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: The Unity and Diversity of Life (MindTap Course List)

14th Edition

ISBN:9781305073951

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter13: Observing Patterns In Inherited Traits

Section: Chapter Questions

Problem 1SQ: A heterozygous individual has a _______ for a trait being studied. a. pair of identical alleles b....

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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?
A heterozygous individual has a _______ for a trait being studied. a. pair of identical alleles b. pair of nonidentical alleles c. haploid condition, in genetic terms
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?
Now assume that the pedigree shown in question 1 shows the inheritance of a rare genetic disease. a) The disease is most likely autosomal dominant b) The disease is most likely autosomal recessive c) The disease is equally likely to be either autosomal dominant or autosomal recessive, but cannot be x-linked d) Cannot be determined from the information given
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?
1. In this pedigree, if individual 3rd generation, 3rd person female mated with an unaffected male and they have 7 kids, what is the probability that their children will have cystic fibrosis? Check the following that apply to indicate that this is an autosomal recessive trait: A. Seen in both male and females equally B. Seen in more females C. Seen only in males D. Skips generations and/or lineages E. No generations and/or lineages are skipped F. All the above
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?
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?
10 cM separates two hypothetical autosomal human genes. The dominant alleles are have complete penetrance and will result to Crossed eyes (e+) and short thumbs (th+). Four children are born to a normal guy and cross-eyed, small-thumbed woman. Two of the children have short thumbs and the other two have crossed eyes. She is carrying her fifth child. What is the probability that this fifth child will be cross-eyed and have short thumbs?
2a) Identify the pattern of Mendelian inheritance represented in the pedigree above. 2b) Explain how you made this determination.
1. A mother with blood type A has a child with blood type O, and she claims that Mister X is the father. Mister X denies that he could possibly be the father, because he has blood type B. If you were the judge presiding over this case, which of the following arguments would you find most convincing? a. Mr. X could be the father if his blood genotype is IBIB, and the mother’s blood genotype is IAIA b. Mr. X cannot be the father because a child with blood type O cannot have a parent with blood type B c. Mr. X could be the father if his blood genotype is IBi, and the mother’s blood genotype is IAi d. Mr. X cannot be the father because a parent with blood type B cannot have a child with blood type O e. Mr. X cannot be the father because the human ABO blood types are sex-linked traits 2. A monohybrid cross between two pea plants producing yellow peas, resulting in an F1 phenotypic ratio of 3:1 (three yellow pea plants to one green pea plant) indicates what about the parental pea plants?…