Introduction To Genetic Analysis
12th Edition
ISBN: 9781319114787
Author: Anthony J.F. Griffiths, John Doebley, Catherine Peichel, David A. Wassarman
Publisher: W. H. Freeman
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 2, Problem 56.9P
Summary Introduction
To determine: The generalities that are assumed regarding the autosomal recessive disease.
Introduction: Autosomal recessive disorders are very rare. Some of the examples of the disorders are sickle cell anemia, cystic fibrosis, and Tay Sachs disease.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
For a recessive condition, two normal heterozygous individuals have children.
What is the likelihood of their children being affected by this condition?
What is the likelihood of their children being carriers without the condition?
What is the likelihood of their asymptomatic children being carriers?
Suppose that an individual with the condition has children with a heterozygous individual, what is the likelihood of their children being carriers?
A WOMAN IS HETEROZYGOUS FOR TWO
HARMFUL RECESSIVE ALLELES IN
DIFFERENT CHROMOSOMES, ONE FOR
PHENYLKETONURIA (PKU) AND THE
OTHER FOR CYSTIC FIBROSIS (CF). SHE
MARRIES AN UNAFFECTED MAN WHO IS
A CARRIER FOR NEITHER DISEASE.
IF SHE HAS A DAUGHTER, WHAT IS THE
PROBABILITY THAT THE CHILD WILL
CARRY NEITHER OF THE RECESSIVE
ALLELES?
EXACTLY ONE?
BOTH?
Hypophosphatemia (vitamin D-resistant rickets) is inherited as a sex-linked dominant trait (H).
A) A normal woman and a man with hypophosphatemia marry. What is the chance of having daughters with rickets? Sons?
B) A heterozygous woman and a normal man marry. Does the mother have rickets? What is the chance of having daughters with rickets? Sons?
Chapter 2 Solutions
Introduction To Genetic Analysis
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
Knowledge Booster
Learn more about
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
- Tarzan and Jane are both normal, but their first child is affected with heroism, an autosomal recessive trait. If they have three more children, what is the probability that all three will be affected?arrow_forwardHuntington disease is a rare dominant disorder that causes neurodegeneration later in life. In a homozygous state, the dominant allele is lethal. A man in his thirties, who already has three children, discovers that his mother has Huntington disease though his father is unaffected. What are the following probabilities? a) The man in his thirties will develop Huntington disease. b) His first child will develop Huntington disease. c) Any one out of three of his children will develop Huntington disease.arrow_forwardAlbinism is an autosomal (not sex-linked) recessive trait. A man and woman are both of normal pigmentation and have one child out of three who is albino (without melanin pigmentation). What are the genotypes of the albino's parents? One parent must be homozygous for the recessive allele; the other parent can be homozygous dominant, homozygous recessive, or heterozygous. Both parents must be homozygous dominant. One parent must be heterozygous; the other parent can be homozygous dominant, homozygous recessive, or heterozygous. Both parents must be heterozygous. One parent must be homozygous dominant; the other parent must be heterozygous. O000arrow_forward
- Annabeth and Percy are concerned about having a child with Niemann - Pick disease, which causes abnormal accumulation of fat in cells. This condition, which is very rare, has affected Annabeth's niece (her brother's daughter) and Percy's aunt (his father's sister ). No one else in either family has the condition. What is the probability Annabeth and Percy's first child will have the disease? What is the probability Annabeth and Percy's first child will have the disease? 1/64 1/32 1/24 1/18 1/12 1/9 1/4 None of these abovearrow_forwardAlbinism is an autosomal (not sex-linked) recessive trait. A man and woman are both of normal pigmentation and have one child out of three who is albino (without melanin pigmentation). What are the genotypes of the albino's parents? Both parents must be homozygous dominant. Both parents must be heterozygous. One parent must be homozygous dominant; the other parent must be heterozygous. One parent must be heterozygous; the other parent can be homozygous dominant, homozygous recessive, or heterozygous. One parent must be homozygous for the recessive allele; the other parent can be homozygous dominant, homozygous recessive, or heterozygous. O O O O Oarrow_forwardYOUR SISTER DIED FROM TAY-SACHS DISEASE, INHERITED AS A RECESSIVE ALLELE (t). you're married and planning to start your family. you're worried about the disease and decide to have genetic testing to see if you or your spouse is a carrier of the tay-sachs allele. the test results show that you're a carrier of the allele, but your spouse isn't. what is the probability that you and your spouse will have a child with tay-sachs disease? show your work.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_forwardTarzan and Jane are both normal, but their first child is affected with heroism, an autosomal recessive trait. What is the probability that their next child will also have the trait?arrow_forwardIn humans, cystic fibrosis is a disease caused by a recessive allele (). Individuals with normal (wildtype) phenotype have the genotype (F). If two heterozygous individuals marry and have 8 children, what is the probability that: a) four will be normal and four will have cystic fibrosis? b) one will have cystic fibrosis and seven will be normal? c) All eight children will have cystic fibrosis?arrow_forward
- Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. Individuals with PKU have two recessive alleles and have very low levels of an enzyme that is needed to properly break down proteins. If a woman and her husband are both carriers and have three children, what is the probability of each of the following? Show your math. (Hint: You can represent your probabilities as fractions or decimals, but probabilities are always between 0 and 1) a. All three children are of normal phenotype. b. One or more of the three children have the disease. c.All three children have the disease.arrow_forwardIn 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.arrow_forwardBob 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?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Biology (MindTap Course List)BiologyISBN:9781305112100Author:Cecie Starr, Beverly McMillanPublisher:Cengage Learning
Human Biology (MindTap Course List)
Biology
ISBN:9781305112100
Author:Cecie Starr, Beverly McMillan
Publisher:Cengage Learning
How to solve genetics probability problems; Author: Shomu's Biology;https://www.youtube.com/watch?v=R0yjfb1ooUs;License: Standard YouTube License, CC-BY
Beyond Mendelian Genetics: Complex Patterns of Inheritance; Author: Professor Dave Explains;https://www.youtube.com/watch?v=-EmvmBuK-B8;License: Standard YouTube License, CC-BY