Answer 14 in some plants a true breeding red flowered

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Answer: 14) In some plants, a true-breeding, red-flowered strain gives all pink flowers when crossed with a white-flowered strain: C R C R (red) × C W C W (white) C R C W (pink). If flower position (axial or terminal) is inherited as it is in peas (see Table 14.1 in your textbook), what will be the ratios of genotypes and phenotypes of the F 1 generation resulting from the following cross: axial-red (true-breeding) × terminal-white? What will be the ratios in the F 2 generation? Answer: Parental cross is AAC R C R × aaC W C W . Genotype of F 1 is AaC R C W , phenotype is all axial-pink. Genotypes of F 2 are 4 AaC R C W : 2 AaC R C R : 2 AAC R C W : 2 aaC R C W : 2 AaC W C W : 1 AAC R C R : 1 aaC R C R : 1 AAC W C W : 1 aaC W C W . Phenotypes of F2 are 6 axial- pink : 3 axial-red : 3 axial-white : 2 terminal-pink : 1 terminal-white : 1 terminal-red. 15) Flower position, stem length, and seed shape were three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as follows: Character Dominant Recessive Flower position Axial ( A ) Terminal ( a ) Stem length Tall ( T ) Dwarf ( t ) Seed shape Round ( R ) Wrinkled ( r ) If a plant that is heterozygous for all three characters is allowed to self-fertilize, what proportion of the offspring would you expect to be as follows? (Note: Use the rules of probability instead of a huge Punnett square.) A) Homozygous for the three dominant traits B) Homozygous for the three recessive traits C) Heterozygous for all three characters D) Homozygous for axial and tall, heterozygous for seed shape
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22 Answer: A) 1⁄64, B) 1⁄64, C) 1⁄8, D) 1⁄32 16) A black guinea pig crossed with an albino guinea pig produces 12 black offspring. When the albino is crossed with a second black one, 7 blacks and 5 albinos are obtained. What is the best explanation for this genetic situation? Write genotypes for the parents, gametes, and offspring. Answer: Albino ( b ) is a recessive trait; black ( B ) is dominant. First cross: parents BB × bb ; gametes B and b ; offspring all Bb (black coat). Second cross: parents Bb × bb ; gametes 1⁄2 B and 1⁄2 b (heterozygous parent) and b ; offspring 1⁄2 Bb and 1⁄/2 bb . 17) In sesame plants, the one-pod condition ( P ) is dominant to the three-pod condition ( p ), and normal leaf ( L ) is dominant to wrinkled leaf ( l ). Pod type and leaf type are inherited independently. Determine the genotypes for the two parents for all possible matings producing the following offspring: A) 318 one-pod, normal leaf and 98 one-pod, wrinkled leaf B) 323 three-pod, normal leaf and 106 three-pod, wrinkled leaf C) 401 one-pod, normal leaf D) 150 one-pod, normal leaf, 147 one-pod, wrinkled leaf, 51 three-pod, normal leaf, and 48 three-pod, wrinkled leaf E) 223 one-pod, normal leaf, 72 one-pod, wrinkled leaf, 76 three-pod, normal leaf, and 27 three- pod, wrinkled leaf Answer: A) PPLl × PPLl , PPLl × PpLl , or PPLl × ppLl . B) ppLl × ppLl . C) PPLL × any of the 9 possible genotypes or PPll × ppLL . D) PpLl × Ppll . E) PpLl × PpLl .
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