Cystic fibrosis is an autosomal recessive disease characterized by two copies of a mutated CFTR gene. If one in 100 (hypothetical scenario, not reality) people in the United States have cystic fibrosis, calculate the p and q frequency for the normal allele (p) and the mutated allele (q). Based on those calculations, what percentage of individuals would be expected to be homozygous dominant?

Cystic fibrosis is an autosomal recessive disease characterized by two copies of a mutated CFTR gene. If one in 100 (hypothetical scenario, not reality) people in the United States have cystic fibrosis, calculate the p and q frequency for the normal allele (p) and the mutated allele (q). Based on those calculations, what percentage of individuals would be expected to be homozygous dominant?

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter5: The Inheritance Of Complex Traits

Section: Chapter Questions

Problem 5QP: As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women....

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In 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.
Cystic fibrosis (CF) is an autosomal recessive trait. A three-generation pedigree is shown below for a family that carries the mutant allele for cystic fibrosis. Note that carriers are not colored in to allow you to figure out their genotypes. Normal allele = F CF mutant allele = f What is the genotype of individual #13? A) ff B) FF C) Ff D) it is impossible to tell
Huntington's disease is a very rare, debilitating disease that affects approximately 1 in 10,000 people. The disease is caused by a particular type of genetic mutation (large number of CAG repeats) in a gene called huntingtin (HTT). Designate the disease-causing allele HTTP and the non-disease- causing allele HTTN. We know that individuals with genotype HTTN will not develop Huntington's Disease, but people with either HTTD/HTTP or HTTN/HTTD will develop the disease. Which allele is dominant--HTTN or HTTP? Explain your reasoning thoroughly, using the correct definition of dominance.
Huntington disease (HD) is a rare dominant condition in humans that results in a slow but inexorable deterioration of the nervous system. HD shows what might be called “age-dependent penetrance,” which is to say that the probability that a person with the HD genotype will express the phenotype varies with age. Assume that 50% of those inheriting the HD allele will express the symptoms by age 40. Susan is a 35-year-old woman whose father has HD. She currently shows no symptoms. What is the probability that Susan will show symptoms in five years?
The following genetic map describes three hypothetical human autosomal genes, each of which exhibits two alleles. Two-factor map distances are shown. A = Artistic (dominant) a = Inartistic (recessive) M = Moral (dominant) m = Immoral (recessive) G = Generous (dominant) g = Greedy (recessive) Assume that these traits exhibit simple Mendelian dominance/recessiveness. The coefficient of coincidence for this map is 0.4. An artistic, moral, generous heterozygous female of genotype AMG/amg marries an inartistic, immoral, greedy homozygous male of genotype amg/amg. What is the probability that their firstborn child will be inartistic, immoral and greedy? What is the probability that their firstborn child will be inartistic, moral and generous? What is the probability that their firstborn child will be artistic, immoral and generous?
A woman diagnosed with early-onset Alzheimer's due to a mutation of the APP genehas children with a man that has no family history of familial Alzheimer's. Give the probability of each possible genotype with corresponding phenotype. (The woman has two possible allele combinations. You must show both possibilities)
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?
In humans, four different blood types (A, B, AB, and O) are encoded by three alleles 1, 1, and i Individuals with both I and I alleles have blood type AB (red blood cells with both A and B antigens). Two copies of the i allele are required for an individual to have blood type O (red blood cells with no antigens). Which of the following correctly indicates the relationship between the I and / alleles for the blood type gene? Select one: OA. I is dominant to / OB. I is recessive to i OC. I and I are co-dominant OD. I and/exemplify incomplete dominance
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).
Briefly describe the assigned trait/disorder. Then, determine the genotype frequency, and allele frequency no. of individuals for each genotype, no. of phenotype for each phenotype (assume Mendelian mode of inheritance) Given the following: N=2348 and q2=0.64; Consider congenital ptosis.
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.
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…