A genetic counselor is examining a family in which both parents are known to be carriers for a recessive CFTR mutation. If the parents are carriers for two different CFTR mutations that are present in the frequency of 0.001 (dad's disease allele) and 0.0001 (mom's disease allele) in human population under Hardy-Weinberg equilibrium, what will be the most prevalent genotype among the three in this population? O heterozygote for two disease alleles homozygote for dad's disease allele O homozygote for mom's disease allele
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- A certain form of congenital glaucoma is caused by an autosomalrecessive allele. Assume that the mutation rate is 10-5 and that peoplewith this condition produce, on the average, only about 80% of theoffspring produced by people who do not have glaucoma.a. At equilibrium between mutation and selection, what will the frequencyof the gene for congenital glaucoma be?b. What will the frequency of the disease be in a randomly matingpopulation that is at equilibrium?Let’s suppose that weight in a species of mammal is polygenic, andeach gene exists as a heavy and light allele. If the allele frequenciesin the population are equal for both types of alleles (i.e., 50%heavy alleles and 50% light alleles), what percentage of individualswill be homozygous for the light alleles in all of the genesaffecting this trait, if the trait was determined by the followingnumber of genes?A. TwoB. ThreeC. FourThe equation p2+ 2pq + q2= 1 representing theHardy-Weinberg proportions examines genes withonly two alleles in a population.a. Derive a similar equation describing the equilibrium proportions of genotypes for a gene withthree alleles. [Hint: Remember that the HardyWeinberg equation can be written as the binomialexpansion (p + q)2.]b. A single gene with three alleles (IA, IB, and i) isresponsible for the ABO blood groups. Individualswith blood type A can be either IA IAor IA i;those with blood type B can be either IB IBor IB i;people with AB blood are IA IB, and type O individuals are ii. Among Armenians, the frequency of IAis0.360, the frequency of IBis 0.104, and the frequencyof i is 0.536. Calculate the frequencies of individuals in this population with the four possible bloodtypes, assuming Hardy-Weinberg equilibrium.In Problems 15–17, you will see that because matingbetween individuals within populations at Hardy-Weinbergequilibrium is random, it is possible to predict…
- The recessive allele that causes Ellis– van Creveld syndrome when homozygous has a frequency of about 0.07 in the Amish population of Pennsylvania, although its frequency is only about 0.001 in the generalpopulation. How many persons out of one thousand in the Amish population would be expected to have the disease? How many out of one million in the general population?Alkaptonuria is a recessive autosomal genetic disorder associated with darkening of the urine. In the United States, approximately 1 out of every 250,000people has alkaptonuria.a. Assuming Hardy-Weinberg equilibrium, estimatethe frequency of the allele responsible for this trait.b. What proportion of people in the U.S. populationare carriers for this trait? In this population, whatis the ratio of carriers to individuals affected byalkaptonuria?c. If a woman without alkaptonuria had a child withthis trait with one husband then remarried, what isthe chance that a child produced by her secondmarriage would have alkaptonuria?d. Alkaptonuria is a relatively benign condition, sothere is little selective advantage to individualswith any genotype; as a result, your assumptionof Hardy-Weinberg equilibrium in part (a) is reasonable. Could you also use the assumption ofHardy-Weinberg equilibrium to estimate the allelefrequencies and carrier frequencies of more severerecessive autosomal conditions…Tay-Sachs disease is caused by loss-of-function mutations in a gene on chromosome 15 thatencodes a lysosomal enzyme. Tach-Sachs is inherited as a autosomal recessive conidition.Among Ashkenazi Jews of central European ancestry, about 1 in 3600 children is born withthe disease. What fraction of the individuals in this population are carriers? (Assume thepopulation is in Hardy-Weinberg equilibrium)
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- In domesticated dogs, size has a high heritability, andthe trait is determined by only a small number ofgenes. In contrast, genetic variation at more than180 QTLs explains only a very small proportion ofthe high heritability for height in humans. What couldexplain the missing heritability in humans, and howcould you test your hypothesized explanations?Consider a rare deleterious recessive allele for a specific gene/locus. In this hypothetical population, the deleterious recessive allele exists at a proportion of 0.01. In an offspring with randomly chosen parents, what is the probability that the offspring will be homozygous for the deleterious recessive allele [q]?Color blindness in humans is caused by an X-linked recessive allele. Ten percent of the males of a large andrandomly mating population are color-blind. A representative group of 1000 people from this population migrates to a South Pacific island, where there are already1000 inhabitants and where 30 percent of the males arecolor-blind. Assuming that Hardy–Weinberg equilibrium applies throughout (in the two original populationsbefore the migration and in the mixed population immediately after the migration), what fraction of malesand females can be expected to be color-blind in the generation immediately after the arrival of the migrants?