Biology: Science for Life with Physiology (5th Edition)
5th Edition
ISBN: 9780321922212
Author: Colleen Belk, Virginia Borden Maier
Publisher: PEARSON
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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 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?
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a. What is the type of inheritance?
b. What is known of the genotype of the male in the above cross?
c. What is known of the genotype of the female in the above cross?
d. Provide map distances if possible.
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?
Chapter 8 Solutions
Biology: Science for Life with Physiology (5th Edition)
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- In 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_forwardUsing the pedigree chart, explain: a) The number of generations seen. b) If all blue-coloured shapes are affected with disease X- how many males are affected? how many females are affected? c) Does this disease have a dominant or recessive inheritance pattern? Justify your answer.arrow_forwardWebbed 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_forward
- 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_forwardThe pedigree below shows a family affected by a disease. Assume that the individuals marked with an asterisk (*) do not carry any allele associated with the affected phenotype, no other mutation spontaneously occurred, and complete penetrance. Answer the following questions below. Use the notation XR for the allele associated with the dominant phenotype and Xr for the allele associated with the recessive phenotype. Q1) Give the genotypes for as many individuals in the pedigree as possible.arrow_forwardIn this figure, the black circles and squares indicate a genetic disease. A. Based on the pedigree in the figure, does this gene appear to be inherited in a dominant or recessive manner? Explain your prediction. B. Use the letter a to indicate a disease allele and the letter A to indicate a normal allele. Predict the genotypes of the original parents. C. What percentage of the offspring of the original parents would you expect to have the disease? D. Predict the genotype of the partner of the diseased daughter in generation II. Justify your prediction.arrow_forward
- Phenylketonuria (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_forwardDuchenne muscular dystrophy is sex linked and usually affects only males. Victims of the disease become progressively weaker, starting early in life.a. What is the probability that a woman whose brother has Duchenne’s disease will have an affected child?b. If your mother’s brother (your uncle) had Duchenne’s disease, what is the probability that you have received the allele?c. If your father’s brother had the disease, what is the probability that you have received the allele?arrow_forward
- Glucose-6-phosphate dehydrogenase (G6PD) deficiency is a disorder that affects the normal function of red blood cells and can eventually lead to anemia. The trait is controlled by a recessive allele found on 4. the X chromosome. An affected son was born to a man and woman who were unaffected. The woman's mother was affected while the father was normal. a. Indicate the gene notation. b. Give the genotypes of the affected boy's parents. Derive the genotypic and phenotypic ratios for the offspring. Show and label your solutions properly. What is the probability that they will have a phenotypically normal daughter as their first child? C. d.arrow_forwardFor the pedigree shown here, the disorder is caused by a recessive (g) allele on the X chromosome. Label each of the following individuals with the correct genotype (XGY, XgY, XGXG, XGXg, XgXg). a. Generation I, number 2 b. Generation II, number 2 c. Generation II, number 5 d. Generation III, number 2 e. Generation III, number 6arrow_forwardConsider the following linked traits in fruit flies: V= vermillion eyes (v+ is red eyes) cv = crossveinless wings (cv+ have crossveins) ct = cut wing margins (ct+ have uncut margins) You perform the following cross: v+v+ cv+cv+ ct+ct+ x vv cvcv ctct You then cross an F1 fly with a fly recessive for all three traits: v+v cv+cv ct+ct x vv cvcv ctct These are the offspring you observed from this cross: Red, crossveins, uncut 580 Vermillion, crossveinless, cut 592 Red, crossveinless, cut 89 Vermillion, crossveins, uncut 94 Red, crossveinless, uncut 45 Vermillion, crossveins, cut 40 Red, crossveins, cut 3 Vermillion, crossveinless, uncut 5 Draw a map of these 3 genes on the fly chromosome they inhabit.arrow_forward
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