ped 2.gif In the following pedigree, the indicated trait is caused by which type of allele? %3D II IV autosomal recessive autosomal dominant O X-linked recessive O X-linked dominant O Y-linked
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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?In addition to codominance, the ABO blood group antigens are also an example of ________. incomplete dominance X-linked recessive inheritance multiple alleles recessive lethal inheritance
- 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?Marian syndrome (Section 13.5) is inherited in an autosomal dominant pattern. What is the chance that a child will inherit the associated allele if one parent does not carry it and the other is heterozygous?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
- Does the phenotype indicated by the red circles and squares in this pedigree show an inheritance pattern that is autosomal dominant, autosomal recessive, or X-linked?Pedigree 3: What is the most likely mode of inheritance of this disease given its pattern on the pedigree? Autosomal Dominant O Autosomal Recessive O X-Linked Dominant O X-Linked RecessivePedigree 2: A. What is the most likely mode of inheritance of this disease? Choose from: autosomal dominant, autosomal recessive, X-linked dominant, X-linked recessive. B State the genotypes of individuals # 1 #4. C If individual #3 has another daughter with the same partner, what is the probability that this daughter will be affected (show the disease)?
- Multiple Alleles- Blood Types 1. Blood Type is controlled by 3 alleles: What are they? allele is recessive alleles are codominant, and in homozygous condition results in O blood group. 2. and 3. What are the genotypes possible for a person who has: A blood? B blood? O blood? AB blood? 4. A man with type AB blood is married to a woman also with type AB blood. Show the cross. What proportion of their children will have: A blood? B blood? O blood? AB blood? 5. A man has type B blood (genotype I"I") is married to a woman with type O (ii) blood. Show the cross. What proportion of their children will have: 1. A blood? 2. B blood? 3. O blood? 4. AB blood? 6. A woman with type A blood (genotype IAi) is married to a type B person (genotype I®i). Show the cross. What proportion of their children will have: A blood? B blood? O blood? AB blood?BB AB AB A. None B.3 с.1 D.2 ВВ OE. The father AB In the pedigree presented above, an autosomal dominant disease which causes significant visual loss and eventual blindness, is segregating in the family. The disease gene causing this sight loss looks as if it may be linked to a marker locus. The alleles of this marker locus that are present in this family are allele A and allele B. Are there any recombinant individuals in this pedigree? AB ВВ AB ВВ AB ВВ AB AB ВВtion 8: below is the pedigree of inheritance of phenylketonuria (PKU). We will designate the letter Caven for the dominant allele and "p" for the recessive allele. 4 The pedigree shows that the pattern of inheritance for the allele for phenylk ylketonuria is: I. II. 1 III. IV. Autosomal dominant Autosomal recessive X-linked dominant X-linked recessive b. The parents in generation I have how many children: I. 3 Boys II. 3 Girls III. IV. 3 Boys and 1 Girl 3 Girls and 1 Boy c. What is the genotype of individual 1 in generation III: I. PP II. pp III. Pp " O 1 III. 50% E III 1 ▬ 2 2 IV. 25% 1 3 IV. Can be PP or Pp ii. Suppose that a man having type AB blood marries a woman having type O blood. What is the probability that their child will have type A blood? I. 100% II. 75% 2 4 3