Category: Science > Biology
Asked by: jennifer1000-ga
List Price: $2.00
06 Oct 2004 12:08 PDT
Expires: 05 Nov 2004 11:08 PST
Question ID: 411141
Brazhyphalangy in man causes shortening of the fingers in the heterozygous condition and is lethal in the homozygous recessive condition. If the fene is homozygous dominant, the hands are normal. A man who has brachyphalangy hands and type A blood has a mother who has normal fingers and type O blood. He is married to a brachyphalangyed wife with type B blood who has a mother with type AB blood and a father with type O blood. How many children would you expect to have shortening of the fingers and type O blood?
Re: Genetic Formula
Answered By: librariankt-ga on 07 Oct 2004 07:05 PDT
HI again! I'm calling the fingers trait F and f (because of the blood type, b would just be confusing!). This question is tricky because there are three rather than two blood alleles - A, O, and B. A and B are equal to each other but dominant to O - so a type O person must be homozygous for O. Man: AO Ff Woman: BO Ff So, this is another 4x4 cross. When you write out each of the possibilities you'll see that 4 children will have type O blood; 1 will have normal fingers (FF), 2 will have short fingers (Ff), and one will die (ff). This is out of 16 possible combinations: ABFF (1) ABFf (2) ABff (1) AOFF (1) AOFf (2) AOff (1) BOFF (1) BOFf (2) BOff (1) OOFF (1) OOFf (2) OOff (1) - Librariankt
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