Determine the patterns of inheritance of the following X-linked dominant cases by giving the results of the crosses between: a) affected male x normal female b) normal male x affected female c) affected male x affected female
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- Hemophilia A is transmitted by an X-linked recessive gene.With an affected father, what is the probability that a childwill have the disease? With an affected father and a carriermother, what is the probability?y 301 Amelogenesis imperfecta is X-linked dominant. Affected XY individuals have extremely thin enamel on the teeth while XX carriers have grooved teeth from uneven deposition of enamel. If an unaffected XY individual were to produce children with a XX carrier partner, a. what would be the expected chance of a XY child being affected with the disease? b. what would be the expected chance of a XY child being affected with the disease?Hemophilia is determined by genes on the X chromosome in humans. Assume that a phenotypically normal woman whose father had hemophilia is married to a normal man. 1) What is the probability that their first son will have hemophilia? 2) What is the probability that their first daughter will have hemophilia?
- Ill.siven the following pedigree below, use Punnett squares for each of the following possibilities: a) X- linked dominant, b) X-linked recessive, c) Autosomal dominant and d) Autosomal recessive in order to determine what is the mode of transmission of this trait. Disease allele = Xª, x², A or a depending on mode of transmission of the disease respectively. Unaffected X chromosome = X *Homozygous unaffected/No 1 *2 carrier=Normal II 1 *4 1 2 3 6. 7 8 a) X-linked dominant 11x12 b) X-linked recessive I 1 x1 2 c) Autosomal dominant 11x12 d) Autosomal recessive I1x12 IV. Based on your analysis what is the mode of transmission for this disease? O+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 bırakFemales with rr genotype are affected, males with rY are affected, females with Rr and RR are unaffected, males with RY are unaffected. a) sex-linked recessive b) autosomal dominant, homozygous lethal c) autosomal recessive d) autosomal dominant
- Consider a couple: a woman who is homozygous for a recessive mutation that causes X-linked colorblindness, and a man with full color vision (he does not carry a copy of the mutation). a) What is the probability that a son of this couple will be colorblind? b) What is the probability that a daughter of the couple will be colorblind?1. A) Apply the concept of sex linkage to explain why color blindness is more prevalent in men than in women. B) Mary is concerned that she may be a carrier for hemophilia, a sex-linked condition located on the X chromosome. Mary is married to John, who doesn't have hemophilia. Assuming Mary is a carrier, what are the genotype ratios expected for Mary and John's kids (specify for boys and for girls)? C) Mary and John have 2 boys, none of them has hemophilia. Can we use this fact as proof that Mary does not carry the allele for hemophilia? Explain your argument.Hemophilia is an X-linked recessive phenotype. Suppose a man who expresses the hemophilia phenotype has children with a woman who has the normal phenotype (and does not have a family history of hemophilia). If the couple have a son, what are the chances that he will have hemophilia? A) 50% B) 25% C) 0% D) 100%
- Inheritance of sickle - cell anemia demonstrates_____ a ) Multiple allele inheritance c ) a dominant genetic disorder d ) incomplete dominance e ) a sex - linked inheritanceA child has a blood type of “B positive”. This child is known to be a heterozygote at the ABO blood type gene and also a heterozygote at the Rh blood type gene. Select all of the choices below that show possible parents of this child.a) Genotypes IA IA X IA IAb) Genotypes IA IA X IA i c) Genotypes IA i X IA id) Genotypes IA IA X IA IBe) Genotypes IA IA X IB if) Genotypes IA IA X i ig) Genotypes IA i X i ih) Genotypes IA i X IB ii) Genotypes IB IB X IB IBj) Genotypes IB IB X IB ik) Genotypes IB IB X IA il) Genotypes IB IB X i im) Genotypes IB i X IB in) Genotypes IB i X i io) Genotypes IB IB X IA IBp) Genotypes IA IA X IB IBq) Genotypes IA IB X IA IBr) Genotypes IA IB X IA is) Genotypes IA IB X IB it) Genotypes IA IB X i iu) Genotypes i i X i iv) Genotypes + + X + +w) Genotypes + + X + -x) Genotypes + + X - -y) Genotypes + - X + -z) Genotypes - - X + -a1) Genotypes - - X - -An individual with 46, XX genotype is diagnosed with Duchenne-type Muscular Dystrophy, a recessive X-linked disorder. Genetic tests confirm that this individual is a heterozygote for this disorder. Briefly, but specifically, explain how it’s possible that they are showing symptoms of this disorder.