Biology: Science for Life with Physiology (6th Edition) (Belk, Border & Maier, The Biology: Science for Life Series, 5th Edition)
6th Edition
ISBN: 9780134555430
Author: Colleen Belk, Virginia Borden Maier
Publisher: PEARSON
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Chapter 8, Problem 10LTB
Summary Introduction
Introduction:
Sex-linked recessive disorders pass down in a family from X and Y sex chromosomes. In such disorders, if only one gene is abnormal, then the disease will not occur. The person carrying on one abnormal gene is the carrier of the disease. Mostly such disorders are X-linked traits.
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The chart below is showing 4 generations of a family that is affected by a hereditary disease.
a. Is the disorder being tracked dominant or recessive? How do you know?
b. There is only one possible genotype for person C. True or False?
c. What are the possible genotypes for person A?
d. What are the possible genotypes for person B?,
e. If two people with the same genotypes as person C's spouse and person A's
spouse had a child, what is the probability that the child will be affected by this genetic disorder?
(draw a Punnett square using the correct genotypes to help you).
% chance offspring will be affected
% chance offspring will not be affected
Match the pattern of inheritance to the appropriate term.
A. Heterozygotes with different alleles of the DTL1 gene survive better than homozygotes B. The disease is usually passed from a mother to all children
C. Pure-breeding pumpkin plants grown on sandy soil have bigger seeds than plants of the same pure-breeding lineage grown on peaty soil.
D. A cross between a true-breeding plant with serrated leaf edges and a true- breeding plant with smooth leaf edges produces an F1 generation with 88% plants with serrated leaves.
E. A cross between two heterozygotes, DdFf, produces offspring in three phenotypic groups; Long wings in offspring with genotype D_F_, short wings in offspring with D_ff, and no wings in offspring with the genotypes ddF_ and ddff.
F. Individuals homozygous for a mutation in SME have a heightened sense of smell and large earlobes.
G. Grey chickens bred together have offspring that are black, grey and white in a 1:2:1 ratio
H.Alleles at locus S…
Hemophilia is a disease caused by a gene found on the X chromosome. Therefore, it is a sex-linked disease which is caused by the recessive allele. Suppose, a man with hemophilia marries a woman who is homozygous dominant for the trait, what is the chance of having children who are hemophiliacs? (XH = normal; Xh= hemophiliac) *
a. 0%
b. 75%
c. 100%
d. 25%
Chapter 8 Solutions
Biology: Science for Life with Physiology (6th Edition) (Belk, Border & Maier, The Biology: Science for Life Series, 5th Edition)
Ch. 8 - What is the relationship between genotype and...Ch. 8 - Add labels to the figure that follows, which...Ch. 8 - Prob. 3LTBCh. 8 - Prob. 4LTBCh. 8 - Prob. 5LTBCh. 8 - Scientists have recently developed a process by...Ch. 8 - What is the physical basis for the independent...Ch. 8 - Prob. 8LTBCh. 8 - Prob. 9LTBCh. 8 - Prob. 10LTB
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- Which of the following must be true about the inheritance the trait depicted in the pedigree diagram below. A. it is recessive B. It is dominant C. It is on the X chromosome D. There is not enough information to determine the mechanism of inheritancearrow_forwardA heterozygous individual is crossed with a homozygous recessive individual. a. Draw a Punnett square to represent this cross. b. What is the probability that an offspring will have a homozygous genotype? c. What is the probability that an offspring will have a dominant phenotype? d. What is the probability that three offspring will be produced that all carry the recessive allele but do not express the recessive phenotype?arrow_forward15. The following pedigree shows inheritance of Huntington's disease, a fatal genetic disorder that causes neurodegeneration. Since signs and symptoms usually do not appear until adulthood, many who are carriers may not realize their risk of passing on the disease-causing allele. The following pedigree represents a family in which some people are affected by Huntington's disease. Reeessive Trit er btmnt be Mec yplicalty Hinheteearrow_forward
- Colorblindness is inherited as an X-linked recessive trait while pattern baldness is controlled by an autosomal gene that is dominant in males but recessive in females. A colorblind man who is also homozygous for baldness has children with a woman who carries normal genes for both traits. What is the probability that any of their child will be: a. Colorblind, bald male b. Colorblind, normal-haired male c. Female with normal sight and baldarrow_forwardTay–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.arrow_forwardUsing the pedigree chart attached: Above is a pedigree for colorblindness. Based on the pedigree, is the disease dominant or recessive and is it sex-linked or autosomal? Why? Furthermore, what is the probability that 18 on this chart is affected but the condition, and what is the probability that 18 is a carrier? Why? Are the probability of being a carrier and an affected individual different? Why?arrow_forward
- Which of the following is indicated by roman numerals in a pedigree? a.Presence of the studied trait b.Generation c.Sex d.Marriage status Which of the following disorders in humans has an autosomal dominant inheritance pattern? a.Albinism b.Hemophilia c.Tay-Sachs disease d.Huntington’s disease For an X-linked recessive allele, what proportion of female offspring will be carriers in the cross of an affected father and a noncarrying mother? a.50 percent b.0 percent c.100 percentarrow_forwardWhich of the following cannot be true about the inheritance the trait depicted in the pedigree diagram below? A. It is recessive B. It is dominant C. It is on the X chromosome D. There is not enough information to determine the mechanism of inheritancearrow_forwardWhat are the genotypes of the mother and father in the pedigree below? The pedigree shows albinism (an autosomal recessive trait). The shaded shapes represent albino individuals. a. mother - homozygous dominant; father - homozygous dominant b. mother - heterozygous; father - homozygous recessive c. mother - homozygous recessive; father - heterozygous d. mother - heterozygous; father - heterozygous e. mother - homozygous dominant; father - homozygous recessivearrow_forward
- A woman with a rare autosomal recessive disorder was told that it was unlikely that her children would have the disorderas her husband did not have it. However, her first child has the disorder. a. What is the most likely explanation? b. Diagram the cross between the woman and her husband using a Punnett square, give the genotypic ratio (GR) and phenotypic ratio (PR) from the Punnett square. c. Based on the Punnett square results, what is the chance that her next child will have the disorder?arrow_forwardRabbits may be classified as agouti, chinchilla, Himalayan, or albino according to coat color. A crossbetween CC^h x C^ch c produced 5 agouti, 3 chinchilla and 2 Himalayans. a. What are the phenotypes of the parent rabbits? b. What are the genotypes of the F1s? c. What mode of inheritance is exhibited? d. If the two F1 agouti genotypes will be crossed, what percentage of their offspring will have the same phenotype? e. What will be the genotypes of the rabbits in (d)?arrow_forwardIn humans, the genes for coloblindedness and hemophilia re both located on the X chromosome with no corresponding gene in the Y. These are both recessive alleles. a. If a man and a woman, both with normal vision, marry and have a colorblind son, draw the Punnet square that illustrates this. b. If the man dies and the woman remarries to a colorblind man, draw a Punnet Square showing the type of children could be expected from hre second marriage. How many/what percentages of each could ne expectedarrow_forward
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