Genetics Smorgasboard of Problems Key 13-14

Genetics Smorgasboard: Practice Problems B NAME: KEY LL 1. What's the probability of a woman with type AB blood and a man with type O blood having a child with type A blood? IA IB IA IBL 2/4 50% probability the child will have type A blood LIAL IBL B 2. A man with type A blood and a woman with type B blood have a child with type O blood. Give genotypes of parents and the child and show a Punnett square for the cross. the 1 I Parents: woman: IL Child: LL man: IBL 3. A wealthy elderly couple dies together in an accident. A man comes forward, claiming that he is their long lost son and is entitled to their fortune. The couple were of blood types AB and O. The man has type O blood. Could he be the heir to the fortune? Show why or why not. I^IBxLL child A I I i I^i IBi No, because the wealthy couple could have only produced offspring with IIB type A or B blood. F: 4. Carnation flowers can exhibit codominance. The allele for red flowers (R) is codominant to the allele for white flowers (W). The hybrid is speckled red and white. Cross a white carnation with a speckled carnation. Show the possible genotypes and phenotypes of the F1 generation. Include ratios, gene: Plovercolor G: c2ccw: 2 cwcw CR W cw P: Øred: 2 speckled: 2 white red + white 5. In radishes, red and white are pure-breeding colors, while hybrids are purple. If a red radish is crossed with a white radish, what will be the phenotype of the F, generation (assuming the F, gene generation self-pollinates)? Show the Punnett Square.kce x alleles: ccred (redtite) speckled: CROW cwow: white F : radish color red: cker F2 C L W CCW x CRCW F2 : I red: 2 purple: I white Purple: Crew white: cwew 6. In Andalusian fowis, black individuals (8) and white individuals (W) are homozygous. A homozygous black bird is crossed with a homozygous white bird. The offspring are all bluish-gray. Show the cross as well as the genotypes and phenotypes of the parents and offspring. FBFBX gene color black: FBFB byish-gray: FF white: FWFw G • OF BF B. 4F BF DF P: FW FW FN FBEN FBEW FB FB FN FOFW FBEW black: 4 bluish gray: white individual? Shoe genotypic and phenotypic ratios if a black individual is crossed with a bluish grays the Punnett Square, FBFBX F 6:2FBFB:2FBFW OFW FW BFW 2 Bluish gray: Owhite If two bluish-gray indivichlals were crossed, what would be the ratios for both phenotype and genotype of the offspring? Show the Punnett Square. FBpWX F8 FW 6. I FBFB: 2FBFW. IFWFW P: 1 black: 2 bluish gray I white FB FW FB FBFB FBPW FW/FSFW FWFW allele: pattern checked: D plain: d allele: color red: R brown: r 9. In pigeons, checkered pattern is dominant, and plain pattern is recessive. In addition, red feathers are dominant and brown feathers are recessive. Write the phenotypes as a ratio for the following cross.   A homozygous red, checkered pigeon x A brown pigeon, that is heterozygous checkered. RRDD x rrDd G: 8RrDd: 8RrDD P: 16 red, checkered (all red + checkered) 10. In cats, the gene for long hair is dominant to the gene for short hair. If two heterozygous cats are crossed, what type of hair will the kittens have (write the phenotypic ratio). alleles: long: H short: h Hh x Hh G: 1 HH: 2 Hh: 1 hh P: 3 long hair: 1 short hair 11. The gene for colorblindness is carried on the X chromosome and is recessive. A man, whose father was colorblind, has a colorblind daughter.   allele: color vision normal: E colorblind: e (him) xey: Yes, because his daughter has two alleles for color blindness, one came from her mom and the other from him. b) Where did he get his gene for colorblindness? From his mom because the y had to have come from his dad. c) Must the fathers of all colorblind girls be colorblind? Why? Yes because a colorblind girl requires 2 alleles for colorblindness, one from mom and the other from dad. Dad only has one X so he must be colorblind. 12. Cross a woman carrier for hemophilia to a hemophiliac man. alleles: normal: H hemophiliac: h XHXh x XhY a) What fraction of the offspring will be carrier females? Hemophiliac females? Normal females? 25%, 25%, 25% b) What fraction will be normal males? Hemophiliac males? 25%, 25%