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 62P
Summary Introduction
To determine: The probability of the child having alkaptonuria.
Introduction: Alkaptonuria, or 'black urine disease, is a sporadic acquired disease that inhibits the body from completely breaking down two protein building blocks that are amino acids, also known as tyrosine and phenylalanine.
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Bob and Joan know from a blood test that they are each heterozygous (carriers) for the autosomal recessive gene that causes sickle cell disease. If their first three children are healthy, what is the probability that their fourth child will have the disease?
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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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- Charles has achondroplasia dwarfism (autosomal dominant) but his wife is normal. What is the likelihood that they will have a child with achondroplasia dwarfism? What is the likelihood that they will have embryos that will die from achondroplasia dwarfism? Please explain.arrow_forwardListen Two parents who do not have a condition have a daughter who has the condition. Which of the following possibilities can be eliminated as inheritance pattern for this condition? a) The condition is autosomal dominant O O c) The condition is sex linked recessive b) The condition is sex linked dominant d) All of the above can be eliminatedarrow_forwardNeither Tim nor Rhoda has Duchenne muscular dystrophy, but their firstborn son does. What is the probabilitythat a second child will have the disease? What is theprobability if the second child is a boy? A girl?arrow_forward
- Mr. and Mrs. Jones have six children. Three of them have attached earlobes (recessive) like their father, and the other three have free earlobes like their mother. What are the genotypes of Mr. and Mrs. Jones and of their numerous offspring?arrow_forwardHemophilia is an X-linked recessive disease. A hemophilic man marries a woman who is not a carrier of the disease. (a) Draw a Punnett square showing the genotypes of their children. (b) What are the chances that their daughters will be carriers of the disease? (c) What percentage of their children are likely to have the disease?arrow_forwardOne form of the bleeding disorder known as von Willebrand disease is an autosomal recessive disease. A man who is a carrier marries a woman who is also a carrier of the disease. (a) What percentage of their children are likely to have a disease phenotype? (b) What percentage of their children are likely to have a normal phenotype? (c) What percentage of their children are likely to be carriers of the disease?arrow_forward
- Mr. and Mrs. Smith, both 40 years of age, are healthy with no knowngenetic conditions. Their newborn child has Down syndrome (trisomy 21).Explain the events that caused this condition.arrow_forwardA couple, who are both normal, have a daughter who is albino and a son who is normal. a) what is the probability that their normal son is a carrier of the albinism gene? Show your work. b) the couple wants to have 3 more children; what is the probability that they will have 3 normal girls? Show your work.arrow_forwardA patient has two parents with Huntington's disease. They may not have inherited this autosomal dominant disorder due to: 1) increased DNA repeats (CAG) during spermatogenesis 2) incomplete penetrance 3) inheriting one recessive allelearrow_forward
- A color-blind man marries a woman with normal vision whose father was color-blind. Remember that color-blindness is an X-linked recessive trait. Hint: see figure 12.7 in book. A) What is the probability that their first child will be a color-blind daughter? B) What is the probability that their first son will be color-blind?arrow_forwardA recently married couple are both heterozygous for cystic fibrosis, a recessive genetic disorder. They would like to start a family and would like to have two children. a.) What is the probability that their second pregnancy would result in a child with cystic fibrosis? b.) What is the probability that their second pregnancy would result in a child with cystic fibrosis?arrow_forwardMark has an autosomal recessive condition called sickle cell anemia, a serious blood disorder that is associated with painful complications. His spouse has not experienced symptoms of sickle cell anemia but their first child developed the disease. What conclusions can be drawn about the mother's genotype and the probability of having a second child with the condition? Answer Choices: A: her genotype is heterozygous and there is a 25% percent chance of having another child with the disorder B: her genotype is heterozygous and there is a 50% percent chance of having another child with the disorder C: her genotype is heterozygous and there is a 75% percent chance of having another child with the disorder D: she lacks the allele for sickle cell anemia in that she lacks symptoms of the disease. There is a 0% percent chance of having another child with the disorderMark has an autosomal recessive condition called sickle cell anemia, a serious blood disorder that is associated with…arrow_forward
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