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a brown mouse is crossed with hertozygous black mouse. if the mother has a litter of four, what are thae chances all of them will be brown?

http://en.wikipedia.org/wiki/Punnett_square

do you have more specifics? because that didn't help a bit

There is a 50% chance the the litter will be brown.

Assuming that the heterozygous mouse has one recessive brown gene (b) and one dominant black gene (B) and the brown mouse is bb, Kyonia is right for one birth.

Thus the brown mouse can give either b or b, while the black mouse can give either b or B. Plug the symbols into the Punnett square to see how we arrived at the conclusion.

However, the question asked for the probability that all of them are brown. To find the probability that all events will occur, you need to multiply the probability of the individual events, in this case, the 4 mice. You can do the calculations.

I hope this helps. Thanks for asking.

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