31. See attachment 1
32. Yes it is possible
33. 1 in 2 or 50%
34. See attachment 2
Explanation:
31. Have a look at attachment 1.
I will refer to each layer of the pedigree as 1, 2, and 3 to make it easier to explain. Since the disease is dominant (DD or Dd will be affected, shown as coloured symbols), then those who are unaffected (white symbols) will be dd. So we can fill those in first.
In order for the parents at layer 1 to have unaffected children in layer 2 (dd), there must be a second d allele from the female (circle). Therefore, she must be Dd. The two affected individuals in layer 2 (coloured squares) have to have received a d allele from their affected father (square) because that is the only allele he has. To be unaffected, they also must have a d from their mother (if they got the D allele, they would be affected).
The same principle works for the affected individuals in level 3, if they are affected, they all must have D from their father (Dd, square), and can only have d from their mother (dd, circle). Therefore, they are all Dd.
32. To have type A blood, an individual must have the genotype AA or AO. To have type O blood, an individual must have the blood type OO. The easiest way to see what genotypes are possible for their children is to draw two punnet squares with each possibility (see attachment 2).
In possibility 1, all of the offspring are AO, so the offspring have type A blood. In possibility 2, 1 in 2 of the offspring would be AO, so type A. 1 in 2 of the offspring would be OO, so type O.
Therefore, it is possible for their offspring to have type O blood, but only if the man’s genotype is AO, rather than AA.
33. Colourblindness is X linked and recessive, meaning it is different from the above examples, because the gender has to be taken into consideration. Females have two copies of the X chromosome, and males only have 1 copy (and 1 Y chromosome).
In order to have the condition, one of two things has to happen, either XX females carry two copies of the colourblind allele, or XY males only have the colourblind allele (and therefore no normal allele).
Let’s designate the normal allele B and the colourblind allele B. There are three possible genotypes for a woman: XBXB, XBXb, or XbXb. There are two possible genotypes for a man XBY and XbY.
A woman with normal colour vision can either be XBXB or XBXb (with XbXb she would be affected). However, in this case we know that this woman’s father was colourblind. This means his genotype has to be have been XbY (if he was XbY, he would be unaffected). Therefore, we know the woman’s genotype must be XBXb. If she has children with a man with normal colour vision (which we know means he must be XBY), what is the probability that their sons will be colourblind? To figure this out we do a simple punnet square. See attachment 3
In this case, we are only interested in the probability of the son’s being colourblind. The possible genotypes for the son are XBY (unaffected) or XbY (colourblind). This means there is a 1:2 or a 50% chance of their son being colourblind.
34. See attachment 4 for diagram. Let's denote the normal allele as "F" and the cystic fibrosis disease allele as "f". The inheritance is recessive, so two copies of f are required to be affected by the disease. Firstly, we know that the grandparents of the parents are all heterozygous (Ff), as are the parents themselves. Therefore, they are unaffected and can be put in the pedigree as such (not filled in black, and denoted Ff). Their oldest child should be to the furthest left of the pedigree. He is affected, so must be ff and his symbol coloured. The second child is normal, which we assume means she is not hetetrozygous or a carrier, so must be FF. The final daughter is a carrier, meaning she is heterozygous but unaffected, Ff.
