Both parents are normal. But, some of the kids have the disease. This indicates that the parents are heterozygous for the disease (let's use Dd). They are phenotypically fine because D is normal. But we know they must both carry the diseased d allele because their kids have the disease.
If you cross Dd x Dd you get DD, Dd, Dd and dd as the possible outcomes. But because each time they have a child they roll the dice as to which of the 4 above happens, they were unlucky as 3 of those times they got dd for their child...so 3 of them have the disease (and are dd). One is phenotypically normal but could still be Dd or DD.
His and her child is also AB cause its the blood type of his/her parents
Answer:
The correct answer will be- Meiosis I, anaphase I
Explanation:
Alleles are the variant alternative forms of a gene which determines the trait of an organism. The trait or phenotype is expressed only when the two alleles controlling a single trait on separate chromosomes are aligned in a complementary position.
During gamete formation, the alleles controlling a trait segregate independently of each other. This segregation of alleles takes place during Anaphase I of meiosis I. During anaphase, the sister chromatids are separated by the mitotic spindles so that the chromosomes can reach the opposite poles.
Thus, Meiosis I, anaphase I am the correct answer.
