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Concept explainers
To analyze:
Describe following on the basis of an autosomal recessive condition:
The chance that the first child will be homozygous recessive for mutation, if both parents are heterozygous carriers of a mutant allele
Parents who are heterozygous carriers for a recessive mutant allele have a normal child. Determine the chance that this child is a heterozygous carrier of the condition.
If parents are heterozygous carriers of a mutant allele, and have the first child with homozygous recessive condition, calculate the probability that the second child of the couple will be homozygous recessive. Also calculate the probability that the second child will be a heterozygous carrier of the condition.
Introduction:
In an autosomal recessive condition, both copies of alleles are responsible for a trait and should be recessive (abnormal) to cause a disease. The diseased condition cannot be observed in an individual with a homozygous dominant or heterozygous dominant allele, but heterozygous dominant are the carrier for the trait.
Phenylketonuria (PKU) is an autosomal recessive disorder and can be observed in individuals who have both the copies of alleles in a homozygous recessive condition.
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Explanation of Solution
Pedigree chart of cross between both heterozygous parents
♂ ♀ |
A | a |
A | AA | Aa |
a | Aa | aa |
The
Homozygous dominant (AA) is
Heterozygous carrier (Aa) is
Homozygous recessive (aa) is
The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is
In this part, they have asked to estimate the probability of a normal child being a carrier for the mutation.
The chance of having normal (AA / Aa) child is
Since the birth of the second child is independent of the birth of first child so, the chance of the second child of the couple will be homozygous recessive (aa) for the mutation is
The chance of the second child of the couple will be heterozygous carrier (Aa) for the mutation is
The chance that the first child of the couple will be homozygous recessive (aa) for the mutation is
The chance of the normal child being a carrier for the condition is
The chance that the second child of the couple will be homozygous recessive (aa) for the mutation is
The chance that the second child of the couple will be a heterozygous carrier (Aa) for the mutation is
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Chapter B Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
- Albinism, lack of pigmentation in humans, results from an autosomal recessive gene designated a. Two parents with normal pigmentation have an albino child. What is the probability that their next child will be albino? What is the probability that the next child will be an albino girl? If the child is normal, what is the probability that it will be a carrier (heterozygous) for the albino gene?arrow_forwardAnswer the following: a. what environmental factors may increase incidence of chromosomal abnormalities? Name at least three. b. for a couple, what is the significance of knowing chromosomal aberrations?arrow_forwardPlease consider the following pedigree. Assume that people who marry in to the family do not carry the allele unless otherwise indicated. Assume complete penetrance. I II 5 6 III 6 IV 1 2 a. Is it possible for the inheritance pattern for the trait illustrated in this pedigree to be as a result of each of the following? Answer yes or no. (i) an autosomal recessive allele (AR) (ii) an autosomal dominant allele (AD) (iii) a X-linked recessive allele (XR) (iv) a X-linked dominant allele (XD) b. Provide a genotype for individual III-6 for the most likely mode of inheritance as determined in (a).arrow_forward
- Cystic Fibrosis (CF) is an autosomal recessive condition. Therefore, heterozygous (Cc) carriers do not display symptoms. Two parents who are carriers plan to start a family and you are a genetic counselor helping to advise them about their chances of having children affected by CF. a) Suppose the couple has 4 children, each one year apart. What is the probability that all 4 children will inherit CF? b) What is the probability that any 3 of their 4 children will not inherit CF, but 1 will be affected? c) What is the probability that their first child will not inherit CF, but the younger 3 children will inherit CF?arrow_forwardReview the process of autosomal dominant inheritance by coloring the following figures. In this instance the father is affected in the mother is unaffected to illustrate the transmission of the disorder from the father to the children color the figures as described.arrow_forwardE. D. A. B. 11. C. Huntington's disease is a degenerative disease of the nervous system that strikes in middle age. The allele that causes the disease (H) is dominant to the allele that results in the normal condition (h). Answer the following questions about the inheritance of this disease. What is the genotype of a man who is normal but whose father had Huntington's disease? What is the genotype of a woman who has Huntington's disease if both of her parents had Huntington's disease? If a man who is heterozygous for Huntington's disease marries a woman who is normal, what would you expect for the genotypes and phenotypes of their children? If a normal man marries a woman who is homozygous for Huntington's disease, what do you expect for the genotypes and phenotypes of their children? Since Huntington's disease is caused by a dominant allele, does this mean it's also the most common allele in the population? F. Since Huntington's disease is caused by a dominant allele, does this mean…arrow_forward
- Answer the following questions given the pedigree below. Please assume that no other mutations are occurring, complete penetrance, and that the individual marked with an asterisk (*) doesn’t carry the allele causing the affected phenotype. Q2) What are the genotypes of the following individuals listed in the table below. Use the uppercase “A” to represent the dominant allele and lowercase “a” for the recessive allele. Individual All possible genotypes I-2 II-1 IV-2 V-2 II-2 II-3arrow_forwardNeurofibromatosis-1 (NF1) is an autosomal dominant disorder where tumours form in the base layer of the skin or in nerve tissues. What is the probability that individuals II-1 and II-2 will have a genetic son with NF1? Find the image attached.arrow_forwardPlease consider the following pedigree. Assume that people who marry in to the family do not carry the allele. Assume complete penetrance. I II III 3 IV 1 2 a. Is it possible for the inheritance pattern for the trait illustrated in this pedigree to be as a result of each of the following? Answer yes or no. (i) an autosomal recessive allele (AR) (ii) an autosomal dominant allele (AD) (iii) a X-linked recessive allele (XR) (iv) a X-linked dominant allele (XD) b. Based strictly on the characteristic patterns of inheritance that define the four different options in (a), give a definitive motivation for the most likely mode of inhertance.arrow_forward
- varrow_forwardColor blindness in humans is controlled by an X-linked completely recessive allele (Xc), while breast cancer is controlled by an autosomal completely dominant allele, B. A color blind male, who is a heterozygote carrier for breast cancer has three children/n with a normal eyed female (whose mother was color blind), who is homozygote recessive for the breast cancer allele. What is the probability that out of three children, 2 will be color blind males, and not show breast cancer, and one will be a color blind female, who shows breast cancer?arrow_forwardRefer to the pedigree below which shows inheritance for achondroplasia (dwarfism), a dominantly inherited trait (denoted as D), which are the darkened circles and squares. Dwarfism (darkened shapes) are dominantly inherited, while normal height is recessively inherited (hh). Based on the pedigree, what is the correct genotype for individual #II-6? Dominant Autosomal Pedigree 2 II 2 3 II 1 2 3 6 9 10 Dd DD DD or Dd ddarrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning
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