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- A. Color-blindness is a recessive, sex-linked disorder in humans. A color- blind man has a child with a woman who is a carrier of the disorder. KEY: X= normal vision XC = color-blindness 2. Illustrate using a Punnett square the probability of having children who will have normal vision and children who will be color-blind. Guide Questions: a. What is the genotype of the male?. b. What is the genotype of the female? c. What is the chance that the child will be color-blind? d. What is the chance that a daughter will be color-blind? e. What is the chance that a son will be color-blind?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…Pretending that Park shin-hye is heterozygous for A blood type marries Choi Tae-joon whose blood type is O. Show the possible blood types of their future children. 1. P1: 2. Punnett Square: 3. Genotype symbols: 4. Genotype in words: 5. Genotypic ratio: 6. Genotypic %: 7. Phenotype symbols: 8. Phenotype in words: 9. Phenotypic ratio: 10. Phenotypic %:
- Cystic Fibrosis (CF) is an autosomal recessive condition. Therefore, heterozygous (Cc) carriers do not display symptoms. Two parents who are carriers plan to start a family and you are a genetic counselor helping to advise them about their chances of having children affected by CF. a) Suppose the couple has 4 children, each one year apart. What is the probability that all 4 children will inherit CF? b) What is the probability that any 3 of their 4 children will not inherit CF, but 1 will be affected? c) What is the probability that their first child will not inherit CF, but the younger 3 children will inherit CF?Sickle cell anemia is an inherited red blood cell disorder in which there are not enough healthy red blood cells to carry oxygen throughout the body. The allele that causes sickle-cell anemia is autosomal recessive (s), and the dominant allele can be represented by S. How many offspring will be affected by the disorder if the mother is a carrier, and the father appears to be normal? (Include the gender) a. b. How many will become carriers? (include the gender) A- 三三三 四 四 II !!Questions a to e are answerable by yes or no. Indicate the possible parental genotypes if your answer is yes.a. Can a man with hairy ears have a hairy-eared daughter?b. Can two normal parents produce a colorblind son?c. Can two normal parents produce a colorblind daughter?d. Can a colorblind woman have a normal son?e. Can a bald man have a nonbald daughter?
- Red-green color blindness is inherited through an X-linked, recessive allele (b). Two parents, Fred and Ginger, have normal vision. They have two daughters, Takiyah and Kelly, who also have normal vision, and a color-blind son, David. 3. Daughter Kelly has a color-blind son, Kevin. Daughter Takiyah has five sons, all with normal vision. What are the genotypes of all the individuals? Show all your work! Fred Ginger David Takiyah Kelly Kevin Takiyah's five sons If Kelly marries a man with normal vision, what is the probability that she'll have a color-blind son? a color-blind daughter?Which of the following rows correctly identifies the relationship between the blood type alleles IA, IB, and i? Select one: a. Relationship between IA and IB Relationship between IB and i Incomplete dominance Codominance b. Relationship between IA and IB Relationship between IB and i Multiple alleles Incomplete dominance c. Relationship between IA and IB Relationship between IB and i Dominant and recessive Multiple alleles d. Relationship between IA and IB Relationship between IB and i Codominance Dominant and recessiveCross a homozygous tall female carrier for hemophilia with a short normal male. Give genotypic and phenotypic ratios of the offspring. Use T = Tall, and t = short for one trait, and H =normal, and h = hemophilia for the second sex-linked trait. The Cross is: XHXhTT X XYtt Question: What is the Ggenotypic ratio? a.) None is correct b.) 3:2:1:4 c.) 4:4:4:4 d.) 1:2:4:4
- The chart below is showing 4 generations of a family that is affected by a hereditary disease. a. Is the disorder being tracked dominant or recessive? How do you know? b. There is only one possible genotype for person C. True or False? c. What are the possible genotypes for person A? d. What are the possible genotypes for person B?, e. If two people with the same genotypes as person C's spouse and person A's spouse had a child, what is the probability that the child will be affected by this genetic disorder? (draw a Punnett square using the correct genotypes to help you). % chance offspring will be affected % chance offspring will not be affectedHemophilia is an X-linked disorder that affects the body’s ability to create blood clots. The allele for normal blood clotting, XH, is dominant over the allele for hemophilia, Xh. An unaffected female that is not a carrier mated with an affected male. Which of the following rows identifies the possible genotypes of the offspring? Select one: a. Female Male XHXH and XHXh XHY and XhY b. Female Male XHXh XHY c. Female Male XHXh XHY and XhY d. Female Male XHXH and XHXh XHYFor the Y-linked gene, use T for hypertrichosis. For X-linked genes, use the letters H for blood trait and E for eye trait. Krisha, a carrier of the genes for hemophilia and colorblindness, is married to Kyle who is color blind, having normal red blood cells. Veronica (Krisha & Kyle's daughter), turned out to be a carrier for both genes. Julian has hypertrichosis and is married to Josie. Both do not exhibit any allele for color blindness and never had any history for hemophilia. Bernard (Julian & Josie's son) will be married to Veronica. What are the complete genotypes of the following persons: a. Krisha b. Kyle c. Veronica d. Julian e. Josie f. Bernard g. Josie & Bernard's baby