Assuming that brown eyes is dominant to blue eyes in an auto
Solution
In your first question, in a normal Bb xBb, the Bb population would be 50% as the ratio is 1:2:1.
Bb X Bb gives four alleles B, b, B, b which gives a population of BB, Bb, bB and bb, so 50% of the population would constitute Bb genotype.
the second part of the question, heterozygote offspring would be also 50% as Bb population is 50%. but if bb is found to be lethal, then the total population is distributed among BB and Bb, then it would constitute to 66.6%. this is because the ratio now would become 1:2 instead of 1:2:1.
third part of question, if BB showed 25% reduced viability which indicates among the 33% of BB individuals, 25% have to be reduced (25% of 33.33 is 8.33%) which shows that 25% (33.33-8.33=25)of the population is BB and remaining population which is 75% is the brown eyed heterozygote offspring.
forth part of the question, if 33% of b penetrates then they possess the phenotype of bb, this can have multiple answers
case1: if it is 1:2:1 ratio then 41.65% are bb if they are not lethal, 25% are BB and 33.35% are heterozygote brown eyed offspring.
case2: if it is 1:2 ratio i.e. if bb is lethal, so the population with heterozygotes would be 75% only
case3: if incomplete penetrance, then they do not exhibit the phenotype but carry the alleles which show that 25% have incomplete penetrance and 50% exhibit brown-eyed phenotype.
Hope it is clear!
