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Current time:0:00Total duration:5:19

AP.BIO:

IST‑1 (EU)

, IST‑1.J (LO)

, IST‑1.J.2 (EK)

we are told the pedigree chart represents the inheritance of colorblindness through three generations and we see this here the standard convention is a square is male circle is female if it's colored in that means that they exhibit the trait in this case it's colorblindness so bill exhibits colorblindness his phenotype is colorblind while bonny does not exhibit colorblindness colorblindness is an x-linked recessive trait if Barbara is expecting another child so this is Barbara right here what is the probability that it will be colorblind so pause this video and see if you can figure that out on your own alright now let's work through this together so they're asking us about their next child here what is the probability that it is going to be colorblind and to help us with that we can try to figure out the genotypes of Tom and Barbara so tom is pretty straightforward he is male we know that because there's a square there so X has an X chromosome and he has a Y chromosome and colorblindness is an x-linked recessive trait and so let me just make clear what's going on so I'll do lowercase C for colorblind color blind and I could do a capital C for the dominant trait which is not colorblind but since they look so similar I'll just use a plus for not colorblind not color not colorblind and so Tom his phenotype he is colorblind and so and he only has one X chromosome where the colorblind what the color Bryant trait is linked to and so that must have the recessive allele right over there so this is Tom's genotype but what about Barbara well we know Barbara is going to have two x chromosomes because Barbara is female and we know that both of them can't be lowercase C because then Barbara would exhibit colorblindness but how can we figure out her actual genotype well we could look at her parents so bill over here is going to have the same genotype as tom at least with respect to colorblindness he is male so he has an X chromosome and a Y chromosome and because he exhibits colorblindness that X chromosome must have the recessive colorblind allele associated with it now Bonnie we do not know she will be ex ex will have two x-chromosomes like Barbara we know that both of these can't have the recessive allele because then Bonnie would be filled in she would exhibit colorblindness but we don't know whether she is a carrier or whether she isn't but let's just think about where Barbara got her chromosomes from one of her X chromosomes comes from her father and the other one comes from her mother so if she got this X chromosome from her father her father only has one x chromosome to give the one that has the colorblind allele so if this is from her father it must have the colorblind allele here and we know that the one from her mother is does not have the colorblind allele because if it was like this then Barbara would be colorblind and she isn't so we know that this must be a plus here it is the dominant non colorblind allele and so now we know both of their genotypes and we can use those to then figure out the possible outcomes for their offspring so for example Tom can contribute a X chromosome that has a colorblind allele or a Y chromosome and Barbara right over here can contribute an X chromosome that has a colorblind allele or an X chromosome that has the non colorblind allele Barbara is a carrier and so let me just draw a little punnett square here and so we have four possible outcomes for their children and they're all equally likely so you can get the X chromosome from Barbara that has the colorblind allele and the X chromosome from Tom that has the colorblind allele you could have the X chromosome from Barbara with a colorblind allele and the Y chromosome from Tom you could have the non colorblind x chromosome Oly or the X chromosome that does not have the colorblind allele on it and get the colorblind X chromosome from Tom or you could have the non colorblind X chromosome and the Y chromosome from the father so there's four equal scenarios and so in how many of these scenarios is the offspring colorblind well here we have a colorblind female she has two of the recessive alleles so that female will be will be colorblind this is a female carrier but they will not show the phenotype of being colorblind this over here is a colorblind male has only one X chromosome and it has the color rind the colorblind allele on it and this is a non colorblind male so out of four equal outcomes two of them have the offspring being colorblind so two out of four that would be a 50% probability that the offspring will be colorblind

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