To curl or not to curl
Earlier I said that the process of flipping coins was a lot like creating and combining gametes. For any given gene, the mother "flips a coin" to determine which of her two copies her egg gets, and the father "flips a coin" to determine which of his two copies each sperm gets, and when you put the two coins together, you know the offspring's genotype, which determines its phenotype.
So, we can use the exact same methods as with coins to figure out genotypes. For example:
Let's consider a gene with two alleles, "C" for straight hair, and "c" for super-curly hair. As you know, the allele with the lower-case letter is recessive, so only people with two "c" alleles will actually have super-curly hair. (Note: this is a completely made-up gene, with no claim to reality).
So imagine a mother and father who both have the genotype Cc, and therefore straight hair. What is the probability that they will have a child with super-curly hair?
This is the same as asking, what is the probability of flipping flipping two coins and getting two tails?
P(coin1=tails AND coin2=tails) = P(coin1=tails) × P(coin2=tails) = 0.5 × 0.5 = 0.25.
Or, to put it in terms of alleles:
P(mother=c AND father=c) = P(mother=c) × P(father=c) = 0.5 × 0.5 = 0.25.
Here are some more problems with the same gene -- find the probability that
Kid has curly hair, assuming mother's genotype is Cc and father's is cc
(To make this problem interactive, turn on javascript!)
- I need a hint ... : if the father has genotype cc,
then P(father=c) = 1 - ...another hint ... : So, P(mother=c AND father=c) =
- ...another hint ... : P(mother=c) * P(father=c) =
- ...another hint ... : 0.5 * 1.0
I think I have the answer: 0.5
Kid has straight hair, assuming mother is homozygous dominant and father is homozygous recessive
(To make this problem interactive, turn on javascript!)
- I need a hint ... : Homozygous means same,
so the mother is CC and the father is cc - ...another hint ... :you can get straight hair either with CC or Cc, so
- ...another hint ... :P((moth.=C AND fath.=C) OR
(moth.=C AND fath.=c) OR
(moth.=c and fath.=C) - ...another hint ... :1.0*0.0 + 1.0*1.0 + 0.0*1.0
I think I have the answer: 1.0
kid has super-curly hair, assuming mother is heterzygous and father is homozygous dominant?
(To make this problem interactive, turn on javascript!)
- I need a hint ... : first, translate the terms:
mother is Cc, father is CC,
and super-curly hair requires cc - ...another hint ... : so P(mother=c AND father=c)
- ...another hint ... : 0.5 *0.0
I think I have the answer: 0.0
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