Help!? 8.7 Red-Green Color BlindnessProblem: Red-Green color blindness is due to the inheritance of an X-linked recessive allele. James and Susan have normal color vision, but Susan'sfather was colorblind (her mother was not). A pedigree chart of their family is provided below. Where squares = males, and circles = females. Openshapes indicate the individual did not inherit the condition, shaded shapes indicate affected individuals. Half-shaded shapes indicate an unaffectedindividual with one recessive allele (a carrier).JamesSusanJimmyStanleySidneyJamieCreated by Paint XPost: Answer the following questions.What are the genotypes of James and Susan?James and Susan raise 4 children; What are the genotypes and phenotypes (normal vision, or RG colorblind) for each child.Finally, what can you deduce about the parents of the children from this information?

Let Xc be the defected allele In X-linked recessive disorders, the male can never be a carrier…

8.7 Red-Green Color Blindness Problem: Red-Green color blindness is due to the inheritance of an X-linked recessive allele. James and Susan have normal color vision, but Susan's father was colorblind (her mother was not). A pedigree chart of their family is provided below. Where squares = males, and circles = females. Open shapes indicate the individual did not inherit the condition, shaded shapes indicate affected indivicuals. Half-shaded shapes indicate an unaffected individual with one recessive allele (a carrier). James Susan Jimmy Stanley Sidney Jamie Created by Paint X Post: Answer the following questions. What are the genotypes of James and Susan? James and Susan raise 4 children; What are the genotypes and phenotypes (normal vision, or RG colorblind) for each child. Finally, what can you deduce about the parents of the children from this information?

8.7 Red-Green Color Blindness Problem: Red-Green color blindness is due to the inheritance of an X-linked recessive allele. James and Susan have normal color vision, but Susan's father was colorblind (her mother was not). A pedigree chart of their family is provided below. Where squares = males, and circles = females. Open shapes indicate the individual did not inherit the condition, shaded shapes indicate affected indivicuals. Half-shaded shapes indicate an unaffected individual with one recessive allele (a carrier). James Susan Jimmy Stanley Sidney Jamie Created by Paint X Post: Answer the following questions. What are the genotypes of James and Susan? James and Susan raise 4 children; What are the genotypes and phenotypes (normal vision, or RG colorblind) for each child. Finally, what can you deduce about the parents of the children from this information?

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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Sex-linked or “X-linked” inheritance Using XG for the normal vision allele and X8 for the color-blind allele, construct the Punnett square that would result from a carrier female (heterozygous = has a normal vision allele on one X chromosome and a color-blind allele on the other X chromosome) mating with a color-blind male (hemizygous recessive = has a color-blind allele on the X chromosome and no allele for vision on the Y chromosome ). Hint: The female genotype is XGX and the male genotype is X&Y. The Y chromosome does not have a gene for normal color vision OR color-blind vision b. If they had a son, what are the chances he will be color-blind? c. If they had a daughter, what are the chances she will be color-blind? d. Is there any way these parents might produce a daughter with normal vision? If so, how? e. Is there any chance these parents might produce a son with normal vision? Why or why not?
SUBJECT: GENETICS Topic: Probability 1. Consider albinism a trait to occur in a human family. If two parents areheterozygous for such trait, the probability of having a normally pigmented child is 3//4 and having an albino child is ¼. What is the probability that 2 will be normal and 1 will be albino?
X-linked Recessive Inheritance A gene is described as X-linked when it occurs on the X chromosome and not the Y. Our convention is to indicate X-linkage by attaching the appropriate gene symbol as a superscript on the letter X. Commonly, the wild-type (+) allele is indicated with only a "+" to avoid having to type a superscript on a superscript. For example, a female that is heterozygous and carrying a recessive mutant allele is indicated as X+Xm. Note the convenience of the shorthand + for m+ in this situation. A mutant male has the genotype XmY. When working with X-linked inheritance, always include the X and Y chromosomes in the descriptions of genotypes, and include the sex (male or female) in the descriptions of the phenotypes (e.g., mutant male, wild-type female, etc.). Here are the genotypes and associated phenotypes for X-linked recessive inheritance: X+X+ Wild-type female X+Xm Wild-type female xmxm Mutant female X+Y xmy Wild-type male Mutant male
Consider the following pedigree. 하 3 10 (5 3 2 (a) What pattern of transmission is most consistent with this pedigree? (1) autosomal recessive, (2) autosomal dominant, (3) X-linked recessive, (4) X-linked dominant. (b) If individual V-2 marries a normal individual, and if the condition has a pene-trance of 85 percent, what is the probability that their second child will express the trait? (c) On the third line, what does the diamond with a 10 in the middle mean?
Explain allelic series. The C-gene expression (Full color, Chinchilla, Himalayan, Albino) with punnet square.
O O c. A person who has a gene allele for a disorder with reduced penetrance (60%) that is transmittedin an autosomal dominant pattern and is usually expressed after age 30 has reached the age of 50 without any manifestations of the disorder. He now states that he wishes that he had decided to have children now that he knows he cannot pass the disorder on to any children he fathered. Is this man's thinking correct? Explain This person is not right because he can still pass the disorder to his offspring. Penetrance is the proportion of individuals carrying a particular variant (or allele) of a gene (the genotype) that also express an associated trait (the phenotype). The individual exhibits signs and symtoms of genetic disorder. He did not develop features of the disorder, the condition is said to have reduced (or incomplete) penetrance. He is a carrier of the allele for this disorder. As carrier he can give the genetic information to his child who could have a complete penetrance and…
In humans, the genes for red-green color blindness (R = normal, r = color blind) and hemophilia A (H = normal, h = hemophilia) are both X-linked and only 3 map units apart. Suppose a woman has four sons, and two are color blind but have normal blood clotting and two have hemophilia but normal color vision. What is the probable genotype of the woman? HR/hr Hr/hr hr/hR Hr/hR HR/Hr
58 A family has two children with an undiagnosed, very rare neurodevelopmental condition illustrated in the pedigree below. Neither parent is affected and there is no family history of this disorder. The following two questions relate to this pedigree. What is the most likely mode of inheritance for this condition? Select one alternative: Dominant O It is not inherited Recessive X-linked recessive
Albinism, lack of pigmentation in humans, results from an autosomal recessive gene designated a. Two parents with normal pigmentation have an albino child. What is the probability that their next child will be albino? What is the probability that the next child will be an albino girl? If the child is normal, what is the probability that it will be a carrier (heterozygous) for the albino gene?
The gene for polydactyly (P) is autosomal and dominant to normal fingers (p). Hemophilia is sex-linked and recessive (X h ). A man and his wife both of whom are polydactylous and have normal red blood cells have a child who is non-polydactylous and with hemophilia. What are their genotypes?
a. b. 1 2 3 Note: Cystic fibrosis in this family is caused by a recessive allele that is found on chromosome 7. A genetic counsellor collected pedigree information regarding the incidence of cyctic fibrosis within this family. The row that indicates the genotypes of individuals 1-1, 1-2, and II-2 is Select one: C. Pedigree of a Family with Cystic Fibrosis PTO 1 2 d. I II I-1 XAY I-1 AA I-1 XAY I-1 Aa 1-2 хаха 1-2 aa 1-2 ΧΑΧΑ 1-2 Aa II-2 XAY 11-2 Aa II-2 хау II-2 aa O
a. The ability to taste the chemical phenylthiocarbamideis an autosomal dominant phenotype, and the inabilityto taste it is recessive. If a taster woman with a nontasterfather marries a taster man who in a previous marriagehad a nontaster daughter, what is the probability thattheir first child will be(1) A nontaster girl(2) A taster girl(3) A taster boyb. What is the probability that their first two childrenwill be tasters of either sex?