10. Examine the pedigree from a family with a genetic disease and answer the questions below: 1:1 1:2 Il:1 I1:2 I1:3 I1:4 I1:5 I1:6 II:1 II:2 II:4 II:5 a) Does this pedigree indicate autosomal dominant, recessive or sex- linked type of inheritance? Give reasons for your choice. b) Assuming that B and b are the normal and mutant alleles respectively, what would be the genotypes of the individuals: П.1, П.2 and II.З ? c) Individual II.3 requested genetic counselling. What is the probability that her child would be affected. Explain why.

10. Examine the pedigree from a family with a genetic disease and answer the questions below: 1:1 1:2 Il:1 I1:2 I1:3 I1:4 I1:5 I1:6 II:1 II:2 II:4 II:5 a) Does this pedigree indicate autosomal dominant, recessive or sex- linked type of inheritance? Give reasons for your choice. b) Assuming that B and b are the normal and mutant alleles respectively, what would be the genotypes of the individuals: П.1, П.2 and II.З ? c) Individual II.3 requested genetic counselling. What is the probability that her child would be affected. Explain why.

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Description: 10. Examine the pedigree from a family with a genetic disease and answer the questionsbelow:1:11:2I1:1II:2I1:3Il:4II:5I1:6II:1III:2II:3I:4III:5II:6a) Does this pedigree indicate autosomal dominant, recessive or sex-linked type of inheritance? Give r

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 3CS: A couple was referred for genetic counseling because they wanted to know the chances of having a...

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1. The figure below shows a pedigree examining the presence of a genetic disorder across multiple generations of a family. Based off this figure, answer the following questions prompted: || IV 3 9 10 5 2 C 6 11 T 7 12 8 15 13 16 14 17 d) What is the genotype for individual #4 in this family pedigree? ||| c) Why are no family members identified as a carrier of this genetic disorder? Figure Key a) Using the figure key provided, identify which members of this family are affected by this genetic disorder in each generation. (Be sure to include sex and generation number for each individual identified). b) Is this an autosomal recessive, autosomal dominant, or sex-linked genetic disorder? Explain. Unaffected Female Unaffected Male Affected Female ||| Affected Male Female Carrier Male Carrier
tion 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
1. The pedigree below shows the incidence of rare, autosomal dominant disorder called Ehlers-Danlos disease. The pedigree covers three generations of a particular family and also shows individual genotypes at a potential marker locus (M). a) Indicate the phase of all gen II and III individuals. DdM1M3 ddM2M6 II DDM3M6 ddM4M5 III DdMзM4 DdMЗМ5 DDM3M4 ddM3M5 DDM3M4 ddM5M6 DDM3M4 ddM4M6 ddM5M6 ddM5M6 b) Which, if any, of the gen III individuals are recombinants? c) Calculate the LOD score as a test of physical linkage between the marker (M) and the disease locus. d) What do you conclude about linkage between D and M?
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. 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 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?
Cystic Fibrosis (CF) is an autosomal recessive condition. Therefore, heterozygous (Cc) carriers do not display symptoms. Two parents who are carriers plan to start a family and you are a genetic counselor helping to advise them about their chances of having children affected by CF. a) Suppose the couple has 4 children, each one year apart. What is the probability that all 4 children will inherit CF? b) What is the probability that any 3 of their 4 children will not inherit CF, but 1 will be affected? c) What is the probability that their first child will not inherit CF, but the younger 3 children will inherit CF?
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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…
II. Given the following pedigree below, use Punnett squares for each of the following possibilities: a) X- linked dominant and b) X-linked recessive in order to determine what is the mode of transmission of this trait. Disease allele = XA or Xª, depending on mode of transmission of the disease respectively. *Unaffected/No carrier-Normal Unaffected X chromosome = X I || III 1 1 2 a) X-linked recessive 9 III 6 genotype (circle one): XX * 1 2 3 11x12 4 ΧΑΧΑ 2 5 xaxa *4 6 7 8 b) X-linked dominant 11 x 12 오 XAX хах
10- A man with X-linked color blindness marries a woman with no history of color blindness in her family. The daughter of this couple marries a normal man, and their daughter also marries a normal man. What is the chance that this last couple will have a child with color blindness? Assume the chance to have a daughter or a son is equal. a) 3/8 b) 1/8 C) 1/16 d) 3/4 02040 e) 1/4 Boş bırak