question- Color blindness is inherited as a sex-linked recessive disease. A normal male marries a female who is heterozygous for the trait. What percentage of their sons will exhibit color blindness?
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question-
Color blindness is inherited as a sex-linked recessive disease. A normal male marries a female who is heterozygous for the trait. What percentage of their sons will exhibit color blindness?
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- This is a modified question from the textbook, Campbell Biology in Focus (3rd edition), question #5 on page 256. A man with normal vision marries a woman with normal vision whose father was red-green color-blind (a recessive, sex-linked condition). Assuming that the woman's mother had normal vision and did not carry the recessive gene, what is the probability that the married couple will have a color-blind daughter? That their first son will be color-blind? That their second son will be color-blind? Include Punnett square for each question in your answer. You may find the below diagram helpful (the orange square means the individual has the phenotype of color-blindness, while the beige means the individual is a carrier of the mutated color-blind gene where XN and Xn represent dominant and recessive allele, respectively).X-linked ichthyosis is an X-linked recessive trait that manifests in part as dry, scaly skin (“ichthy-” = fish or fish like). Suppose a couple are considering having a child together. Parent A is heterozygous for the ichthyosis allele while Parent B is hemizygous negative for the ichthyosis allele. What is the probability their child would be unafflicted with ichthyosis but be a carrier of the ichthyosis-causing allele? a.0% b.25% c.50% d.75% e.100%My Question is what is the probability their first child will have hemophilia and drawn pedigrees for family members with genotypes. My explantion so far: A man has both X and Y chromosomes as sex chromosomes in his body. Here, though the brother of the man is hemophiliac, a man can’t be a carrier of hemophilia. So, it can be said that his chromosome is “XnY”.Here, the “n” stands for “normal”.Though the paternal uncle is hemophiliac, a man cannot be a carrier of hemophilia, his niece will not be a career. So it can be said that the woman is also not a carrier and has the “XnX” chromosome.So, as the mother is not a carrier, their first child does not have a chance of having hemophilia. This can be determined as it is known that there is no hidden carrier of hemophilia in the family.
- The autosomal (not X-linked) gene for brachydactyly, short fingers, is dominant to normal finger length. Assume that a female with brachydactyly in the heterozygous condition is married to a man with normal fingers. What is the probability that(a) their first child will have brachydactyly?(b) their first two children will have brachydactyly?(c) their first child will be a brachydactylous girl?Neurofibromatosis-1 (NF1) is an autosomal dominant disorder where tumours form in the base layer of the skin or in nerve tissues. What is the probability that individuals II-1 and II-2 will have a genetic son with NF1? Find the image attached.QUESTION 10 Consider the following pedigree. With no other information provided, what is the most likely mode of transmission (or mode of inheritance) of the allele for the unusual trait? autosomal dominant sex-linked recessive autosomal recessive O sex-linked dominant
- Question 9 Consider that a certain gene is a maternal effect gene and that the allele for dark brown pigment is incompletely dominant to the allele for no pigment (white). The incomplete dominant phenotype is light tan. If a heterozygous female is crossed with heterozygous male, what will be the phenotypic ratio of the progeny? A) 1 dark brown: 2 light brown: 1 white. B) 3 dark brown: 1 white C). all dark brown D) all light tan E) cannot be determined CS Scanned with CamScannerThe allele for color-blindness is carried on the X chromosome. Making color blindness (a recessive trait) an X - linked trait. A colorblind make and a carrier female for color blindness. (But is not colorblind herself) have a child. Show your work! A) what is the % chance that their son will be color blind? B) what is the % chance that their daughter will be color blind?152 Phenylketonuria (PKU) is a disorder caused by a recessive allele. Two carrier individuals have progeny. Answer the following questions in order and show solutions whenever relevant. If they have a normal child, what is the probability that he or she will be heterozygous? If they have three children, what is the probability of having 2 affected children and one normal child?
- Analysis of X-Linked Dominant and Recessive Traits As a genetic counselor investigating a genetic disorder in a family, you are able to collect a four-generation pedigree that details the inheritance of the disorder in question. Analyze the information in the pedigree to determine whether the trait is inherited as: a. autosomal dominant b. autosomal recessive c. X-linked dominant d. X-linked recessive e. Y-linkedAssume no Bombay allele for this entire question.) Consider a cross between someone with blood type B+ with someone with blood type A+ (and you don’t know their genotype). a. Can they have a child with blood type O-? If not, why not? If so, give the genotypes of the parents that would allow this. (Remember to include the genotypes for the A/B/O gene as well as the Rh gene. ) Give two different sets of genotypes for these parents (B+ and A+) that would NOT allow them to have an O- child. Let’s say these two parents have a child with blood type B-. Give the genotypes of the parents and the child.The next four questions are all related to this problem: Polydactyly (PD) is an autosomal dominant trait (polydactyly - P; wildtype - p). Cystic fibrosis (CF) is an autosomal recessive trait (cystic fibrosis - f; wildtype - F). A PD woman, otherwise normal in phenotype, marries a healthy normal man. Their 4 children are: 1) normal, 2) PD, 3) CF, 4) CF + PD. You will walk through a series of steps to answer this question: What is the probability that their 5th child will have at least one of these conditions? Here is the first step: 1. What is the cross? (Hint: You can use the 4 existing children to determine the genotypes of the parents.) O PpFF (female) x ppFF (male) O PoFf (female) x ppft (male) O pof female) x PPFI (male) O PPFF Ifemale) x ppFf (male) 2. What is/are the target genotypes? O pof OP.F. O pott O P.M 3. What is the probability the child will have PD AND cystic fibrosis? Answer to two decimal places (eg. 0.88). 4. What is the probability that their 5th child will have at…
