2. The following questions refer to the pedigree chart in the figure for a family, some of whosemembers exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.2IwwWw2 3456IIIIV123A. What is the genotype of individual II-4?B. What is the genotype of individual II-5?C. If individual II-2 and his partner have a second child, what is the probability that their secondchild will be a normal, unaffected male? Show your work:3. Phenylketonuria is an inherited disease caused by a recessive autosomal allele. If a woman and herhusband are both carriers, what is the probability that their first child will be a phenotypicallynormal female? First, draw the pedigree for this family. Then, calculate the probability (show yourwork):34. A man who is heterozygous for PTC tasting marries a woman who is homozygous recessive. Theyhave two children. Their oldest daughter is a taster, and their youngest boy is a non-taster. What isthe probability that their third child will be a taster boy? Draw a pedigree for this family:

To answer these genetic questions, I'll break down each scenario, analyze the pedigree charts, and…

Answered: 2. The following questions refer to the…

Concepts of Biology

1st Edition

ISBN:9781938168116

Author:Samantha Fowler, Rebecca Roush, James Wise

Publisher:Samantha Fowler, Rebecca Roush, James Wise

Chapter7: The Cellular Basis Of Inheritance

Section: Chapter Questions

Problem 9RQ: The genotype XXY corresponds to: a. Klinefelter syndrome b. Turner syndrome c. Tripto-X d. Jacob...

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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 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 single allele gives rise to the Hbs form of hemoglobin. Individuals who are homozygous for the allele (HbS/HbS) develop sickle-cell anemia (Section 9.6). Heterozygous individuals (HbA/HbS) have few symptoms. A couple who are both heterozygous for the HbS allele plan to have children. For each of the pregnancies, state the probability that they will have a child who is: a. homozygous for the HbS allele b. homozygous for the normal allele (HbA) c. heterozygous: HbA/HbS
The genotype XXY corresponds to Klinefelter syndrome Turner syndrome Triplo-X Jacob syndrome
1. Study the given alleles. Write the correct phenotype for each genotype. X– normal Genotype XC – Color-blind Phenotype XX XY XCXC www m XX www w ww w 2. Study the given alleles. Write the correct genotype for each phenotype. X- normal Genotype хн- Hemophiliaс Phenotype Hemophiliac female Hemophiliac male Normal female carrier of the gene Normal male Normal female
IX. The blood of several members of a particular family was determined, and the results are tabulated below. Individual Blood Type Genotype Father Mother B Child 1 AB Child 2 B. Child 3 Grandfather (mother's side) Grandmother (father's side) AB Assuming that no adultery or adoption took place for all generations, answer the following questions: a. Give the most probable genotype for each individual. Supply your answers on the table above. b. What is/are the most probable genotype(s) for the grandmother on the mother's side? c. If the father and the mother decide to have two more children, what is the probability that the first would be female type A and the second male type AB?
97. This family is (picture above) affected with blindness. Individual (IV.1) is clinically unaffected. What is the chance that he is homozygous for the normal allele? A-½ B-% C- almost 0 D-1/3 E- almost 100% ANS: IV TO O 560-556
2: Chart v Add-ins Media Links Comment Footer Page Numb Header Screenshot v Cystic fibrosis is a hereditary illness causing respiratory issues and recurrent lung infections. It is a recessive disorder (c). Using the details below, fill in the Punnett square to determine probability of an offspring having cystic fibrosis. Mother Father -does not -does not have have cystic cystic fibrosis fibrosis genotype Cc -genotype Cc Mother Father Probability of offspring having cystic fibrosis: ada)
1. The pedigree chart in Figure 5.29 shows the inheritance of haemopiu family. Study the pattern of inheritance in the pedigree chart, and then answer the questions that follow. о 5 6. 3 8 9 10 11 Key Unaffected male Haemophiliac male О Unaffected female Fig. 5.29 Pedigree chart of a family affected by haemophilia a) What is the genotype and phenotype of individuals 2 and 4? b) (i) How many of the unaffected family members are definitely carriers of the recessive allele? (ii) How are you able to tell which of the family members are carriers? (4) (1) (3) c) (i) If Individual 11 marries a carrier female, what percentage of their sons is likely to be haemophiliacs? (1) (ii) Use a genetic diagram to show how you worked out your answer in i, (6) 2. Why is haemophilia never passed from father to son, even though it is most common in males? (4) 3. Can a mother pass on a sex-linked gene to her daughter? (1) 4. Sipho has red-green colour blindness. One of his grandfathers was also. colour…
A. HUMAN PEDIGREE CASE ANALYSIS1. One couple has three children with the following sexes and ages: one son (40 y.o.) and two daughters (35 y.o. and 33 y.o.), all of them have normal pigmentation. Another couple has a son (35 y.o) and a daughter (20 y.o.) and all of them also have normal pigmentation. Both couples have normal pigmentation. The younger daughter from the first couple married the son of the second couple and they had three children. Their eldest daughter (5 y.o.) has normal pigmentation while their only son (3 y.o.) and one daughter (1 y.o.) have albinism. a. Draw the pedigree of this family. Follow protocols in making a pedigree. Provide the genotype of all individuals in the pedigree. Please provide also the gene notation. b. What is the mode of inheritance of this trait? c. Justify your answer in letter (b).d. For their normal daughter, what is the probability that she is a carrier? Show solution. e. If they will have a fourth child, what is the probability that the…
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 I³I®) 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 I^i) is married to a type B person (genotype Iºi). Show the cross. What proportion of their children will have: A blood? B blood? О blood? AB blood?

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