Question 2:Consider the following pedigree, which traces the inheritance of a single-gene hereditary disease.Only individuals that are either affected or normal are shown. In other words, potentialheterozygotes are NOT indicated. Assume that there is complete penetrance and no variableexpressivity, and no new mutations.2IIIIIIVCharacterize each of the following modes of inheritance as impossible or possible. Justify youranswers.a.Autosomal recessiveb.Autosomal dominantC.X-linked recessiveX-linked dominantY-linkedSex-limitedSex influenced, dominant in malesSex influenced, dominant in femalesMitochondrial inheritanced.e£.ůh.1.

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Description: Question 2:Consider the following pedigree, which traces the inheritance of a single-gene hereditary disease.Only individuals that are either affected or normal are shown. In other words, potentialheterozygotes are NOT indicated. Assume that there i

Introduction :- A genetic trait or condition can be passed down from parent to child via autosomal…

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Question 2: Consider the following pedigree, which traces the inheritance of a single-gene hereditary disease. Only individuals that are either affected or normal are shown. In other words, potential heterozygotes are NOT indicated. Assume that there is complete penetrance and no variable expressivity, and no new mutations. I II 5 IV Characterize each of the following modes of inheritance as impossible or possible. Justify your answers. Autosomal recessive b. Autosomal dominant C. X-linked recessive X-linked dominant Y-linked Sex-limited Sex influenced, dominant in males Sex influenced, dominant in females d. e £. 2 h. i. Mitochondrial inheritance

Question 2: Consider the following pedigree, which traces the inheritance of a single-gene hereditary disease. Only individuals that are either affected or normal are shown. In other words, potential heterozygotes are NOT indicated. Assume that there is complete penetrance and no variable expressivity, and no new mutations. I II 5 IV Characterize each of the following modes of inheritance as impossible or possible. Justify your answers. Autosomal recessive b. Autosomal dominant C. X-linked recessive X-linked dominant Y-linked Sex-limited Sex influenced, dominant in males Sex influenced, dominant in females d. e £. 2 h. i. Mitochondrial inheritance

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter12: Genes And Cancer

Section: Chapter Questions

Problem 2CS: Mike was referred for genetic counseling because he was concerned about his extensive family history...

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question- Color blindness is inherited as a sex-linked recessive disease. A normal male marries a female who is heterozygous for the trait. What percentage of their sons will exhibit color blindness?
Question 1: In tomatoes, round fruit (O) is dominant to oblong fruit (o). Write as much as possible of the genotypes of: (A) A plant from a homozygous round-fruited stock (B) A plant from an oblong strain (C) A round-fruited plant which resulted from a testcross (D) A oblong-fruited plant which resulted from the cross of two round-fruited ones.
Question 6: If individuals IV-4 and IV-5 mate and have a child, what is the probability that the child will have the genetic condition? Assume people mating into the pedigree do not carry the abnormal allele unless information given in the pedigree suggests otherwise. Write the probability as a fraction without any spaces, answers given as a decimal will be marked incorrect. (For example write your answer as 5/72 – note that is NOT the correct answer).
Question 6 Albinism is caused by an autosomal recessive allele that interferes with skin pigmentation in mammals. Two normally pigmented human parents already have an albino boy. They plan to continue to have children until they get a girl. Some or all of this information is important for each of the questions below. a) What is the probability that their next child (currently unborn) will be a girl with albinism? Explain your reasoning. b) What is the probability their first female child will be albino? Explain your reasoning. c) The answer to part (b) is different (and, yes, the answer is different) from the answer to part (a). Explain why. (Hint: it has something to do with the underlined words.)
A couple who are about to get married learn from studying their family histories that, in both their families, theirunaffected grandparents had siblings with cystic fibrosis(a rare autosomal recessive disease).a. If the couple marries and has a child, what is theprobability that the child will have cystic fibrosis?b. If they have four children, what is the chance that thechildren will have the precise Mendelian ratio of 3:1 fornormal:cystic fibrosis?c. If their first child has cystic fibrosis, what is theprobability that their next three children will be normal?
Neurofibromatosis-1 (NF1) is an autosomal dominant disorder where tumours form in the base layer of the skin or in nerve tissues. What is the probability that individuals II-1 and II-2 will have a genetic son with NF1? Find the image attached.
Question #1 a. Explain why there is a non-Mendelian 2:1 yellow/agouti phenotypic ratio in this cross. b. What is the Raly gene? What does it encode? How does it relate to the lethality of Ay? c. A heterozygous mouse AAy does not produce enough pigment and it is yellow. Homozygous individuals die during development, but AAy heterozygous do not. Why? You may use any graphic aid to explain this, if necessary. This means that you may use a Punnett square or a forking segregation diagram, or computations with fractions.
Question:- Based on your selected mode of inheritance, show the genotypes for the following individuals. [Use these symbols for alleles: if it is autosomal, then use the symbols B - dominant, b - recessive (e.g. BB, bb etc.) if it is X-Linked, then X(B) - dominant, X(b) - recessive, and Y for Y-chromosome (e.g. X(B)X(B), X(B)Y etc.) ] I-1 I-2 II-7 II-8 III-10 III-11 III-12 IV-8 IV-9
Question 1 refers to the pedigree chart in Figure 11.2 for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. 1) What is the genotype of individual II-5? What are the genotypes of individuals II-2, 3, and 4 and II-6, 7, and 8?
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. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
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?