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- Answer the following questions. 1. Construct a map for the genes d,e,f. Assume that: d and e = 3%; e and f = 5%. Give 2 arrangements of the genes/maps. 2. If d and f = 2%, what is the correct arrangement of the genes d,e,f? 3. Consider the fourth gene "g". if g and e = 1.5%, give two possible arrangements. 4. If d and g = 1.5 % give the correct order of the four genes %3DA research team interested in mapping human genes discovered a new restriction length polymorphism (RFLP). a. First, they test the RFLP in a family with the pedigree below. The individuals colored blue are affected by hemophelia, a blood clotting disease. Dad Mom daughter1 daughter2 son1 son1 On what chromosome is the RFLP found? Explain your answer. b. Next, they genotyped (determined the RFLP polymorphism, type 1 or type 2) of daughterl's family with the results shown below: husband daughter1 %3D Are the RFLP and the hemophilia gene linked? Explain your reasoning.A couple consulted a genetic counselor and they found out the following: Basha: A blood type, blue eye color, normal skin condition, carrier of cystic fibrosis gene, and carrier of hemophilia gene Popoy: B blood type, has brown eye color, normal skin condition, carrier of cystic fibrosis gene and negative to hemophilia. Clues for the table: • Blood Type: The probability that they’ll have an offspring with blood type the same as Popoy is 25% only • Eye Color: The probability that they’ll have an offspring with brown eye color is 75%. • Skin condition: The probability that their offspring will be an albino is 25% What is the probability that the offspring will have/ be: A. a girl with all the same phenotype as her mother. B. 2 girls with neither the blood type of the parents. C. 2 boys with normal skin color and carrier of cystic fibrosis gene; 1 girl with normal skin color and negative to cystic fibrosis. D. 2 girls carrier of both cystic fibrosis gene and hemophilia, a boy † positive…
- 1. Study the given alleles. Write the correct phenotype for each genotype. X – normal Gen otype xC - Color-blind Phenotype XX XY XXC xCY 2. Study the given alleles. Write the correct genotype for each phenotype. xH - Hemophiliac Phenotype X- normal Gen otype Hemophiliac female Hemophiliac male Normal female carrier of the gene Normal male Normal female 28 3. Determine the genotype and phenotype of the offspring. A color-blind mother (XCx) married a normal sighted (XY) father. Genotype: Phenotype: Genotype: Phenotype: Genotype: Genotype: Phenotype: Phenotype: a. There are b. There are c. There are d. There are % normal sons. % normal daughters. % color-blind sons. % color-blind daughters. % normal female, carrier of the disorder. or or or or e. There are or9. F. A. B. E. C. D. The allele for normal skin pigmentation (A) is dominant to absence of pigmentation (a), albinism. Answer the following questions about the inheritance of this disease. What is the genotype of a woman with normal skin pigmentation who has a baby that is an albino? What is the genotype of the baby in A above? did not have the Can 2 people with normally pigmented skin have a baby that is an albino? Explain your answer. If 2 heterozygous people marry, what is the chance that their first child is an albino? Explain your answer. for the the If the first child from E above is an albino, what is the chance the second child is normal? Explain your answer. Can 2 people who are both albinos have a child who is not an albino? Explain your answer. 40Mutations in the genes for clotting factor VIII and IX cause hemophilia A and B, respectively. A woman may be heterozygous for mutations in both genes, with a mutated factor VIII allele on one X chromosome, and a mutated factor IX allele on the other. All of her sons should have either hemophilia A or B. However, on rare occasions, one of these women gives birth to a son who does not have hemophilia, and his one X chromosome does not have either mutated allele. Explain.
- An individual has the following genotype. Gene loci (A) and (B) are 15 m.u. apart. What are the correct frequencies of some of the gametes that car be made by this individual? Bl a O A. Ab = 7.5%; AB = 42.5% B. ab = 25%; aB = 50% O C. AB = 7.5%; aB = 42.5% O D. aB = 15%; Ab = F0% E. aB = 70%; Ab = 15% Reset Selection OMark for Review What's This?Blood types in humans are caused by a combination of two of three possible alleles - IA, IB, and i. The i allele is recessive and individuals homozygous for this allele have blood types O. Blood type A may be due to either homozygous condition or the IA i Blood type AB is caused by having a copy of both the IA and IB alleles. Rh factor is another “marker” on human red blood cells and is either positive (dominant—RR or Rr) or negative (recessive—rr). Franks has blood type A-negative. Frank’s father is B-positive and his mother is AB-positive. Frank marries Susan, a woman that is B-positive. Susan’s mother is A-negative and her father is AB-positive. Give the genotype of all of the individuals mentioned above. b. Give the genotypes and phenotypes of the possible offspring from this marriageSeveral genes in humans in addition to the ABO gene () give rise to recognizable antigens on the surface of the red blood cells. The Rh marker is determined by positive (R) and negative alleles () of gene R, where R is completely dominant to r. The presence of M and N surface proteins are controlled by two codominant alleles of gene L (LM and LM. For each mother-child pair, choose the father of the child from among the males in the right column. (Assume that all mothers and fathers are HH; there is no involvement of the Bombay phenotype.) Paternal genotypes maybe used once, more than once, or not at all. Each mother-child pair matches with one or more than one paternal genotype. Maternal phenotype: Child phenotype: Paternal genotype: Reset A, M, Rh(neg) O, M, Rh(pos) B, N, Rh(neg) O, N, Rh(pos) O, M, Rh(neg) A, MN, Rh(pos) A, N, Rh(pos) AB, MN, Rh(pos) B, N, Rh(pos) A, MN, Rh(neg) Genotypes of possible fathers AiLMLN rr BiLMLN RR ii LNLN rr ii LMLM rr AALMLN RR
- Duchenne muscular dystrophy is sex linked and usuallyaffects only males. Victims of the disease become progressively weaker, starting early in life.a. What is the probability that a woman whose brotherhas Duchenne’s disease will have an affected child?b. If your mother’s brother (your uncle) had Duchenne’sdisease, what is the probability that you have receivedthe allele?c. If your father’s brother had the disease, what is theprobability that you have received the allele?α-thalassemia is another blood disorder in which abnormal hemoglobin molecules are formed, leading to anemia. One mutant allele known to cause α-thalassemia occurs in α-hemoglobin and is called Constant Spring. Normal α-hemoglobin is 141 amino acids long, while the Constant Spring protein is 172 amino acids long. Include answers to both A and B in your response. A maximum of 2 sentences per part. A) Explain how a frameshift mutation in the coding region of α-hemoglobin could result in the Constant Spring protein. B) Explain how a single base-substitution mutation in the coding region of α-hemoglobin could result in the Constant Spring protein.Basha: A blood type, blue eye color, normal skin condition, carrier of cystic fibrosis gene, and carrier of hemophilia gene Popoy: B blood type, has brown eye color, normal skin condition, carrier of cystic fibrosis gene and negative to hemophilia. What is the probability that the offspring will have/ be: A. a girl with all the same phenotype as her mother. B. 2 girls with neither the blood type of the parents.