PEDIGREE ANALYSIS— please answer all of them, they are all connected 3. This pedigree is for the autosomal dominant trait achondroplasia (use letter A as allele/s), a rareform of dwarfism.II2IIIVa. What is the most convincing genotype of individual (I-2)?b. What is the most convincing genotype of individual (II-1)?c. What is the probability of having a child with achondroplasia IF the couple l-1 and I-2 will have anotherchild?d. What is the probability of individual (IV) of having achondroplasia?e. If the couple (II-1 and II-2) will be having another offspring, what is the probability of having a son withachondroplasia?2.

(According to Bartleby guidelines, only the first three have been answered. Kindly post the…

3. This pedigree is for the autosomal dominant trait achondroplasia (use letter A as allele/s), a rare form of dwarfism. II 2 II IV a. What is the most convincing genotype of individual (I-2)? b. What is the most convincing genotype of individual (III-1)? c. What is the probability of having a child with achondroplasia IF the couple l-1 and I-2 will have another child?

3. This pedigree is for the autosomal dominant trait achondroplasia (use letter A as allele/s), a rare form of dwarfism. II 2 II IV a. What is the most convincing genotype of individual (I-2)? b. What is the most convincing genotype of individual (III-1)? c. What is the probability of having a child with achondroplasia IF the couple l-1 and I-2 will have another child?

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 5QP: Achondroplasia is a rare dominant autosomal defect resulting in dwarfism. The unaffected brother of...

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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
Achondroplasia is a rare dominant autosomal defect resulting in dwarfism. The unaffected brother of an individual with achondroplasia is seeking counsel on the likelihood of his being a carrier of the mutant allele. What is the probability that the unaffected client is carrying the achondroplasia allele?
The genotype XXY corresponds to: a. Klinefelter syndrome b. Turner syndrome c. Tripto-X d. Jacob syndrome
Klinefelter syndrome (XXY) can most be easily diagnosed by _______. a. pedigree analysis. b. aneuploidy c. karyotyping d. phenotypic treatment
Please answer all of them, they are all connected. PEDIGREE ANALYSIS and SYMBOLOGY: Examine the pedigree which has X linked Dominant inheritance of disorder. Use letter X* (asterisk denotes disorder) as genotype of the individuals which can be XX, XY, X*X*, X*X and X*Y. a. What is the genotype of IV-6? b. What is the genotype of III-6? c. What is the genotype of II-3? d. What is the genotype of III-8? e. If couple I-1 and I-2 will have a son, what is the probability of having the disorder? f. If couple III-8 and III-9 will have another child, what is the probability of having the disorder? g. Theoretically, if individual IV-3 and individual IV-5 will marry and will have a child, what is the probability of having a child without the X-linked disorder?
9. Make a pedigree for each of the following situations. For each individual, write the individual's genotype (when possible) next to the individual's symbol (e.g. O xty, I Gg): a. Two parents do not have cystic fibrosis and they have a daughter with cystic fibrosis and a son who does not have cystic fibrosis. The daughter grows up and she mates with a male who does not have cystic fibrosis. Their only child is a boy and he has cystic fibrosis. b. A man with hemophilia mates with a female without hemophilia. They have one son and one daughter. The daughter has hemophilia and the son does not have hemophilia. The son grows up, and he marries and mates with a female. Their only child is a boy, and he has hemophilia.
1. Why did the royal family have an issue with the hapsburg jaw/lip? 2. Was it an Autosomal or X-linked Genetic Disorder? 3. Dominant or Recessive?(Explain) Using a Punnett Square predict what the Genotypes are for the persons II-3 & II-(* C/C or C/A mean that they married and had children with a blood cousin).
Pedigree 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)?
IV 1 = female - male 2 3 || 2 4 1 3 ? 4 2 5 5 3 4 6 6 The disease trait in this pedigree is autosomal recessive and causes a severe hearing defect. What is the probability that the 2 people indicated in IV (individuals IV-4 and IV-5) will have a child with the disease trait? OA. The probability is 1 in 2 OB. The probability is 1 in 12 OC. The probability is 1 in 4 OD. The probability is 1 in 8 OE. The probability is 1 in 6
re ||| E 6. Label the genotypes for this pedigree of an X-linked recessive disorder (red- green colorblindness). (a) 2 2 3 1 3 a. How do you know? b. Label the genotypes. 4 O To 2 4 5 5 6 6 7. Is the following pedigree autosomal recessive, autosomal dominant or X-linked recessive? 2 8 O T 58