Concept explainers
A
dominant neuromuscular disorder spinocerebellar ataxiatype
Does either group of lod scores indicate statisticallysignificant odds in favor of genetic linkage? Explainyour answer.
What is the maximum value for each set of lod scores?
Based on the available information, is
Based on available information, is
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Genetic Analysis: An Integrated Approach (2nd Edition)
- Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior history of the disease. Consider the following pedigree (the darkly colored symbols represent affected individuals): a. Circle the individual(s) in which the mutation most likely occurred. b. Is the person who is the source of the mutation affected by retinoblastoma? Justify your answer. c. Assuming that the mutant allele is fully penetrant, what is the chance that an affected individual will have an affected child?arrow_forwardAs it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progenyarrow_forwardA 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?arrow_forward
- 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?arrow_forwardA 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?arrow_forwardA woman who sought genetic counseling is found to be heterozygousfor a chromosomal rearrangement between the second andthird chromosomes. Her chromosomes, compared to those in anormal karyotype, are diagrammed on the next page:(a) This woman is phenotypically normal. Does thissurprise you? Why or why not? Under what circumstancesmight you expect a phenotypic effect of such arearrangement?arrow_forward
- Figure 19-18a shows a plot of P values (represented bythe dots) along the chromosomes of the dog genome.Each P value is the result of a statistical test of association between a SNP and body size. Other than the clusterof small P values near IGF1, do you see any chromosomalregions with evidence for a significant association between a SNP and body size? Explainarrow_forwardA genetic disorder is caused by a LOF mutation (Bm) and epigenetic imprinting of gene B. Through pedigree analysis of many families, researchers have observed the following results: Female carrier B+Bm x B+B+ --> B+Bm and Mother's genotype Father's genotype Children's genotype Children's phenotype Cross 1: B+Bm B+B+ B+Bm , B+B+ some affected, some unaffected Cross 2: B+B+ B+Bm B+Bm , B+B+ always unaffected Based on these results, gene B is imprinted on the: a. maternal b. paternalarrow_forwardCampomelic dysplasia (CMD1) is a congenital humansyndrome featuring malformation of bone and cartilage.It is caused by an autosomal dominant mutation of agene located on chromosome 17. Consider the followingobservations in sequence, and in each case, draw whateverappropriate conclusions are warranted.(a) Of those with the syndrome who are karyotypically46,XY, approximately 75 percent are sex reversed,exhibiting a wide range of female characteristics.(b) The nonmutant form of the gene, called SOX9, isexpressed in the developing gonad of the XY male,but not the XX female.(c) The SOX9 gene shares 71 percent amino acid codingsequence homology with the Y-linked SRY gene.(d) CMD1 patients who exhibit a 46,XX karyotypedevelop as females, with no gonadal abnormalities.arrow_forward
- Niemann Pick Type C disease is a recessive disorder that causes the accumulation of cholesterol and other lipids in lysosomes, ultimately affecting both the liver and the nervous system. Below are the genotypes and phenotypes of offspring of a family with a history of Niemann Pick. 7 NN ( all normal phenotype) 3 Nn (all normal phenotype) 4 nn (1 early onset dementia, 1 mid-life onset dementia, 2 late-onset dementia). From this information, Niemann-Pick disease is an example of: A) variable expressivity B) incomplete dominance C) incomplete penetrance D) variable expressivity and incomplete penetrance E) multiple allelesarrow_forward10 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?arrow_forwardA DNA variant has been found linked to a rare autosomal dominant disease in humans and can thus beused as a marker to follow inheritance of the diseaseallele. In an informative family (in which one parentis heterozygous for both the disease allele and the DNA marker in a known chromosomal arrangementof alleles, and his or her mate does not have the samealleles of the DNA variant), the reliability of such amarker as a predictor of the disease in a fetus is related to the map distance between the DNA markerand the gene causing the disease.Imagine that a man affected with the disease(genotype Dd) is heterozygous for the V1and V2forms of the DNA variant, with form V1on the samechromosome as the D allele and form V2on the samechromosome as d. His wife is V3V3dd, where V3isanother allele of the DNA marker. Typing of the fetusby amniocentesis reveals that the fetus has the V2andV3variants of the DNA marker. How likely is it thatthe fetus has inherited the disease allele D if thedistance…arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning