Question

Mendel crossed true-breeding plants with wrinkled and green peas to true breeding plants with round and yellow peas. Then he crossed the F1 plants to true-breeding plants with wrinkled, green peas and observed in the progeny 31 plants with round, yellow peas, 26 plants with round, green peas, 27 plants with wrinkled, yellow peas, and 26 with wrinkled, green peas.

Required:
a. What proportion of these four phenotypes would be expected?
b. Calculate the appropriate chi-square and determine if it is consistent with Mendelian expectations (be sure to indicate the probability range)?

Answer 1

See the answer below

Step-by-step explanation:

Let us assume that the genes for texture (wrinkle or round) and color (green or yellow) is represented by s(S) and y(Y) respectively.

True breeding wrinkled and green = ssyy

True breeding round and yellow = SSYY

crossing the two:

SSYY x ssyy

F1 progeny: all SsYy

Then, he crossed the F1 progeny with another ssyy

SsYy x ssyy

Progeny:

4 SsYy - round and yellow

4 Ssyy - round and green

4 ssYy - wrinkled and yellow

4 ssyy - wrinkled and green

a. Proportion of the phenotype expected = 1:1:1:1

b. Chi square =
`((O - E)^2)/(E)`, where O = observed frequency, and E = expected frequency.

phenotype O E Chi square

round and yellow 31 1/4 x 110 = 27.5
`((31 - 27.5)^2)/(27.5)` = 0.45

round and green 26 1/4 x 110 = 27.5
`((26 - 27.5)^2)/(27.5)` = 0.08

wrinkled and yellow 27 1/4 x 110 = 27.5
`((27 - 27.5)^2)/(27.5)` = 0.01

wrinkled and green 26 1/4 x 110 = 27.5
`((26 - 27.5)^2)/(27.5)` = 0.08

Total Chi square = 0.45 + 0.08 + 0.01 + 0.08 = 0.62

Degree of freedom = 4 -1 = 3

Chi square tabulated (95% probability level) = 7.815

The calculated Chi square value is less than the tabulated Chi square value, hence, the outcome from the experiment is consistent with Mendelian expectations at 95% probability level.