Introduction to Genetic Analysis
11th Edition
ISBN: 9781464109485
Author: Anthony J.F. Griffiths, Susan R. Wessler, Sean B. Carroll, John Doebley
Publisher: W. H. Freeman
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Chapter 2, Problem 44.3P
Summary Introduction
To determine: Whether the information of a particular problem can be demonstrated in the form of a branch diagram or not.
Introduction: Branch diagram is a different approach for obtaining both genotypic as well as
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Salim and Sara are contemplating having children, but Salim’s brother has galactosemia and Sara has a sister who has three children, none of whom is affected. Sara dad has no history in his family of any sign of the diease and it is assumed to be homozygous normal. What is the probability that salim and saras first child will have galactosemia?
Kate and her husband are both heterozygous for galactosemia gene. If Kate and her husband have four children, how many of their children are likely to have galactosemia?
Salim and Sara are contemplating having children, but Salim’s brother has galactosemia and Sara has a sister who has three children, one of whom is affected. Sara dad has no history in his family of any sign of the diease and it is assumed to be homozygous normal. What is the probability that salim and saras first child will have galactosemia?
Chapter 2 Solutions
Introduction to Genetic Analysis
Ch. 2 - Prob. 1PCh. 2 - Prob. 2PCh. 2 - Prob. 3PCh. 2 - Prob. 4PCh. 2 - Prob. 6PCh. 2 - Prob. 7PCh. 2 - Prob. 8PCh. 2 - Prob. 9PCh. 2 - Prob. 10PCh. 2 - Prob. 11P
Ch. 2 - Prob. 12PCh. 2 - Prob. 13PCh. 2 - Prob. 14PCh. 2 - Prob. 15PCh. 2 - Prob. 16PCh. 2 - Prob. 17PCh. 2 - Prob. 18PCh. 2 - Prob. 19PCh. 2 - Prob. 20PCh. 2 - Prob. 21PCh. 2 - Prob. 22PCh. 2 - Prob. 23PCh. 2 - Prob. 24PCh. 2 - Prob. 25PCh. 2 - Prob. 26PCh. 2 - Prob. 27PCh. 2 - Prob. 28PCh. 2 - Prob. 31PCh. 2 - Prob. 32PCh. 2 - Prob. 33PCh. 2 - Prob. 34PCh. 2 - Prob. 35PCh. 2 - Prob. 36PCh. 2 - Prob. 37PCh. 2 - Prob. 38PCh. 2 - Prob. 39PCh. 2 - Prob. 40PCh. 2 - Prob. 41PCh. 2 - Prob. 42PCh. 2 - Prob. 43PCh. 2 - Prob. 44PCh. 2 - Prob. 44.1PCh. 2 - Prob. 44.2PCh. 2 - Prob. 44.3PCh. 2 - Prob. 44.4PCh. 2 - Prob. 44.5PCh. 2 - Prob. 44.6PCh. 2 - Prob. 44.7PCh. 2 - Prob. 44.8PCh. 2 - Prob. 44.9PCh. 2 - Prob. 44.10PCh. 2 - Prob. 44.11PCh. 2 - Prob. 44.12PCh. 2 - Prob. 44.13PCh. 2 - Prob. 44.14PCh. 2 - Prob. 44.15PCh. 2 - Prob. 45PCh. 2 - Prob. 47PCh. 2 - Prob. 48PCh. 2 - Prob. 49PCh. 2 - Prob. 50PCh. 2 - Prob. 51PCh. 2 - Prob. 52PCh. 2 - Prob. 53PCh. 2 - Prob. 56PCh. 2 - Prob. 57PCh. 2 - Prob. 58PCh. 2 - Prob. 59PCh. 2 - Prob. 60PCh. 2 - Prob. 61PCh. 2 - Prob. 62PCh. 2 - Prob. 63PCh. 2 - Prob. 64PCh. 2 - Prob. 65PCh. 2 - Prob. 66PCh. 2 - Prob. 67PCh. 2 - Prob. 68PCh. 2 - Prob. 69PCh. 2 - Prob. 70PCh. 2 - Prob. 71PCh. 2 - Prob. 72PCh. 2 - Prob. 73PCh. 2 - Prob. 74PCh. 2 - Prob. 75PCh. 2 - Prob. 76PCh. 2 - Prob. 77PCh. 2 - Prob. 78PCh. 2 - Prob. 79P
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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
- The accompanying pedigree shows a family in which one child (II-1) has an autosomal recessive condition. On the basis of this fact alone, provide the following information. 1) What is the chance that among the three children in generation II who have the dominant phenotype, one of them is AAAA and two of them are AaAa? (Hint: Consider all possible orders of genotypes.) Express your answer to two decimal places.arrow_forwardGalactosemia is an autosomal recessive human disease that is treatable by restricting lactose and glucose in the diet. If two individuals who are both heterozygous for the recessive galactosemia allele have three children, what is the probability that two of the children will have galactosemia and one will not? Type your answer as a fraction (not a decimal), with no spaces (e.g. 1/16).arrow_forwardSalim and Sara are contemplating having children, but Salim’s brother has galactosemia and Sara’s great-grandmother also had galactosemia. Sara has a sister who has three children, none of whom is affected.a. Construct a genetic pedigree for this family showing the possible genotype of each member.b. What is the probability that Salim and Sara’s first child will have galactosemia? Explain your calculations.arrow_forward
- *Cystic fibrosis is a rare autosomal recessive condition. phenotypically normal man whose father had cystic fibrosis marries a phenotypically normal woman from outside the family.² a) Draw the pedigree as far as described. b) If the frequency of heterozygotes in the general population is 1/50, what is the probability that the couple's first child will have cystic fibrosis? c) If the first child does have cystic fibrosis, what is the probability that the second child will be normal?arrow_forwardAlong with the trait in the pedigree, individual IV-6 and the woman are also both heterozygous for the autosomal dominant allele causing Huntington's disease. If they have a child, what is the probability that it will be affected by at least one of these traits? Remember to include both the trait in the pedigree and Huntington's disease in your calculations. Enter your answer to two decimal places (e.g., 0.55).arrow_forward152 Phenylketonuria (PKU) is a disorder caused by a recessive allele. Two carrier individuals have progeny. Answer the following questions in order and show solutions whenever relevant. If they have a normal child, what is the probability that he or she will be heterozygous? If they have three children, what is the probability of having 2 affected children and one normal child?arrow_forward
- A couple seeks advice from a genetic counselor because they know that they are both carriers for cystic fibrosis (which is autosomal recessive), and the woman is a carrier for Duchenne muscular dystrophy (which is sex-linked recessive). For each of their future sons and daughters (calculate separately), what is the probability that they will be affected for: Both cystic fibrosis and nuscular dystrophy If the couple’s first born son has both cystic fibrosis and muscular dystrophy, what is the probability that their second born son will have both diseases?arrow_forwardThe three genes X, Y, and Z are linked on an autosomal chromosome in humans (X to Y is 15 cM, and Y to Z is 18 cM). If an individual that is heterozygous at all three loci (XYZ/xyz) has children with an individual that is homozygous recessive at all three loci (xyz/xyz), what is the probability that they will have a child that is phenotypically identical to either parent (X-Y-Z- or xxyyzz)? Assume there is no genetic interference to double crossover events at this site.arrow_forwardThe autosomal (not X-linked) gene for brachydactyly, short fingers, is dominant to normal finger length. Assume that a female with brachydactyly in the heterozygous condition is married to a man with normal fingers. What is the probability that(a) their first child will have brachydactyly?(b) their first two children will have brachydactyly?(c) their first child will be a brachydactylous girl?arrow_forward
- Cystic fibrosis in humans is caused by mutations in a single gene and is inherited as an autosomal (non-sex chromosome) recessive trait. An unaffected couple has two children. The first child has cystic fibrosis, and the second child is unaffected. What is the probability that the second child is a carrier (heterozygous) for the mutation that causes the disease? 2/3 1 1/2 3/4arrow_forwardThis is a pedigree for a dominant trait caused by gene A in humans. Shaded symbols show individuals affected with the trait; non-shaded individuals are normal (aa). Among the progeny arising from the marriage of individual III-1, what proportion would be expected to show the trait? Among the progeny arising from the marriage of individual III-6, what proportion would be expected to show the trait?arrow_forwardPhenylketonuria (PKU) is a disorder caused by a recessive allele. Two carrier individuals have progeny. If they have a normal child, what is the probability that he or she will be heterozygous? Show solutions.arrow_forward
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