* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Name __________________________ Period _____ Advanced Genetics Practice Problems – Beyond Mendel Part 1: Incomplete Dominance 1. 2. 3. What is incomplete dominance? How does this affect the phenotype of the heterozygotes? In traits with incomplete dominance, neither allele is completely dominant over the other. In heterozygotes, the phenotype appears as a blend of the two alleles. Japanese Four O’ clock Flowers show an incomplete dominance for Color. When an offspring has both the Red and the White allele, it appears Pink (Rr). RR = Red Rr = pink rr = white a. Show a cross between a Purebred Red (RR) and a Purebred White (rr) flower. What is the Genotype and Phenotype ratio? Cross: RR x rr R R Genotypic Ratio: 100% Rr Phenotypic Ratio: 100% pink r Rr Rr r Rr Rr b. Show a cross between two Pink Japanese Four O’clock Flowers. What is the Genotype and Phenotype Ratio? Cross: Rr x Rr R r Genotypic Ratio: 25% RR: 50% Rr: 25% rr Phenotypic Ratio: 25% red: 50% pink: 25% white R RR Rr r Rr rr In Horses the valuable Palomino (Gold) color is the result of a cross between a purebred Chestnut (brown) and a purebred Cremello (off-white). BB = chestnut (brown) Bb = palomino (gold) Bb = cremello (off-white) a. Use a Punnett Square to show a cross between 2 Palomino Horses. What is the Genotype and Phenotype Ratios of this cross? Cross: Bb x Bb B b Genotypic Ratio: ¼ RR: ½ Rr: ¼ rr Phenotypic Ratio: ¼ Chestnut:: ½ Palomino: ¼ Cremello B BB Bb b Bb bb b. What are the Genotype and Phenotype Ratios for a cross between a Cremello horse and a Palomino horse. Cross: bb x Bb b b Genotypic Ratio: ½ Bb: ½ bb Phenotypic Ratio: ½ Palomino: ½ Cremello B Bb Bb b bb bb c. If you were a horse breeder that wanted only Palomino offspring, what parent cross would ensure this result? Use a Punnett Square to justify your answer. To produce only palomino horses, I would cross a chestnut horse with a cremello horse. Cross: BB x bb B B b Bb Bb b Bb Bb Part 2: Co-Dominance 4. 5. What is co-dominance? How does this affect the phenotype of the heterozygotes? In co-dominance traits, both alleles are dominant. Both versions of the trait are seen in the phenotype of the heterozygotes. Roan (RW) is a color of cattle in which both Red (R) and White (W) hairs are present due to Co dominance. RR = Red RW = Roan WW = White a. What are the phenotype and genotype ratios of offspring produces by a Roan bull and a Red cow? Cross: RW x RR b. c. d. e. R W Genotypic Ratio: 2 RR: 2 RW Phenotypic Ratio: 2 red: 2 roan R RR RW R RR RW What are the phenotype and genotype ratios of offspring produced by a Roan bull and a White cow? Cross: RW x WW R W Genotypic Ratio: 2 RW: 2 WW Phenotypic Ratio: 2 roan: 2 white W RW WW W RW WW What are the phenotype and genotype ratios of offspring produced by a Roan bull and a Roan cow? Cross: RW x RW R W Genotypic Ratio: 1 RR: 2 RW: 1 WW Phenotypic Ratio: 1 red: 2 roan: 1 white R RR RW W RW WW What are the phenotype and genotype ratios of offspring produced by a Red bull and a White cow? Cross: RR x WW R R Genotypic Ratio: 4 RW Phenotypic Ratio: 4 roan W RW RW W RW RW If you had a Roan bull, what are all the possible genotypes from this bull’s parents? The parents could RW x RR (5a), RW x WW (5b), RW x RW (5c) or RR x WW (5d). Part 3: Multiple Allele Traits 6. 7. 8. What are multiple allele traits? Genes of multiple allele traits have more than 2 variations of the gene. Blood Type is determined by 3 different alleles A, B and O. A and B are co-dominant, while O is recessive. Possible blood types are A, B, AB and O. A man with O type blood marries a woman with AB type blood. What are the possible blood types of their offspring? Cross: ii x IAIB i i Genotypic Ratio: 0.5 IAi: 0.5 IBi A A A Phenotypic Ratio: 0.5 Type A: 0.5 Type B I I i I i IB IB i IB i Vincent has type A blood and his mother has type O blood. Christine has type B blood and her father has type O blood. Vincent and Christine are not related. IA = A marker IB = B marker IA IB = A & B marker ii = O marker (no marker) a. b. c. What are Vincent and Christine’s genotype? Vincent’s genotype is IAi. Since his mother is type O, Vincent must have inherited the recessive allele from her. Christine’s genotype is IBi. Since her father is type O, Chrisine must have inherited the recessive allele from him. What are the possible genotypes of Christine’s mother? Vincent’s father? Christine’s mother had to have an IB allele, so she could have any of the following genotypes: IA IA, IAi or IA IB Vincent’s father had to have an IA allele, so he could have any of the following genotypes: IB IB, IBi or IA IB Suppose Christine and Vincent got married. What is the probable phenotype ratio for their offspring? Cross: IAi x IBi IA i Genotypic Ratio: 1 IA IB: 1 IAi: 1 IB i: 1 ii B A B B Phenotypic Ratio: 1 Type AB: 1 Type A: 1 Type B: 1 Type O I I I I i i IA i ii Part 4: Sex-Linked Traits 9. What is a sex-linked trait? A sex-linked trait is controlled by a gene that is located on a sex chromosome instead of an autosome. 10. In drosophila (fruit flies) white eye is recessive X-linked trait (Xr). Red eyes are normal and dominant (XR). A white eyed female is crossed with a red eye male. XR = red eyes Xr = white eyes a. What phenotypic ratios would be obtained in the F1 generation (include sex in phenotype)? Cross: XrXr x XRY Xr Xr Genotypic Ratio: ½ XRXr: ½ XrY R R r R r Phenotypic Ratio: ½ Red-eyed females: ½ white-eyed males X X X X X r r Y XY XY b. If a male and female from the F1 generation in problem #1 mate, what would be the F2 phenotype ratio? Cross: XRXr x XrY Genotypic Ratio: ¼ XRXr: ¼ XrXr: ¼ XRY: ¼ XrY R r X X Phenotypic Ratio: ¼ Red-eyed females: ¼ white-eyed females: ¼ Red-eyed males: ¼ white-eyed males Xr XRXr XrXr R r Y X Y XY 11. Red-green color blindness is caused by a sex-linked recessive allele. A colorblind man marries a woman whose father was colorblind. What is the probability of them having a colorblind girl? A colorblind boy? XN = normal vision Xn = color blind Cross: XnY x XNXn Xn Y XN XNXn XNY Xn XnXn XnY Genotypic Ratio: ¼ XNXn: ¼ XnXn: ¼ XNY: ¼ XnY Phenotypic Ratio: ¼ normal vision females: ¼ color-blind females: ¼ normal vision males: ¼ color-blind males The probability that they will have a color-blind girl is ¼ and the probability that they will have a color-blind boy is ¼. 12. Hemophilia (a recessive, sex-linked condition) is a disorder that results in the inability to form blood clots. XN = normal Xn = hemophiliac a. A normal man marries a woman whose dad was hemophiliac. What is the probability of this couple having a hemophiliac son? Daughter? Cross: XNY x XNXn Genotypic Ratio: ¼ XNXN: ¼ XNXn: ¼ XNY: ¼ XnY N X Y Phenotypic Ratio: ½ normal females:: ¼ normal males: ¼ hemophiliac males XN XNXN XNY n N n n X X X XY The probability that they will have a hemophiliac son is ¼. The probability that they will have a hemophiliac daughter is 0/4. b. A normal female (homozygous dominant) marries a normal man. What is the probability of their children having hemophilia? Cross: XNXN x XNY Genotypic Ratio: ½ XNXN: ½ XNY N X Y Phenotypic Ratio: ½ normal females:: ½ normal males XN XNXN XNY N N N N X X X X Y The probability of this couple having a child with hemophilia is 0%. c. A man with hemophilia has a daughter of normal phenotype (XNXn). She marries a man who is normal for the trait. What is the probability that a daughter of this mating will be a hemophiliac? A son? If the couple has four sons, what is the probability that all four will be born with hemophilia? The probability that they will have a hemophiliac daughter is 0/4. The probability that they will have a hemophiliac son is ¼ (see 5a for the Punnett square). The probability that four sons would be born with hemophilia is (1/4)4 = 1/256. 13. Pseudohypertrophic muscular dystrophy is a disorder that causes gradual deterioration of the muscles. It is seen only in boys born to apparently normal parents and usually results in death in the early teens. Is pseudohypertrophic muscular dystrophy caused by a dominant or recessive allele? Is its inheritance sex-linked or autosomal? How do you know? Explain why this disorder is seen only in boys and never in girls? Pseudohypertrophic muscular dystrophy (PMD) is most likely caused by a recessive, X-linked gene. It is recessive because parents without the trait may have children with the trait (they are carriers). It is X-linked because it is seen only in boys who receive only one X-chromosome (and are more likely to display X-linked recessive traits). It is never seen in girls because boys who have PMD do not survive to sexual maturity and so are not able to reproduce and pass the gene on. The following questions (#14 & 15) are being offered as extra credit (3 pts each). Due with your HW packet. 14. White eye color and lozenge eye shape are both x-linked recessive trait in Drosophila (fruit flies). a. b. A homozygous normal female is crossed with a white, lozenge eye male. Assuming no crossing over takes place, what is the F1 and F2 phenotype ratios A remarkable white eyed Drosophila female undergoes non-disjunction of the “X” chromosome 50% of the time. She is mated with a normal mate and 200 zygotes are produced. What would be the number and phenotype (include sex) of the offspring? Zygotes with no X chromosome die. 15. In Drosophila, white eye is recessive X-linked trait. Vestigial wing is a recessive autosomal (not sex linked) trait. What will be the phenotype and ratios of the F1 and F2 generations produced from a cross of a white eyed, homozygous normal wing female and a red eyed (normal), vestigal winged male? Include sex in phenotype.