Both hemophilia (h) and favism (gd) are inherited as X-linked recessive traits. Hemophilia is an inherited disorder of blood clotting, and favism is an inherited hemolytic anemia caused by absence of the enzyme glucose-6-phosphate dehydrogenase. A phenotypically normal woman is known to have the X chromosome genotype h + / + gd. The frequency of recombination between h and gd is 16%. What proportion of sons born to this woman are expected to be phenotypically normal with respect to both hemophilia and favism?
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Both hemophilia (h) and favism (gd) are inherited as X-linked recessive traits. Hemophilia is an inherited disorder of blood clotting, and favism is an inherited hemolytic anemia caused by absence of the enzyme glucose-6-phosphate dehydrogenase. 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?Given the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the phenotype of this individual be?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?Three autosomal genes are linked along the same chromosome.The distance between gene A and B is 7 mu, the distance betweenB and C is 11 mu, and the distance between A and C is 4 mu. Anindividual that is AA bb CC was crossed to an individual that is aaBB cc to produce heterozygous F1 offspring. The F1 offspring werethen crossed to homozygous aa bb cc individuals to produce F2offspring.Where would a crossover have to occur to produce an F2 offspringthat was heterozygous for all three genes?Three autosomal genes are linked along the same chromosome.The distance between gene A and B is 7 mu, the distance betweenB and C is 11 mu, and the distance between A and C is 4 mu. Anindividual that is AA bb CC was crossed to an individual that is aaBB cc to produce heterozygous F1 offspring. The F1 offspring werethen crossed to homozygous aa bb cc individuals to produce F2offspring.If we assume that no double crossovers occur, what percentageof F2 offspring is likely to be homozygous for all three genes?
- Three autosomal genes are linked along the same chromosome.The distance between gene A and B is 7 mu, the distance betweenB and C is 11 mu, and the distance between A and C is 4 mu. Anindividual that is AA bb CC was crossed to an individual that is aaBB cc to produce heterozygous F1 offspring. The F1 offspring werethen crossed to homozygous aa bb cc individuals to produce F2offspring.Draw the arrangement of the alleles on the chromosomes in theparents and in the F1 offspring.Congenital hypertrichosis (CH) is a very rare X-linked dominant inherited condition. CH is characterized by the growth of dark hair over the body. CH is so rare, only 50 cases have been identified since the Middle Ages. The incidence of this condition is considerably higher in a small Mexican village (from which the partial pedigree below is derived) than the rest of the human population. I II III Use the following information to answer the two questions. IV D II-4 8 9 IV-6 0=10~ 11 1. Using appropriate nomenclature, identify the genotypes for the following 2 individuals: 12 13 your response must include an appropriate legend/key to identify allele symbols. 2. Show how a Punnett square (using the allele symbols from the previous question) is used to determine the probability in percent of individuals III-11 and III-12 next offspring has CH?Hemophilia is caused by an X-linked recessive mutation in humans. If a man whose paternal uncle (father's brother) was a hemophiliac marries a woman whose brother is also a hemophiliac, what is the probability that their first child will have hemophilia? (Assume that no other cases of hemophilia exist in the pedigree.) 1/3 0 1/8 0 1/4 1/2
- In Drosophila, the white gene located on the X chromosome affects eye color; an autosomal gene, wingless, is on an autosomal chromosome. Use the following allele symbols: Xw+ _ , Xw+Y = wild type red eyes; X-linked dominant allele Xw Xw , XwY = white eyes; X-linked recessive allele Y = Y sex chromosome vg+ = wild type wings; autosomal dominant vg = wingless; autosomal recessive Predict ratios/proportions of genotypes and phenotypes of offspring from the following cross, of a white-eyed male with wild type wings and a wild type red eyed female with wild type wings: indicate sex of offspring along with phenotypes. XwY vg+ vg x Xw+Xw vg+vgA tomato geneticist attempts to assign five recessivemutations to specific chromosomes by using trisomics.She crosses each homozygous mutant (2n) with each ofthree trisomics, in which chromosomes 1, 7, and 10 takepart. From these crosses, the geneticist selects trisomicprogeny (which are less vigorous) and backcrosses themto the appropriate homozygous recessive. The diploidprogeny from these crosses are examined. Her results, inwhich the ratios are wild type:mutant, are as follows:Which of the mutations can the geneticist assign towhich chromosomes? (Explain your answer fully.)Bar (B) is a dominant sex-linked mutant of D. melanogaster. Held-out wings and ebony body are recessive, autosomal mutants on chromosome III, mapping 12 map units apart. A Bar-eyed, ebony-bodied male was crossed to a held-out female, and the resulting F1 progeny WERE INBRED to produce the F2. At what frequency do you expect wild-type flies from this cross? (Reminder: Crossing over does not occur in the male). Based on the information in Problem 3, how often would you get a bar ebony but not held-out progeny?