Reptile & Amphibian Forums

will triple het pairs produce

3 way crosses.
2way crosses 66% het for third trait.
all three single traits 66% het for other two and 33% het for both.
normal appearing 66% het for any individual trait, 33% het for any two and 16% triple het???
completely normal offspring.

did i miss anyhting?
am i even close?
it sounds right, but sounds to easy also.
thanx

Yes,there could be single homozygous that are 66% het for the other two recessive traits,and double homozygous that are 66% het for the last gene.(I would forget those 33% and 16% labels though.)

I think there is a 1 in 256 chance of producing triple homozygous...a 1 in 16 chance of producing any double homozygous...and of course a 1 in 4 chance of any single homozygous...(these odds are per egg).
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Charles

thank you for that info. is it really 256 to 1 on the triple visual? that really sounds like too much to me. how did you get that number? punnet square or some other way.

hrmm i might be wrong...I was thinking along the lines of 1/4 x 1/4 = 1/16 for a double homozygous...So perhaps it is 1/64 for a triple,and 1/256 for a quadruple.
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Charles

Taking a pair of triple hets (Aa Bb Cc) and getting the probability of being homozygous at all three loci is hard to figure out. You could have any combination of homozygous mutant with homozygous normal. After all, AA bb cc, aa BB cc, aa BB CC, etc. are all triple homozygous. However, you probably want to know the probability of being homozygous for all three mutants (aa bb cc). That is 1/64, and the probability of being triple homozygous normal (AA BB CC) is also 1/64.

The probability of a baby being homozygous for at least one mutant is easy to calculate. The probability is 1 - (3/4 x 3/4 x 3/4) = 1 - 27/64 = 37/64 = 0.578.

The probability of a baby being aa bb but not cc (homozygous aa and homozygous bb) is 1/4 x 1/4 x 3/4 = 3/64.

The probability of a baby being aa cc but not bb (homozygous aa and homozygous cc) is 1/4 x 3/4 x 1/4 = 3/64.

The probability of a baby being bb cc but not aa (homozygous bb and homozygous cc) is 3/4 x 1/4 x 1/4 = 3/64.

So the probability of a baby being homozygous for two mutants (without specifying which two) is 3/64 plus 3/64 plus 3/64 = 9/64 = 0.141. Adding the probability of a triple homozygote gives the probability of getting a baby that is homozygous for at least 2 mutants = 9/64 plus 1/64 = 10/64 = 0.156.

These calculations assume that homozygosity is what matters, as is the case with three recessive mutants. Throwing in one or more dominant or codominant mutants would probably change the starting questions.

Hope this helps.

Statistics is like a bikini. What it reveals is enlightening; what it conceals is vital.

Paul Hollander

okay, thank you. great breakdown.
thanks again.

Err....wow, Nice breakdown...I think I will stick to straight single recessive or co-doms...otherwise I may have a brain anerism trying to figure out double or triple hets....
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PHLdyPayne