## Nms Team 81 Homework

None of the children showed the symptom and thus none of them are recessive homozyogous. The probability of one or more to be carriers (heterozygotes) = 1 – (The probability of none of them are carriers). The probability of none of them are carriers = 1/3 (dominant homozygous) X 1/3 X 1/3 X 1/3 = 1/81 The probability of one or more to be carriers = 1 – 1/81 = 80/81 3. (1point) Sweet yellow tomatoes with a pear shape bring a high price per basket to growers. Pear shape, yellow color, and terminal flower position are recessive traits produced by alleles, f, r , and t , respectively. The dominant phenotypes for each trait – full shape, red color, and axial flower position – are the product of dominant alleles F, R , and T . A farmer has two pure-breeding tomato lines. One is full, yellow, terminal and the other is pear, red, axial. Design a breeding experiment (crosses) that will produce a line with tomato that is pure-breeding for pear shape, yellow color, and axial flower position. 1) Cross the parents to produce FfRrTt , trihybrids ( FFrrtt X ffRRTT ) 2) Self fertilize F1 to create F2 population that will include all possible phenotypes and genotypes ( FfRrTt X FfRrTt ) 3) Among F2, 3/64 should produce yellow, pear-shaped tomatoes and have axial flowers. You can at least know genotypes for recessive traits ( ffrrT_ ). 4) To determine which of these plants are TT , self fertilize them identify the plants that breed true (pure) – or not segregating for T ) for axial flower position.

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