30.03.2023 - 04:20

In peas, a tall, yellow-seeded plant is crossed with a homozygous short, green-seeded plant and yields 203 tall, green-seeded plants, 199 short, green-seeded plants, 207 tall, yellow-seeded plants, and 192 short, yellow-seeded plants. The most likely geno

Question:

In peas, a tall, yellow-seeded plant is crossed with a homozygous short, green-seeded plant and yields {eq}203 {/eq} tall, green-seeded plants, {eq}199 {/eq} short, green-seeded plants, {eq}207 {/eq} tall, yellow-seeded plants, and {eq}192 {/eq} short, yellow-seeded plants. The most likely genotype of the tall, yellow-seeded parent is:

a. TtYy.

b. TTYY.

c. ttyy.

d. ttYY.

e. none of the above.

Answers (1)
  • Joella
    April 8, 2023 в 17:53
    The most likely genotype of the tall, yellow-seeded parent is TtYy (option a). This can be determined through a series of genetic crosses and Punnett squares. First, we can assume that the tall, yellow-seeded parent is heterozygous for two traits: height and seed color. This means that the genotype is TtYy. If we cross TtYy with a homozygous recessive short, green-seeded plant (ttyy), we can make a Punnett square to determine the possible offspring genotypes: | | t | t | |----|-----|-----| | Yy | tYy | tYy | | Yy | tYy | tYy | From this cross, we expect all offspring to be heterozygous for height and seed color, resulting in tall, yellow-seeded plants (TtYy). Next, we can use the information given in the problem to make additional Punnett squares for the other three possible crosses: | | T | t | |----|-----|-----| | Yy | TYy | tYy | | Yy | TYy | tYy | | | T | t | |----|-----|-----| | yy | ty | ty | | yy | ty | ty | | | T | t | |----|-----|-----| | yy | tY | ty | | yy | tY | ty | From these crosses, we can calculate the expected number of each genotype and phenotype using the laws of probability. By comparing these expected numbers to the actual numbers given in the problem, we can determine that the most likely genotype of the tall, yellow-seeded parent is TtYy.
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