The family tree shows the inheritance of a recessive condition. Which of the following statements is/are correct?
1 Individuals 1 and 2 are both heterozygous for this condition.
2 Individuals 3 and 4 are both heterozygous for this condition.
3 Individuals 5 and 6 each have a 50% chance of being heterozygous for this condition.

Answer:

25) E

Please remember that the condition is RECESSIVE. Therefore, an individual needs to be homozygous recessive to have the condition. One offspring out of the four offspring of 1 and 2 has the condition. This is only possible when both 1 and 2 are heterozygous. If even one was homozygous dominant, it would not be possible for any offspring to have the condition. If one was heterozygous and one was homozygous recessive, half of the offspring would have the condition. The only way one out of the 4 offspring could have the condition is if 1 and 2 are heterozygous.
3 and 4 also have to be heterozygous. They do not have the condition, so they are definitely not homozygous dominant. Their father has the condition (homozygous recessive). Therefore, both of them have at least one recessive allele, and since they are not homozygous recessive, they must be heterozygous.
Parents of 5 and 6 do not have the condition = parents are not homozygous recessive
5 and 6 have a brother with the condition = brother is homozygous recessive
Since parents of 5 and 6 have an offspring with the condition, both parents have to be heterozygous (only possible combination)
Draw the punnett square for parents of 5 and 6.
Ratio (homozyg dom: heterozygous : homozygous recessive) = 1:2:1
Since 5 and 6 do not have the condition, they can either be homozygous dominant or heterozygous. We only focus on ratio of 1: 2
Percentage chance of 5 and 6 each being heterozygous = 2/3 x 100 = 66.66%