the pedigree shows the inheritance of a rare kidney disease how is this trait likely inherited if persons 1 and 2 marry what is the probability that their child will have the disease so let's try and answer the first question how is this trait likely inherited and we have to look at the hallmarks of different types of traits in this case we see that there are several individuals like this one and this one who are affected even though the parents are not affected and that means this is clearly uh or not clearly but most likely a recessive trait furthermore we see that equal numbers of men and women are affected or males and females are affected and that makes it likely that this is an autosomal recessive and not an x-linked recessive the second part of the problem asks what is the probability that the child of one and two will have the disease now since it's a recessive disease it means that we are computing the probability that if one and i'm going to draw one out here separately and two mate and have a child what is the probability that the child has the disease we also see that the the father of individual two so there's the mother and here's the father is affected whereas for individual one the mother and father are not affected however the grandmother is affected so as before the the first step in these problems is to write down as many genotypes as possible and in order to do that we should pick some symbols for the dominant and recessive alleles of the trait and so i'm going to say that over a1 is normal a2 over a2 is disease and a1 over a2 the head must be normal since we inferred in in response to the first question that this is a recessive trait and so the heterozygote must be normal must not have the disease and so we can immediately write down that this individual the grandmother of individual number one must be a2 over a2 whereas the father of individual number 2 is also a2 over a2 what about individual 2 well individual 2 at the very least is a 1 over dash since are phenotypically normal and the same can be said for individual number one by the same token all these other individuals who are unaffected can be labeled as a1 over dash at the very least we can say a little bit more about individual number two since they are they're a parent their father had the disease and was homozygous for a2 over a2 they must have at least one a2 allele and therefore they must be a heterozygote similarly the father of individual number one must also have at least one a2 allele their mother had the disease or was affected and therefore they must also be ahead can we narrow down the genotype of individual one like we were able to narrow down the genotype of individual two as it turns out we can't individual one's mom has an a1 allele and individual one's father is a heterozygote therefore individual one can be both a1 over a1 or a1 over a2 which is represented by the a1 over dash notation now having worked out all the different genotypes as best as we could let's go about computing the probability that the this child of individual 1 and 2 will have the disease now the probability of having the disease is the same as the probability of this individual being homozygous for the a2 allele um now three things must be true or there must be three events that have to happen for this individual to be homozygous for the a2 allele the first thing is that his mother must be a heterozygote okay so we can say that number two is ahead the second event that must happen is that this individual's father must also be a heterozygote if the father was homozygous for the a1 allele there would not be a second a2 allele for this individual to end up as a2 over a2 therefore in order for this individual to be homozygous for the a2 allele the father must also be a heterozygote and the third thing that must happen is that when these two individuals mate they the the progeny must be homozygous for the a2olil so we're looking at a monohybrid cross and we are saying that this individual has to be homozygous for the a2 allele from the monohybrid cross now all these three events are independent of each other and we are interested in all three events occurring together so we have the and clause we can apply the product rule to compute the probability of this individual being recessive or having being homozygous for the a2le next we can start computing these probabilities what is the probability that this child's mother individual number two is heterozygote well she is already a heterozygote when we worked out the genotypes we we inferred that she was a heterozygote and therefore the first probability is one what about the third probability the the probability that this individual is uh homozygous for the a2 allele given that the parents are both heterozygotes and we know that that probability is a quarter finally we have to compute the probability that individual number one is a one over a two and not a one over a a1 we know that individual one's father is already a heterozygote in order to compute the probability that this individual is a 1 over 8 a 2. so individual number 1 is a1 over a2 we must know what is the exact genotype of their mother but based on the information that we've been provided we in fact do not know what the genotype of the mother is however we can make a reasonable assumption based on the fact that this is a rare kidney disease and the fact that the individual one's mother is marrying in from outside the family that it's highly unlikely given that it's a rare disease that she would be a carrier and therefore using this assumption of a rare disease we say that his mother is a homozygous for the a1 allele and in this case the probability that this individual will be a1 over a2 is half because um there's uh half his father's sperm carry the a1 allele and half of the sperm carry the a2 allele okay and then we can just multiply these fractions to obtain the probability of this child having this rare kidney disease assuming that this child's grandmother or on his father's side the paternal grandmother is a homozygous for the a1 allele because we were told that this disease is in fact rare we conclude that the probability that this child is has this kidney disease is one eighth now as an exercise let's also work out the situation in which individual one's mother is is a heterozygote even though that's highly unlikely if individual one's mother is a heterozygote what would be the probability of this child having the kidney disease and as before we need to work out the probability that individual 1 will be a head so let's do a punnett square where we write down the gametes of individual ones father and individual one's mother the father produces half a one gametes and half a2 gametes since they are heterozygous similarly the mom also produces half a a1 gametes and half a2 gametes and this is a monohybrid cross and therefore we have a quarter a1 over a2 over here and here and a quarter a1 over a1 and a quarter a2 over a2 now we want the probability that this individual is heterozygous and we may be tempted as we have done in monohybrid process before to state that the probability that this individual is a 1 over a 2 is a quarter plus quarter or half that would be incorrect and the reason is we already know that this individual has the dominant phenotype they do not have the disease therefore this possibility where individual one is a one a two over a two is not possible now our genotypic proportions are a quarter a1 over a1 is to a half a one over a two or one a one over a one is to two a one over a2 and this means that the probability of a1 over a2 is two-thirds since the heterozygous genotype is twice as likely as the homozygous a1 genotype so if the mother of individual 1 is a heterozygous individual instead of being homozygous for the a1 allele the probability that individual one is a head is two two-thirds instead of being half and if you multiply the the fractions through this means that the probability that this uh their child individual one an individual two's child will be um uh homozygous for the recessive trait and that is it'll have the disease is one sixth