punnett square

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mercuryy
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punnett square

Post by mercuryy » Sat Apr 08, 2006 2:32 am

Here is a questoin from the past National Bio Competition: A geneticist crossed YYRR (yellow-round) peas with yyrr (green-wrinkled) peas and self-pollinated the F1 to produce F2 offspring. In the F2 generation what proportion of the yellow-round individuals were pure-breeding (homozygous for both genes)?

The answer is 1/9, but I am not sure how you get that. isn't it suppose to be 9/16 since both yellow and round allell are dominant?

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Post by kiekyon » Sat Apr 08, 2006 7:40 am

your answer is correct if you are looking for the ratio of yellow green pea.
however look at your question closely again,

In the F2 generation what proportion of the yellow-round individuals were pure-breeding (homozygous for both genes)?


among the 9 yellow round pea, only 1 is homozygous, hence the answer=1/9

mercuryy
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Post by mercuryy » Wed Apr 26, 2006 12:44 am

Thanks. i got that question now. here's another of similar type. but i am not sure how are u suppose to draw the diagram for this one.

A couple are both carriers of the autosomal recessive allele for albinism. They have two children.
What is the probability that both children will be phenotypically identical with regard to skin colour?

The answer is suppose to be 5/8 . Can anyone show me how to do that?
Last edited by mercuryy on Wed Apr 26, 2006 11:27 pm, edited 1 time in total.

mercuryy
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Post by mercuryy » Wed Apr 26, 2006 9:36 pm

help anyone? ?????????:?

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Khaiy
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Post by Khaiy » Thu Apr 27, 2006 12:32 am

Here's how this kind of problem works.

Each parent are carriers for albinism, which means that they have the allele for the trait, but don't express it (which means that their genotype would be Aa).

This means that each parent has a one in two chance of passing on the albino allele to any offspring (gender doesn't matter since the trait is autosomal).

The question asks what the odds are that both children will have an identical phenotype (to each other) in terms of skin color. So, you'll need to figure out the odds of the children both being albinos, and both having regular skin color.

For albinism, each parent has a 1/2 chance of passing on the allele. The multiplication rule says that this would be (1/2)*(1/2), which equals a 1/4 chance of any one child of theirs being an albino. The chance of them having both children being albinos would be 1/16 (Multiply the odds of one child by the odds of a second child in a row being an albino, again thanks to the multiplication rule).

For a single child with normal skin color, each parent again has a 1/2 chance of passing on the normal pigment allele. However, there are more combinations of alleles that will equal normal pigment compared to albinism (draw the punnett square to see). The result of this is that any child of theirs has a 3/4 chance of having normal skin pigment. The odds of two children in a row having normal pigment is, according to the multiplication rule again, (3/4)*(3/4)=9/16.

Now, since there are two different ways that you could fulfill the requirements of the question (that both children be either normal or albino), you have to add the odds of each different way together, in accordance with the addition rule. So, you have a (1/16) chance of both children being albino, plus a (9/16) chance of both children expressing the normal skin color. Added together, you get 10/16, which reduces down to 5/8.

I may have skipped some information, so if you have any questions please feel free to ask.

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Post by mercuryy » Thu Apr 27, 2006 1:26 am

Thanks A LOT. :lol: i have no problem understanding it. i did the question myself up to the part where i got stuck on the 2 possible criteria. and u explain it . Thanks again.

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