2. The following questions refer to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. 2 I ww Ww 2 34 5 6 III IV 1 23 A. 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 second child 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 her husband are both carriers, what is the probability that their first child will be a phenotypically normal female? First, draw the pedigree for this family. Then, calculate the probability (show your work): 3 4. A man who is heterozygous for PTC tasting marries a woman who is homozygous recessive. They have two children. Their oldest daughter is a taster, and their youngest boy is a non-taster. What is the probability that their third child will be a taster boy? Draw a pedigree for this family:

2. The following questions refer to the pedigree chart in the figure for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. 2 I ww Ww 2 34 5 6 III IV 1 23 A. 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 second child 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 her husband are both carriers, what is the probability that their first child will be a phenotypically normal female? First, draw the pedigree for this family. Then, calculate the probability (show your work): 3 4. A man who is heterozygous for PTC tasting marries a woman who is homozygous recessive. They have two children. Their oldest daughter is a taster, and their youngest boy is a non-taster. What is the probability that their third child will be a taster boy? Draw a pedigree for this family:

Biology: The Unity and Diversity of Life (MindTap Course List)

14th Edition

ISBN:9781305073951

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter13: Observing Patterns In Inherited Traits

Section: Chapter Questions

Problem 11GP: A single allele gives rise to the Hbs form of hemoglobin. Individuals who are homozygous for the...

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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
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
5. A woman with type AB blood marries a man with type homozygous type B blood. What are the phenotype and genotype ratios of the offspring? 6. Hemophilia is a sex linked recessive trait. If a carrier woman marries a normal man. The couple gives birth to a boy. What are the chances that this child will have hemophilia? Make sure use XX or XY here) 7. The height of a certain plant is a result of polygenic inheritance involving two pairs of genes. Draw a Punnett square to show the result of a cross between a tall (AABB) plant and a short (aabbi. What is the phenotype ratio? 8. What is a mutation? 9. What is sickle cell anemia? 10. What is hemophilia? 11. What is a pedigree? 12. Draw a pedigree to represent the following. Shaded symbols should represent individuals with hemophilia: A man with hemophilia marries a normal woman. They have 3 children, two normal boys and a girl who is a carrier. The girl marries a normal man and they have two boys and two girls. One of the girls is a carrier…
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…
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?
8. what is the correct answer?
Double click on the image below to complete Question 5. 5. Examine the pedigree below. Write the genotype for each individual in the family. 6. The two children of the original couple are both carriers. How do you know?