EBK BIOLOGY
5th Edition
ISBN: 9780100667976
Author: Maier
Publisher: YUZU
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Chapter 8, Problem 9LTB
Summary Introduction
Introduction:
The pedigree analysis referred to a diagrammatic representation of the genotypes and the
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The pedigree shows inheritance of an autosomal recessive disease in an extended family. Assume unrelated individuals marrying into the
family do not carry the disease, unless there is reason to believe otherwise.
What is the chance that IV-3 and IV-4 will have a child with the disease?
Individuals I-1, Il-5, III-5 and III-16 have the disease.
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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.
Rabbits may be classified as agouti, chinchilla, Himalayan, or albino according to coat color. A cross between CC h x C ch c produced 5 agouti, 3 chinchilla and 2 Himalayans.
a. What are the phenotypes of the parent rabbits? Copy the genotypes then write the corresponding phenotype beside each.
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 theiroffspring will have the same phenotype?
e. What will be the genotypes of the rabbits in (d)?
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- 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_forward1 II 1 4 6. II 1 2 4 5 6. 8. IV The above image shows a pedigree for a monogenic inherited disease. Although this trait is only observed in males in this family, the pattern of inheritance of this disease is autosomal recessive. Use the pedigree to explain why the inheritance of this disease cannot be autosomal dominant. If this trait is X-linked recessive, what would be the genotypes of the people in Row I? 2. 2. 3. 2. 3.arrow_forwardThe 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 affectedarrow_forward
- A 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_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_forwardA 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_forward
- Rabbits 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_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_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_forward
- Webbed fingers is inherited as an X-linked disease An unaffected male marries an affected female. a. Draw a Punnett square of the possible offspring. b. List the phenotypes of the possible children c. Draw a pedigree that displays the inheritance in you Punnett squarearrow_forwardA man who has color blindness and type O blood has children with a woman who has normal color vision and type AB blood. The woman’s father had color blindness. Color blindness is determined by an X-linked gene, and blood type is determined by an autosomal gene. a. What are the genotypes of the man and the woman? b. What proportion of their children will have color blindness and type B blood? c. What proportion of their children will have color blindness and type A blood? d. What proportion of their children will be color blind and have type AB blood?arrow_forwardWhich 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_forward
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