) causes short tails in heterozygotes (t+/t). Thesame mutation acts as a recessive lethal that causeshomozygotes (t/t) to die in utero. In a populationconsisting of 150 mice, 60 are t+/t+ and 90 areheterozygotes.a. What are the allele frequencies in this population?b. Given that there is random mating among mice, nomigration, and no mutation, and ignoring the effects of random genetic drift, what are the expectednumbers of the different genotypes in this nextgeneration if 200 offspring are born?c. Two populations (called Pop 1 and Pop 2) of micecome into contact and interbreed randomly. Thesepopulations initially are composed of the followingnumbers of wild-type (t+/t+) homozygotes andtailless (t+/t) heterozygotes:Pop 1 Pop 2Wild type 16 48Tailless 48 36What are the frequencies of the two genotypes in thenext generation?arrow_forwardBased on this information (picture) A. What is the probability that a randomly sampled individual from the population has two copies of the a allele (that is, that it has an aa genotype)? B. What is the probability that both members of a randomly sampled married couple (man and woman) are aa at the asparagus-smelling gene? C. What is the probability that both members of a randomly sampled married couple (man and woman) are heterozygotes at this locus (meaning that each person has one allele A and one allele a)? D. Consider the type of couple described in (c). What is the probability that the first child of such a couple also has one A allele and one a allele (is a heterozygote)? Remember that the child must receive exactly one allele from each parent.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_forward
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