Consider two purebreeding lines of grapes P1 and P2 The aver

Consider two pure-breeding lines of grapes, P1 and P2. The average diameter of P1 grapes is 1.2 cm and the variance is 0.30 cm2. The average diameter of P2 grapes is 1.6 cm and the variance is 0.56 cm2. The two pure-breeding lines are crossed to produce F1 progeny. Based on this description, which of the following statements is false?

a) V_G for F₁ is > 0

b) V_P for P₁ < V_G for P₂

c) VE for F₁ is > 0

d) V_G for P₁ = 0

e) V_G for P₂ = 0

Multiple people on Chegg have posted this question and received the answer that option b was false, which was noted as incorrect. If you can back up your answer with an explanation, that would be great!

Solution

Answer

average diameter of P1 grapes = 1.2 cm

average diameter of P2 grapes = 1.6 cm

variance of P1 grapes = 0.30 cm²

variance of P2 grapes = 0.56 cm²

since,

the usual equation is V_P = V_G + V_E

V_P is the phenotypic variance

V_G is the genetic variance

V_E is the environmental variance

As mentioned in the question the P1 and P2 are pure breeding lines, this states that the diameter of the seed in F1 was about halfway between that in the two parents and the variance in the F1was similar to that seen in the parents,

Since, the parents are pure-breeding lines, the genetic variance will not exists and it clearly signifies that V_G is 0 , therefore

V_P = V_E ; V_g = 0

To get the value of V_E we need to calculate the average of two pure breeding lines -

(0.30 + 0.56 /2 ) = V_E

0.86 /2 = 0.43 cm²

V_E = 0.43 cm²

F1 genetic variance can be calculated by using the formula

V_P = V_G + V_E

from the above the value of V_E=V_P which is equal to 0.43 cm²

Therefore the value of Vg can be calculated as

0.43=VG+0.43

VG=0.43-0.43

V=0

Therefore the statement VG for F1 is >0 is false.

Hence, the correct answer is option a.