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Any help please!!! I can not figure this out. made Punnett squares for all and only...

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amayabrianna | Student, Undergraduate | eNotes Newbie

Posted November 10, 2009 at 3:53 AM via web

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Any help please!!! I can not figure this out. made Punnett squares for all and only one that I can get is Tt, Tt. Not an option in my book though!!!

A recessive allele t is responsible for a condition called distornia.  A man who has this condition marries a woman who doesn't.  One of their 4 children has the condition.  What are the possible genotypes of the man and the woman? (to help answer this question, create a Punnet square for each possibility)

The father is Tt; the mother is TT

The father is tt; the mother is TT

Both parents are TT

The father is tt; the mother is Tt

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pohnpei397 | College Teacher | (Level 3) Distinguished Educator

Posted November 10, 2009 at 4:41 AM (Answer #1)

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Your answer of Tt, Tt can't be right because the man has to have the condition.  If he has it, he must be tt.

The correct answer here is that the father is tt and the mother is Tt.  Here's why:

We know he has to be tt so that he can have the condition.  But if she is TT, no children can possibly have the condidion.  So therefore, she must be Tt.

I asssume you didn't like that answer because your Punnett squares tell you that two of their kids should be tt if the are tt,Tt.  But the thing is that two of them don't HAVE to have it.  Punnet squares are just a probability thing, not a for sure thing.  Each of their kids had a 50% chance of getting the condition, but only 25% of the kids did get it.

So, the other three options are impossible, but the fourth is at least possible so it must be the right answer.

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codigodavida | College Teacher | (Level 1) Adjunct Educator

Posted May 18, 2010 at 8:55 AM (Answer #2)

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For one child to have distornia it has to be tt. So both father and mother must transmit t. The father is tt because he has the condition, and the mother must be Tt because she is normal but is nevertheless able to transit a t.

By the way, what is the need for a Punnett square to solve this? I reckon that you were tricked by this silly demand and the 3:1 "rule" - yes, to get 1/4 tt you would cross Tt with Tt; but you must not forget that a family of 4 children is statistically very small: the theoretical value of 1/2 (for tt to come from a cross between tt and Tt, as in the example given) is a weak predictor of how many children actually would be tt. While the tendency is 1/2, the actual number in a 4-member offspring could be anything from 0 to 4. You can relate to this by recalling that the odds for a boy in any family are 1/2, but you could have 3 girls and 1 boy, right?

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compare-the-meerkat | Student , Undergraduate | eNotes Newbie

Posted January 3, 2010 at 1:04 AM (Answer #3)

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The answer is the last one. tt and Tt can make the 4 combinations of Tt, Tt, Tt and tt:

  T      t

t Tt     tt

t Tt     tt

Just because only one of the four children has the disease, doesnt mean that the chances are 1/4, it could be 1/2 and only one child has it! By 1/4 it means that each time a child is born it has a 1/4 chance of getting the condition, the chance does not diminish when a child with the condition is born.

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