ACHIEVE:INTRO TO GENETIC ANALYSIS 1TERM
12th Edition
ISBN: 9781319401399
Author: Griffiths
Publisher: MAC HIGHER
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Chapter 2, Problem 56.3P
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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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, 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?
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
ACHIEVE:INTRO TO GENETIC ANALYSIS 1TERM
Ch. 2 - Prob. 1PCh. 2 - Prob. 2PCh. 2 - Prob. 3PCh. 2 - Prob. 4PCh. 2 - Prob. 5PCh. 2 - Prob. 6PCh. 2 - Prob. 7PCh. 2 - Prob. 8PCh. 2 - Prob. 9PCh. 2 - Prob. 10P
Ch. 2 - Prob. 11PCh. 2 - Prob. 12PCh. 2 - Prob. 13PCh. 2 - Prob. 14PCh. 2 - Prob. 15PCh. 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. 29PCh. 2 - Prob. 30PCh. 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. 45PCh. 2 - Prob. 46PCh. 2 - Prob. 47PCh. 2 - Prob. 48PCh. 2 - Prob. 49PCh. 2 - Prob. 50PCh. 2 - Prob. 51PCh. 2 - Prob. 52PCh. 2 - Prob. 53PCh. 2 - Prob. 54PCh. 2 - Prob. 55PCh. 2 - Prob. 56PCh. 2 - Prob. 56.1PCh. 2 - Prob. 56.2PCh. 2 - Prob. 56.3PCh. 2 - Prob. 56.4PCh. 2 - Prob. 56.5PCh. 2 - Prob. 56.6PCh. 2 - Prob. 56.7PCh. 2 - Prob. 56.8PCh. 2 - Prob. 56.9PCh. 2 - Prob. 56.10PCh. 2 - Prob. 56.11PCh. 2 - Prob. 56.12PCh. 2 - Prob. 56.13PCh. 2 - Prob. 56.14PCh. 2 - Prob. 56.15PCh. 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. 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. 79PCh. 2 - Prob. 80PCh. 2 - Prob. 81PCh. 2 - Prob. 82PCh. 2 - Prob. 83PCh. 2 - Prob. 84PCh. 2 - Prob. 85PCh. 2 - Prob. 86PCh. 2 - Prob. 87PCh. 2 - Prob. 88PCh. 2 - Prob. 89PCh. 2 - Prob. 90PCh. 2 - Prob. 91PCh. 2 - Prob. 1GSCh. 2 - Prob. 2GSCh. 2 - Prob. 3GS
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- Galactosemia 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_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_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
- 152 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_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_forwardThe 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_forward
- The 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_forwardCystic 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_forwardThe family below has the autosomal dominant trait of dimples. In this family, I-1 and I-2 are known to be heterozygous. It is not known whether any other individuals in this family are heterozygotes. If Narayan and Pritya have a child, what is the chance it will NOT have dimples? A) 1/64 B) 1/6 C) 1/4 D)4/9 E) 1/8 F) 1/2 G) 1/9 H) 1/12 I) 1/16arrow_forward
- This 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_forwardA couple who are about to get married learn from studying their family histories that, in both their families, theirunaffected grandparents had siblings with cystic fibrosis(a rare autosomal recessive disease).a. If the couple marries and has a child, what is theprobability that the child will have cystic fibrosis?b. If they have four children, what is the chance that thechildren will have the precise Mendelian ratio of 3:1 fornormal:cystic fibrosis?c. If their first child has cystic fibrosis, what is theprobability that their next three children will be normal?arrow_forwardShown below is a pedigree for Phenylketonuria (PKU), an autosomal recessive metabolic disorder. The characteristic feature of PKU is severe mental retardation A) What is the probability that individual II-1 is heterozygous for this gene? B) What is the probability that individual III-4 is heterozygous for this gene? C) If individuals III-3 and III-4 were to marry, what is the probability that their child would express PKU?arrow_forward
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