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# A2 Biology Genetics Question AQA

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1. This is from the unit 4 biology paper i did in 2016... the answer to 3ci is 0.0625 but i have no idea how to calculate that... help please. thanks in advance
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2. You multiply the probabilities of two separate events happening in order to find the probability of them happening together. In this case, there is a 25% chance that the individual will be colour blind, and a 25% chance that the individual will be a non-tongue roller. 0.25*0.25=0.0625, which is the probability that the child will be both colour blind and a tongue roller.

Both parents are heterozygous for tongue rolling, so both have Tt genotypes.
Tt x Tt gives TT, Tt, Tt, and tt. 1/4 are tt (non tongue rolling), or 0.25.

The father (11) is not colour blind, and therefore doesn't have the colour blindness allele. He can't have two copies of the allele for colour blindness, as he has only one X chromosome. He can't have one copy of the allele for colour blindness, because if he did, he'd be colour blind. As he produced a colour blind child with 10, who has colour vision, she must be a carrier.
X(B)X(b) x X(B)Y gives X(B)X(B), X(B)X(b), X(B)Y, X(b)Y. 1/4 are X(b)Y (and none are X(b)X(b), which is the other combination that would result in colour blindness), or 0.25 probability of being colourblind.
3. (Original post by oopqoo)
You multiply the probabilities of two separate events happening in order to find the probability of them happening together. In this case, there is a 25% chance that the individual will be colour blind, and a 25% chance that the individual will be a non-tongue roller. 0.25*0.25=0.0625, which is the probability that the child will be both colour blind and a tongue roller.

Both parents are heterozygous for tongue rolling, so both have Tt genotypes.
Tt x Tt gives TT, Tt, Tt, and tt. 1/4 are tt (non tongue rolling), or 0.25.

The father (11) is not colour blind, and therefore doesn't have the colour blindness allele. He can't have two copies of the allele for colour blindness, as he has only one X chromosome. He can't have one copy of the allele for colour blindness, because if he did, he'd be colour blind. As he produced a colour blind child with 10, who has colour vision, she must be a carrier.
X(B)X(b) x X(B)Y gives X(B)X(B), X(B)X(b), X(B)Y, X(b)Y. 1/4 are X(b)Y (and none are X(b)X(b), which is the other combination that would result in colour blindness), or 0.25 probability of being colourblind.
Thank you so much

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