Consider the following pedigree showing transmission of an unusual trait: a. What is the most likely mode of transmission? b. Explain your reasoning for your answer to part (a) c. Using the letters R and r to represent the alleles, what is the proband's (the individual marked "p") genotype? d. If the proband (marked "p") marries a normal male and they decide to have 2 children, what is the probability that they will have 2 boys who are both normal? Show your work.

Consider the following pedigree showing transmission of an unusual trait: a. What is the most likely mode of transmission? b. Explain your reasoning for your answer to part (a) c. Using the letters R and r to represent the alleles, what is the proband's (the individual marked "p") genotype? d. If the proband (marked "p") marries a normal male and they decide to have 2 children, what is the probability that they will have 2 boys who are both normal? Show your work.

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 3CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...

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A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What if the couple wanted prenatal testing so that a normal fetus could be aborted?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?
Dihybrid Cross Practice In a breed of dog called a Doberman, black fur is dominant to brown fur and floppy ears are dominant to straight ears. These letters represent the genotypes and phenotypes of the dogs: EE = floppy ears Ee = floppy ears bb = brown fur ee = pointed ears BB = black fur %3D Bb = black fur 1. A female dog with the genotype BBee is crossed with a male dog with the genotype bbEe The square is set up below. Fill it out and determine the phenotypes and proportions in the offspring. Ве Ве Ве Ве bE How many out of 16 have: black fur and floppy ears? be black fur and pointed ears? brown fur and floppy ears? bE brown fur and pointed ears? be
. An RFLP analysis of two pure lines A/A • B/B and a/a • b/bshowed that the former was homozygous for a long RFLPallele (l) and the latter for a short allele (s). The two werecrossed to form an F1, which was then backcrossed to thesecond pure line. A thousand progeny were scored asfollows:a. What do these results tell us about linkage?b. Draw a map if appropriate.c. Incorporate the RFLP fragments into your map.
Considering the genetics cross andobserved phenotypes pictured:a. Which genetic cross(es) led to themost number of healthy progeny?b. What type of model organism islikely being used in these experiments?c. Explain what is likely causing the embryonic lethality based on the observed percentages of embryoniclethality.d. What form of genetic material (maternal, paternal, or both) is leading to the embryonic lethalphenotype?e. Why do you think the second bar has the highest “n” number? What does the “n” value indicate?
LABORATORY EXERCISES IN GENETICS a. On the basis of this pedigree, what do you think is the most likely mode of inheritance for Duchenne Muscular Dystrophy? are most likly made op intentence for Duchenne Miscular Rysty Uale b. If couple III-7 and IlI-8 have another child, what is the probability that the child will have Duchenne Muscular Dystrophy? c. If IlI-2 and IlI-7 mated, what is the probability that one of their children would have the disease?
dd-ons Help в I U A Calibri 12 三三1 |:三 6. Consider a guinea pig with a homozygous genotype and a white fur color phenotype. a. What is the probability this parent will produce a gamete with the dominant allele? b. What is the probability this parent will produce a gamete with the recessive allele? C. If 31 sperm cells are collected from this guinea pig, how many would you expect to have the recessive allele (as determined by sequencing the gene)? !!!
An a cross between two true-breeding plants, Parent #1 with yellow, round peas and Parent #2 with green, wrinkled peas, 1. a. What is the genotype of each plant in the cross. What proportion of the F1 progeny will have the same phenotype as Parent #2. After you self the F1s, what proportion of the progeny will have the same phenotype as Parent #1. C. What proportion of the progeny will have the same genotype as Parent # 1. d. b.
The accompanying pedigree shows a very unusual inheritance pattern that actually did exist. All progeny areshown, but the fathers in each mating have been omittedto draw attention to the remarkable pattern.a. Concisely state exactly what is unusual about thispedigree.b. Can the pattern be explained by Mendelianinheritance?
Purple flowers are dominant to white flowers. Identify the phenotypefor the following genotype Ff, FF, ff and determine if the genotype is heterozygous or homozygous. * For each row, you should select two columns. Purple flowers White flowers Heterozygous Homozygous Ff FF ff Brown eyes are dominant to blue eyes. Identify the phenotype for the following genotype BB, Bb, bb and determine if the genotype is heterozygous or homozygous. * 口ロ口

9. Make a pedigree for each of the following situations. For each individual, write the individual's genotype (when possible) next to the individual's symbol (e.g. O xty, I Gg): a. Two parents do not have cystic fibrosis and they have a daughter with cystic fibrosis and a son who does not have cystic fibrosis. The daughter grows up and she mates with a male who does not have cystic fibrosis. Their only child is a boy and he has cystic fibrosis. b. A man with hemophilia mates with a female without hemophilia. They have one son and one daughter. The daughter has hemophilia and the son does not have hemophilia. The son grows up, and he marries and mates with a female. Their only child is a boy, and he has hemophilia.
PEDIGREE ANALYSIS ANSWER THE FOLLOWING QUESTIONS 1. The pedigree below shows a family's pedigree for colorblindness. Which sex can be carriers of colorblindness and not have it? 2. With this in mind, what kind of trait is colorblindness? 3. Why dies individual IV-7 have colorblindness? 4. Why do all the daughters in generation II carry the colorblind gene? 5. Name two (2) IV generation colorblind males.