1.Two albinos marry and have four normal children. Illustrate this mating with appropriate symbols and expla

Answer 1

A - albionic allele
a - normal allele

So if both parents are albino and have 4 normal ofspring both their genotype would have to be (Aa) as the larger A is dominant over the smaller a it hides the albionic trait in the physical appearence. However if the above parents with the genotype Aa mate the offspring have a 25% chance that the offspring are abionic. So the four children are all chance events that fell within the 25% quarter of being albionic. There you have it, I hope you get the idea.

Have a look at the link below

Answer 2

Difficult to show on a Punnett square, because skin color is a trait that is known to be due to at least three genes, and some hypothesize it is due to as many as six or more genes. However, oculocutaneous albinism can be caused by a mutation
in any one of four genes: the OCA2, SLC45A2, TYR, and TYRP1 genes. Mutations in the MC1R gene modify the course of oculocutaneous albinism. This appears to be one of the rare cases where a genetic mutation turns out to be dominant.

So let's say (for simplicity) that skin color is caused by only three genes, which I'll designate as X, Y, and Z. Xa, Ya, and Za are the mutated versions of the genes that code for albinism. If the mother has genes X, Y, and Za, and the father has genes X, Ya, Z, it's possible that the children may inherit X, Y and Z from the parents, and won't carry any of the abnormal genes.

Answer 3

X Y

x xx xy

y yx yy


they both carry the gene but have a 25 percent chance of passing it down to their children in their case they beat the odds of having four non albino children

Answer 4

Look in your science/biology book for the section about Punnet Squares.

Answer 5

This can't happen