Problem 6: The accompanying pedigree was obtained for a rare kidney disease.$ 8 8 00a. Deduce the inheritance of this condition, stating your reasons. b. If individuals 1 and 2 marry,what is the probability that their first child will have the kidney disease?Problem 7: Consider the accompanying pedigree of a rare autosomal recessive disease, PKU.11IIIIVB4a. List genotypes of as many of the family members as possible.b. If individuals A and B marry, what is the probability that their first child will have PKU? Iftheir first child is normal, what is the probability that their second child will have PKU? If theirfirst child has the disease, what is the probability that their second child will be unaffected?

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Description: Problem 6: The accompanying pedigree was obtained for a rare kidney disease.$ 8 8 00a. Deduce the inheritance of this condition, stating your reasons. b. If individuals 1 and 2 marry,what is the probability that their first child will have the kidne

Introduction: Pedigree: In terms of genetics, a pedigree is a diagram that shows how a trait or…

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Problem 6: The accompanying pedigree was obtained for a rare kidney disease. $ 8 8 0 0 2 a. Deduce the inheritance of this condition, stating your reasons. b. If individuals 1 and 2 marry, what is the probability that their first child will have the kidney disease?

Problem 6: The accompanying pedigree was obtained for a rare kidney disease. $ 8 8 0 0 2 a. Deduce the inheritance of this condition, stating your reasons. b. If individuals 1 and 2 marry, what is the probability that their first child will have the kidney disease?

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 1CS: 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?
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?
Punnet square problems A=Codominant; B=Codominant; O=Recessive Mary is homozygous for type A blood. Steve is homozygous for type O blood. If they have children, what are the possible phenotypes and genotypes of their children, and what is the probability of each? Mary and Steve have a son, Brad. Brad’s wife, Samantha is heterozygous for type B blood. If they have children, what are the possible phenotypes and genotypes of their children, and what is the probability of each? Stella loves roses and decides to cross her red rose with her white rose. All of the resulting offspring of this cross are pink roses. What can you say about the red and white alleles as a result of this cross? Stella decides to cross two of the pink roses. What are the possible genotypes and phenotypes of the offspring and the probabilities of each? DNA replication, Transcription and Translation problems It is S phase of the cell cycle, and time to replicate the cell’s DNA. Using the following strand of DNA…
Sample problems related to non Mendelian inheritanceRead, analyze and answer completely the following problems: 1. Explain why it is possible for the proband in the following pedigree to have children of blood types A, B, and AB. Considering epistatic genes, what are the possible genotypes of II-2? 2. Agouti (A) is wild type and produces alternating bands of pigment on each hair. Black (a) is recessive to agouti. A mutation on gene B (recessive b) can eliminate all color. In a cross between agouti (AABB) and albino (aabb) mice, what genotypes, phenotypes, and proportions are expected in the offspring in F1 and F2 generations?
Need help, please. What are the ratios for the progeny phenotype(s)?
Practice Pedigree Problem help. I am confused so please show. Label Phenotypes and genotypes as you go about the Pedigree (and whatever else might be required)! Thank you again for your help, these questions confuse me). Hair or fur length in cats is controlled by a single, autosomal gene; the short hair-allele is dominant to the allele for long hair. Hair color is produced by a different gene which is located on the X chromosome. One allele for this sex-linked gene produces yellow, while an alternate allele produces black fur color; individuals which are heterozygous for these alleles are calico or tortiseshell in color. a). If a long-haired, black male is mated with a calico female homozygous for short hair, what kind of kittens will be produced in the F1generation? Give both genotypes and phenotypes; express the genotypes both symbolically and in words.
Unpacking Problem 731. Define homozygous, mutation, allele, closely linked, recessive, wild type, crossing over, nondisjunction, testcross, phenotype, and genotype.2. Does this problem concern sex linkage? Explain.3. How many chromosomes does Drosophila have?4. Draw a clear pedigree summarizing the results of crosses1, 2, and 3.5. Draw the gametes produced by both parents in cross 1.6. Draw the chromosome 4 constitution of the progeny ofcross 1.7. Is it surprising that the progeny of cross 1 are wild-typephenotype? What does this outcome tell you?8. Draw the chromosome 4 constitution of the male testerused in cross 2 and the gametes that he can produce.9. With respect to chromosome 4, what gametes can thefemale parent in cross 2 produce in the absence of nondisjunction? Which would be common and which rare?10. Draw first- and second-division meiotic nondisjunctionin the female parent of cross 2, as well as in the resultinggametes.11. Are any of the gametes from part 10 aneuploid?12.…
As attached, is a pedigree related question. I have a question, is it correct that individual I 1 's phenotype is Ff (heterozygous) whereas individual I 2 is ff (recessive homozygous)? That would mean that the ratio of Widow's Peak: Straight hair(no widow's peak) would be 1:1 ? So the chances of getting a Widow's Peak is 50% ? Is this correct? Am I doing it right.
Hi, I'm having trouble with my study guide for my upcoming genetics exam. If someone could please help with work shown and an explanation it would help so much! Thank you!! 2a. The pedigree below represents inheritance of rare condition. What pattern of inheritance is most consistent with the data? Assign alleles to all individuals to support your answer. If an allele is unknown, assign it a ? symbol. NOTE: Individuals whose phenotype or genotype cannot be determined are assumed to be unaffected and homozygous, unless otherwise indicated. 2b. In addition to the alleles you’ve indicated, describe 2 overall features of the pedigree that make it consistent with your chosen form of inheritance. 2c. Based on your mode of inheritance, what is the probability that the child of couple IV-4 x IV-5 will be affected? Show your work. attached is the pedigree
Klinefelter syndrome (XXY) can most be easily diagnosed by _______. a. pedigree analysis. b. aneuploidy c. karyotyping d. phenotypic treatment
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?