Black fur (B) is dominant to white fur (b) in guinea pigs. An F1 cross was conducted and resulted in 100% heterozygous guinea pi
1 answer:
Answer:
D. 1 BB : 2 Bb : 1 bb
Explanation:
This question involves a single gene coding for fur color in guinea pig. Black fur (B) is dominant over white fur (b). This means that, as stated in the question, if a black fur parent (BB) is crossed with a white fur parent (bb), a 100% heterozygous offspring (Bb) with black fur will result.
If two heterozygous guinea pigs are crossed i.e. Bb × Bb, the following gametes will be produced by each heterozygous parent:
Bb = B and b
Using these gametes in a punnet square (see attached image), offsprings with the following genotypic ratio will be produced:
1 BB : 2 Bb : 1 bb
BB and Bb = black fur guinea pigs
bb = white fur guinea pigs
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Answer:
Thus, the minimum number of photons per second is 77.34
Explanation:
Light intensity, =
Pupil has a diameter, d = 8.5 mm
= 8.5 x m
Radius of the eye, r = 4.25 x m
∴ Area of the eye, A =
=
=
Let be the minimum number of photons.
Therefore, =
x A
= x
= W
Thus the minimum number of photons is given by
where E =
=
=
Therefore, =
= 77.34 photons per second
Thus, the minimum number of photons per second is 77.34