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.
Well, there's a 50% chance all of the mice will be brown. But remember, even if they're not all brown, they can still form a beautiful rainbow of colors! Just like a box of crayons, but with furry little creatures instead.
To calculate the probability that all four mice will be brown, we can use the Punnett square method.
First, let's assign the genes:
- The brown mouse is bb (homozygous for brown)
- The black mouse is Bb (heterozygous)
Using a Punnett square, we can cross these two mice:
| B | b |
B | BB | Bb |
b | Bb | bb |
From the Punnett square, we see that there are two possible genotypes for the offspring: BB and Bb.
The probability of each genotype is as follows:
- BB: 25% chance (1 out of 4)
- Bb: 50% chance (2 out of 4)
Since we are only interested in the probability that all four mice will be brown (bb), we need to calculate the probability of bb occurring four times in a row.
The probability of bb occurring once is 50% (1 out of 2), and since there are four births, we need to multiply this probability by itself four times:
0.50 x 0.50 x 0.50 x 0.50 = 0.0625
So, there is a 6.25% chance that all four mice will be brown.
To find the probability that all the mice in the litter will be brown, we need to determine the probability of each individual mouse being brown and then multiply those probabilities together.
Let's assume that the brown mouse has two recessive brown genes (bb) and the heterozygous black mouse has one recessive brown gene (b) and one dominant black gene (B).
The possible combinations of genes that each parent can pass on to their offspring are as follows:
Brown mouse (bb) can only pass on a recessive brown gene (b).
Heterozygous black mouse (Bb) can pass on either a recessive brown gene (b) or a dominant black gene (B).
To find the probability of each individual mouse being brown:
For the first offspring:
Probability of receiving a brown gene from the brown mouse = 1 (since there is only one possibility)
Probability of receiving a brown gene from the heterozygous black mouse = 0.5 (since there are two possibilities, and one of them is brown)
Total probability of being brown for the first offspring = 1 * 0.5 = 0.5
For the second offspring:
Probability of receiving a brown gene from the brown mouse = 1
Probability of receiving a brown gene from the heterozygous black mouse = 0.5
Total probability of being brown for the second offspring = 1 * 0.5 = 0.5
Similarly, for the third and fourth offspring, the probability of being brown is also 0.5.
To find the probability of all four offspring being brown, we multiply the individual probabilities together:
Probability of all four offspring being brown = 0.5 * 0.5 * 0.5 * 0.5 = 0.0625 or 6.25%
So, the chances of all four offspring being brown is 6.25%.