In a population of weasels, black (B) and white (W) are co-dominant traits. The heterozygous genotype produces black and white spotted weasels. Consider the genetic cross shown in Punnett square. If there are 40 baby weasels born as a result of the cross, how many should be black?

Responses
A 00
B 55
C 1010
D 20

1 answer

To determine how many weasels should be black, we first need to understand the genetics of the traits involved.

In a cross involving black (B), white (W), and spotted (BW) weasels, we can set up a Punnett square. Here are the possible gametes from two parents:

  • Suppose one parent is homozygous black (BB) and the other parent is homozygous white (WW). The Punnett square would look like this:
   B   B
-----------
W | BW | BW |
-----------
W | BW | BW |

In this case, all offspring (100%) would be black and white spotted (BW), so if we consider this cross, no offspring would be solid black.

Alternatively, if we consider a cross between two heterozygous parents (BW x BW), the Punnett square would look like this:

   B   W
-----------
B | BB | BW |
-----------
W | BW | WW |

The results would then be:

  • BB (black) = 1
  • BW (spotted) = 2
  • WW (white) = 1

So the ratio would be 1:2:1 (1 black : 2 spotted : 1 white).

Now, if there are 40 baby weasels from this cross, we apply the 1:2:1 ratio:

  • Total parts = 1 (black) + 2 (spotted) + 1 (white) = 4 parts.
  • Each part represents \( \frac{40}{4} = 10 \) weasels.

From this ratio, the number of black weasels (BB) would be:

  • Black (BB) = 1 part = 10 weasels.

Therefore, from this genetic cross, if 40 baby weasels are born, we can conclude that the number of black weasels would be 10.

Given the options:

A: 00
B: 55
C: 1010
D: 20

None of the options correctly match the derived answer of 10. Please double-check the options provided.