Question

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) VG for F1 is > 0
b) VP for P1 < VG for P2
c) VE for F1 is > 0
d) VG for P1 = 0
e) VG for P2 = 0

Answer 1

Answer:

a)

Explanation:

From the given information:

The avg diameters:

For P1 grapes =1.2 cm

For P2 grapes = 1.6 cm

Their vairance is also given as follows:

For P1 = 0.30 cm²

For P2 = 0.56 cm²

Recall that:

the phenotypic variance VP = genetic variance (VG) + environmental variance VE

As stated that P1 and P2 lines are pure breeding lines, which means that the diameter of the seed in F1 was around halfway between the two parents' and the difference in the F1 was close to the parents'.

This implies that, provided the parent lines exist as pure breeding;

The genetic variation would be absent, indicating that VG is zero.

∴

VP = VE  and VG = 0

Now, averaging the results of two pure breeding lines, we have:

$V_E = \Big (\dfrac{0.30 + 0.56}{2} \Big) \\ \\ V_E = \dfrac{0.86}{2} \\ \\ V_E = 0.43 \ cm^2$

The formula used in estimating F1 genetic variance can be expressed as:

VP = VG + VE

Recall that

VE = VP = 0.43 cm²

∴

0.43 = VG + 0.43

VG = 0.43 - 0.43

VG = 0

Thus, The statement that is false is in Option a.

VG for F1 is > 0