36. Look at this pedigree and circle each row yes/no. Any assumption (that is not outrageous) can be made), for example, can assume someone is a carrier. a. Could this trait be inherited as a simple autosomal recessive? YES oo 8995 ON b. Could this trait be inherited as a simple autosomal dominant? YES ON C. Could this trait be inherited as a simple x linked recessive? YES ON d. Could this trait be inherited as a simple x linked dominant? YES ON e. Could this trait be inherited as a Y linked? YES ON

36. Look at this pedigree and circle each row yes/no. Any assumption (that is not outrageous) can be made), for example, can assume someone is a carrier. a. Could this trait be inherited as a simple autosomal recessive? YES oo 8995 ON b. Could this trait be inherited as a simple autosomal dominant? YES ON C. Could this trait be inherited as a simple x linked recessive? YES ON d. Could this trait be inherited as a simple x linked dominant? YES ON e. Could this trait be inherited as a Y linked? YES ON

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter16: Human Genetics And The Human Genome

Section: Chapter Questions

Problem 10TYU: Examine the following pedigrees. Which is the most likely mode of inheritance of each disorder? (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 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?
Examine the following pedigrees. Which is the most likely mode of inheritance of each disorder? (a) autosomal recessive (b) autosomal dominant (c) X-linked recessive (d) a, b, or c (e) a or c 10.
Analysis of X-Linked Dominant and Recessive Traits As a genetic counselor investigating a genetic disorder in a family, you are able to collect a four-generation pedigree that details the inheritance of the disorder in question. Analyze the information in the pedigree to determine whether the trait is inherited as: a. autosomal dominant b. autosomal recessive c. X-linked dominant d. X-linked recessive e. Y-linked
3) You have sought a job at a human genetic counseling clinic. They gave you an aptitude test shown below. Answer the yes/no questions concerning each of the following pedigrees. Pedigree A: YES or NO Could this trait be inherited as a simple autosomal recessive? Could this trait be inherited as a simple autosomal dominant? Could this trait be inherited as a simple X-linked recessive? Could this trait be inherited as a simple X-linked dominant? Pedigree B: YES or NO Could this trait be inherited as a simple autosomal recessive? Could this trait be inherited as a simple autosomal dominant? Could this trait be inherited as a simple X-linked recessive? Could this trait be inherited as a simple X-linked dominant?
Retinitis pigmentosa, a form of blindness in man, maybe caused either by a dominant autosomal gene R, or a recessive autosomal gene a. An afflicted man whose parents are both normal marries a woman with genotype AaRr. a. What proportion of the children are expected to suffer from this affliction if R and A are inherited independently? b. If this couple want to have normal children only, what isthe probability of having normal children?
Consider the following pedigree. 하 3 10 (5 3 2 (a) What pattern of transmission is most consistent with this pedigree? (1) autosomal recessive, (2) autosomal dominant, (3) X-linked recessive, (4) X-linked dominant. (b) If individual V-2 marries a normal individual, and if the condition has a pene-trance of 85 percent, what is the probability that their second child will express the trait? (c) On the third line, what does the diamond with a 10 in the middle mean?
. The pedigree below was obtained for a rare kidney disease.a. Deduce the inheritance of this condition, stating your reasons. b. If persons 1 and 2 marry, what is the probability that their first child will have the kidney disease?
The next four questions are all related to this problem: Polydactyly (PD) is an autosomal dominant trait (polydactyly - P; wildtype - p). Cystic fibrosis (CF) is an autosomal recessive trait (cystic fibrosis - f; wildtype - F). A PD woman, otherwise normal in phenotype, marries a healthy normal man. Their 4 children are: 1) normal, 2) PD, 3) CF, 4) CF + PD. You will walk through a series of steps to answer this question: What is the probability that their 5th child will have at least one of these conditions? Here is the first step: 1. What is the cross? (Hint: You can use the 4 existing children to determine the genotypes of the parents.) O PpFF (female) x ppFF (male) O PoFf (female) x ppft (male) O pof female) x PPFI (male) O PPFF Ifemale) x ppFf (male) 2. What is/are the target genotypes? O pof OP.F. O pott O P.M 3. What is the probability the child will have PD AND cystic fibrosis? Answer to two decimal places (eg. 0.88). 4. What is the probability that their 5th child will have at…
The next four questions are all related to this problem: Polydactyly (PD) is an autosomal dominant trait (polydactyly = P; wildtype = p). Cystic fibrosis (CF) is an autosomal recessive trait (cystic fibrosis = f; wildtype = F). A PD woman, otherwise normal in phenotype, marries a healthy normal man. Their 4 children are: 1) normal, 2) PD, 3) CF, 4) CF + PD. You will walk through a series of steps to answer this question: What is the probability that their 5th child will have at least one of these conditions? Here is the first step: 1. What is the cross? (Hint: You can use the 4 existing children to determine the genotypes of the parents.) PFF (female) x ppFF (male) PpFf (female) x ppFf (male) O ppff (female) x PPFF (male) O PPFF (female) x ppFf (male)
Color 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?
15. 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 Hinhetee