Answer first 3 Inbox (5,344)- hattangaw6@gmail.comFor the following questions use the pedigree shown below:II36Assume that individual I-2 is not a carrier.Show your work on your scratch paper.The genotype for individual 1-1 isV[Choose]AAX-linked recessive traitX-linked dominant traitThe genotype for individual II-3 isAutosomal dominant trait-3 married Il-6: what is the probabilof having a child that showed the trait?1/2Aa3/4XaXaIf -3 married I-6: what is the probabilityof having a daughter that showed the traitAa or aa1/8AA or Aa-3 maried -6 and they had a daughte116Y-inked traitXAXA or XAXathat then married -2: What is theprobability that they had a daughter thathad the traitXAXAXAXaXAYThis pedigree most likely representsAutosomal recessive trait1/4

For the following questions use the pedigree shown below: %3D 2 3 4 5 Assume that individual 1-2 is not a carrier. Show your work on your scratch paper. The genotype for Individual 1-1 is V[Choose) AA X-linked recessive trait The genotype for individual -3 is X-linked dominant trait Autosomal dominant trait -3 marred -6 what is the probabil of having a child that showed the trait? 12 Aa 3/4 -3 married -6: what is the probability of having a daughter that showed the trait XaXa Aa or aa

For the following questions use the pedigree shown below: %3D 2 3 4 5 Assume that individual 1-2 is not a carrier. Show your work on your scratch paper. The genotype for Individual 1-1 is V[Choose) AA X-linked recessive trait The genotype for individual -3 is X-linked dominant trait Autosomal dominant trait -3 marred -6 what is the probabil of having a child that showed the trait? 12 Aa 3/4 -3 married -6: what is the probability of having a daughter that showed the trait XaXa Aa or aa

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter12: Genes And Cancer

Section: Chapter Questions

Problem 3CS: Mike was referred for genetic counseling because he was concerned about his extensive family history...

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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?
Mike was referred for genetic counseling because he was concerned about his extensive family history of colon cancer. That family history was highly suggestive of hereditary nonpolyposis colon cancer (HNPCC). This predisposition is inherited as an autosomal dominant trait, and those who carry the mutant allele have a 75% chance of developing colon cancer by age 65. Mike was counseled about the inheritance of this condition, the associated cancers, and the possibility of genetic testing (on an affected family member). Mikes aunt elected to be tested for one of the genes that may be altered in this condition and discovered that she did have an altered MSH2 gene. Other family members are in the process of being tested for this mutation. Once a family member is tested for the mutant allele, is it hard for other family members to remain unaware of their own fate, even if they did not want this information? How could family dynamics help or hurt this situation?
Mike was referred for genetic counseling because he was concerned about his extensive family history of colon cancer. That family history was highly suggestive of hereditary nonpolyposis colon cancer (HNPCC). This predisposition is inherited as an autosomal dominant trait, and those who carry the mutant allele have a 75% chance of developing colon cancer by age 65. Mike was counseled about the inheritance of this condition, the associated cancers, and the possibility of genetic testing (on an affected family member). Mikes aunt elected to be tested for one of the genes that may be altered in this condition and discovered that she did have an altered MSH2 gene. Other family members are in the process of being tested for this mutation. Is colon cancer treatable? What are the common treatments, and how effective are they?
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?
Mike was referred for genetic counseling because he was concerned about his extensive family history of colon cancer. That family history was highly suggestive of hereditary nonpolyposis colon cancer (HNPCC). This predisposition is inherited as an autosomal dominant trait, and those who carry the mutant allele have a 75% chance of developing colon cancer by age 65. Mike was counseled about the inheritance of this condition, the associated cancers, and the possibility of genetic testing (on an affected family member). Mikes aunt elected to be tested for one of the genes that may be altered in this condition and discovered that she did have an altered MSH2 gene. Other family members are in the process of being tested for this mutation. Seventy-five percent of people who carry the mutant allele will get colon cancer by age 65. This is an example of incomplete penetrance. What could cause this?
Consider this pedigree showing an autosomal dominant rare disorder. What is the degree of penetrance? Show your work. na оп 16 19 fa 16 R 9X
d/1n5NtidRwTwUzcDkDPi5Z9P SHPZ9IA-XH-pfftLbhNc/edit 1) @ Is Add-ons Help Last edit was 15 minutes ago | Calibri в I UA 12 + 3I | II 6 1 I 7. Construct a Punnett square for a cross between two heterozygous pea plants with violet flower color. a. What genotypes would you expect in the offspring? b. What percentage or ratio of each genotype would you expect in the offspring? !!!
PEDIGREES: Problem (continued) This pedigree shows the inheritance of cystic fibrosis in this family. I • QUESTIONS ••. 5. What is the genotype of individual II-3? Use the letter "f" to 1 2 represent the disease allele. II 1 2 3 6. Individuals II-I and II-2 are sisters. Explain how it is possible for one sister to have cystic fibrosis but NOT the other. III 1 2 3
Pedigree attached shows an autosomal recessive genetic disease. G is the normal allele and g is the disease-causing allele. Individual 1’s father is heterozygous (*) and his mother is homozygous dominant. Other individuals in the pedigree may be carriers, but are not marked. The question mark (?) indicates that you do not yet know anything about this individual’s phenotype with regard to the disease. part a) What is the probability that individuals 1 and 2 will have a child (5) who is a boy with the disease (the child is unborn and the sex is not yet known)? a)1/8 b)1/4 c)0 d)1/16 part b) What is the probability that the daughter (6) that individual 3 and 4 just had will have the disease? a)1/8 b)1/6 c)1/4 d)1/12
FAlpQLSfiOhfAvlhxzCSiUll_6rt-nU5b0WI73UmWOxkOw8OCwk01ng/formResponse B 1 2 Bb x Bb b 4 The fur in both parents in this cross is * 1 B B Bb x Bb b 3 4 brown black O homozygous dominant homozygous recessive 3. 近

/d/1n5NtidRwTwUzcDkDPi5Z9P_SHPZ91A-XH-pfftLbhNc/edit (1) O pols Add-ons Help Last edit was seconds ago BIUA ミ: 12 + ext Calibri I|1 6 I 2 Section 5: Trihybrid cross and Laws of probability For a trihybrid cross, in which inheritance of alleles for three genes is tracked, drawing a Punnett square that combines all three genes may not be practical. Instead the laws of probability may be used. The product law of probabilities says that when alleles for separate genes segregate independently, we can figure out the probability of a particular combined genotype by multiplying the probability of the alleles for each gene. 13. We cross a homozygous tall pea plant with yellow, round seeds to a homozygous dwarf pea plant with green, wrinkled seeds. All the F1 offspring are all tall plants with yellow, round seeds. a. What are the expected F2 ratios (use fractions) of tall and dwarf plants? b. What are the expected F2 ratios (use fractions) of yellow and green seeds? C. What are the expected F2…
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