Biology: The Unity and Diversity of Life
15th Edition
ISBN: 9780357093795
Author: STARR
Publisher: CENGAGE LEARNING (CUSTOM)
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Chapter 14, Problem 4SQ
Summary Introduction
Introduction: Tay–Sachs disease is caused due to mutations in the HEXA gene that encodes lysosomal enzyme. It follows autosomal recessive inheritance, that is, two copies of mutated gene are required for the occurrence of the disease.
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A man and a woman have a child with cystic fibrosis. Neither parent has cystic fibrosis. How could this happen?
a) Both parents carry the cystic fibrosis allele, and each passed that allele to their child.
b) The child had a spontaneous mutation on both copies of their CFTR alleles, leading to cystic fibrosis.
c) One parent gave the child two copies of the cystic fibrosis CFTR allele.
c) One parent gave the child a wild type CFTR allele, and the other parent gave them a cystic fibrosis CFTR gene.
Tay–Sachs disease is caused by recessive alleles on anautosome. In which case(s) could two parents with anormal phenotype have a child with Tay–Sachs?a. Both parents are homozygous for a Tay–Sachs allele.b. Both parents are heterozygous for a Tay–Sachsallele.c. One parent is homozygous for a Tay–Sachs allele,and the other is heterozygous.
In a trait that follows Mendelian rules of inheritance, what is the only way an organism can have the recessive phenotype?
A. By having the homozygous recessive genotype
B. By being heterozygous for that genotype
C. It is not possible to get the recessive phenotype in Mendelian genetics
D. If one parent is homozygous recessive for that trait, any offspring are guaranteed the recessive phenotype
E. By being homozygous dominant for that genotype
Chapter 14 Solutions
Biology: The Unity and Diversity of Life
Ch. 14 - Prob. 1DAACh. 14 - Prob. 2DAACh. 14 - Prob. 3DAACh. 14 - Prob. 4DAACh. 14 - Prob. 5DAACh. 14 - Prob. 1SQCh. 14 - Pedigree analysis is necessary when studying human...Ch. 14 - A recognized set of symptoms that characterize a...Ch. 14 - Prob. 4SQCh. 14 - A trait that is present in a male child but not in...
Ch. 14 - Choose the statement that is incorrect. a. A son...Ch. 14 - Prob. 7SQCh. 14 - Prob. 8SQCh. 14 - Klinefelter syndrome (XXY) can most be easily...Ch. 14 - Prob. 10SQCh. 14 - Does the phenotype indicated by the red circles...Ch. 14 - G6PD deficiency is an X-Linked recessive disorder....Ch. 14 - Marian syndrome (Section 13.5) is inherited in an...Ch. 14 - Duchenne muscular dystrophy, which is inherited in...Ch. 14 - Human females have two X chromosomes (XX); males...Ch. 14 - A mutation on an autosome causes a particular...Ch. 14 - Expression of the SRY gene on the Y chromosome...Ch. 14 - The somatic cells of most individuals with Down...Ch. 14 - Mutations in the genes for clotting factor VIII...
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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
- A pedigree lists a father as the proband for a genetic disorder that he inherited from his mother. Out of his 3 children, his son inherits the condition but his two daughters do not. What can you hypothesize about this disorder? a) It is an X-linked recessive disorder. b) His wife is also a carrier of an affected X-chromosome, too. c) His son had to inherit a defective X allele from his mother since he obviously received a Y from Dad. d) Although his daughters inherited the father's mutant X allele, a wt X allele from their mother prevented them from inheriting the disorder. e) All the answers could be correct.arrow_forwardConsider a couple: a woman who is homozygous for a recessive mutation that causes X-linked colorblindness, and a man with full color vision (he does not carry a copy of the mutation). a) What is the probability that a son of this couple will be colorblind? b) What is the probability that a daughter of the couple will be colorblind?arrow_forwardAlleles for Tay-Sachs disease are inherited in an autosomal recessive pattern. Why would two parents with a normal phenotype have a child with Tay-Sachs? a. Both parents are homozygous for a Tay-Sachs allele. b. Both parents are heterozygous for a Tay-Sachs allele. c. New mutations gave rise to Tay-Sachs in the child. d. b or carrow_forward
- You have a true-breeding strain of miniature-winged fruit flies, where this wing trait is recessive to the normal long wings. How would you show whether the miniature wing trait is sex-linked or autosomal?arrow_forwardPhenylketonuria (PKU) is a disease that results from a recessive gene.Suppose that two unaffected parents produce a child with PKU. a. What is the probability that a sperm from the father will contain the PKU allele?b. What is the probability that an egg from the mother will contain the PKU allele?c. What is the probability that their next child will have PKU?d. What is the probability that their next child will be heterozygous for the PKU gene?arrow_forwardPhenylketonuria (PKU) is a disease that results from a recessive gene. Suppose that two unaffected parents produce a child with PKU. a. What is the probability that a sperm from the father will contain the PKU allele? b. What is the probability that an egg from the mother will contain the PKU allele? c. What is the probability that their next child will have PKU? d. What is the probability that their next child will be heterozygous for the PKU gene?arrow_forward
- What is the difference between a monohybrid cross and a dihybrid cross? A. A monohybrid cross involves a single parent, whereas a dihybrid cross involves two parents. B. A monohybrid cross produces a single progeny, whereas a dihybrid cross produces two progeny. C. A monohybrid cross involves organisms that are heterozygous for a single trait, whereas a dihybrid cross involves organisms that are heterozygous for two traits. D. A monohybrid cross is performed only once, whereas a dihybrid cross is performed twice.arrow_forwardFruit flies are very useful model organisms that have been used to study genetics. One mutant recessive trait in fruit flies is called "eyeless" because it causes flies to have no eyes. Cross an eyeless fly with a homozygous normal fly. a. Draw the Punnett square. b. What is the genotyoe ratio of the offspring? c. What is the phenotype ratio of the offspring? d. What genetic problem is this? (monohybrid, dihybrid or multiple allele)arrow_forwardThe pedigree below shows the inheritance of fur color in rabbits. What is the mode of inheritance for the black fur color trait? a) x-linked dominant b)autosomal recessive c)autosomal dominant d) x-linked recessivearrow_forward
- Locus heterogeneity means that a genetic disordera. has a heterogeneous phenotype.b. is caused by mutations in two or more different genes.c. involves a structural change in multiple chromosomes.d. is inherited from both parentsarrow_forwardHow is it possible Jonathan has the dominant disorder, Huntington’s Disease, if none of his family had it? a) The mutant allele remained ‘hidden’ in previous generations. b) He is homozygous for mutant huntingtin alleles unlike his grandparents, who are clearly heterozygous for the condition. c) The appearance of Huntington Disease in him might be the result of a new mutation appearing during his lifetime. d) His parents and grandparents were ashamed to admit they had it.arrow_forwardBecause red-green colour-blindness is an X-linked recessive condition, which of the following situations is not possible? a) a colourblind father passes the condition to this daughter b) a colourblind father passes the condition to his son c) a heterozygous mother passes the condition to her daughter d) a heterozygous mother passes the condition to her son Red-green colour blindness is a deficiency of colour vision so that a person affected by it cannot tell the difference between red and green. This is an X-linked recessive condition. Which statement is correct? a) The allele is written as X^r, and an affected female is heterozygous. b) The allele is written as X^r; an affected male is X^rY and a heterozygous female is X^RX^r c)The allele is written as X^r and a male with genotype X^RY is affected d) The allele is written as X^R; a normal male is X^RY and a homozygous recessive female is X^rX^r.arrow_forward
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