Holstein cattle suffers from the condition citrullinemia in which homozygus recessive calves die within a week of birth bc they can not break down ammonia that is produced when amino acids are metabolized. If a cow that is heterozygous for the citrullinemia gene is inseminated by a bull that is homozygous dominant, ehat is the probability that a calf inherits citrullinemia ( autosomal recessive)
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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?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?
- An allele responsible for Marfan syndrome Section 13.4 is inherited in an autosomal dominant pattern. What is the chance that a child will inherit the allele if one parent does not carry it and the other is heterozygous?Albinism is caused by an autosomal recessive allele of a single gene. An albino child is born to phenotypically normal parents. However, the paternal grandfather is albino. Exhaustive analysis suggests that neither the mother nor her ancestors carry the allele for albinism. Suggest a mechanism to explain this situation.Both Duchenne muscular dystrophy and color blindness are caused by recessive alleles. DMD, unlike color blindness, nearly always occurs in males. Explain why.
- Redgreen color blindness is an X-linked recessive disorder in humans. Your friend is the daughter of a color-blind father. Her mother had normal color vision, but her maternal grandfather was color-blind. What is the probability that your friend is color-blind? (a) 1 (b) (c) (d) (e) 0In humans, ABO blood types refer to glycoproteins in the membranes of red blood cells. There are three alleles for this autosomal gene: IA, IB, and i. The IA allele codes for the A sugar, The IB allele codes for the B sugar, and the i allele doesn't code for any sugar. IA and IB are codominant, and i is recessive to both IA and IB. If an individual with type AB blood has a child with an individual with type O blood, what blood types could their children possibly have?In humans, the gene for albinism (a) is recessive to the gene for normal skin pigmentation (A). If two heterozygous persons have children, what is the probability they will have a child who is an albino? What is the probability they will have a child that is a carrier of the recessive gene?
- Another type of monohybrid inheritance involves the expression of both phenotypes in the heterozygous situation. This is called codominance. One of the best-known examples of codominance occurs in the blue Andalusian chicken. "Blue" birds are heterozygous (BW) and result from the mating between a black bird (BB) and a white bird (WW). Blue birds do not really have blue feathers, instead having a mixture of black and white feathers that reflects light to appear blue. Thus, the "blue" coloration is not a consequence of blending of the pigments (after all, the mix is not gray) but rather the result of both colors existing on the same bird. That is both phenotypes occur on the same individual. a. If a blue Andalusian hen is mated with a white rooster, what will be the genotypic and phenotypic ratios in the F1 generation? Genotypic ratio- Phenotypic ratio- b. List the parental genotypes of crosses that could produce at least some: White offspring- Black offspringA dominant genetic factor P is required for the formation of an enzyme needed to convert phenylalanine to tyrosine. The recessive allele “p” result in lack of the enzyme and as a consequence phenylketonuria (PKU) can ensure. A woman who was spared the dire effect of PKU as a result of a special diet during infancy and childhood is married to a man who does not have PKU but does carry the recessive allele. During her pregnancy the woman is placed on a special diet once more, since excess phenylalanine and other toxic product can cross the placenta in a PKU women and resulted in mental retardation in the new born. What geneotype are to be expected among the off spring in this case and what would be their phenotype?A man, Penoy, whose sister died in early childhood from a recessive lethal disease marries a woman Esmae, who has the same family history. Because Penoy has survived beyond childhood, he does not have the disease, but he may be a carrier (i.e. heterozygous, as may also be the case with Esmae). What is the probability that their first child will suffer from the disease? [Hint: first calculate the probability that Penoy is heterozygous; then determine the probability that both parents are carriers. Remember that he has survived to adulthood when calculating this probability].