Brachydactyly, inherited as a dominant trait, is a skeletal
abnormality resulting in unusually short fingers. A brachydactylous
woman(heterozygous) mates with a normal man. Could they have
brachydactylous children? What proportion of the children would be
expected to have the trait? Show progression of answer through use of
Punnett square.

Hi newcomer-ga, and welcome back to Google Answers.

First, let's assign letters to the brachydactyly gene alleles.  Let
"B" be the dominant allele and "b" be the recessive (normal) allele.
Each person has two alleles for any given gene.

For the woman, she is Bb, meaning that she has one dominant (B) allele
and one recessive normal (b) allele.

She mates with a normal man.  Because the brachydactyly gene is
dominant, a normal person (male or female) must not have a (B), or
they would have brachydactyly.  So, the male must be (bb).

Now, we use the Punnett square (with the alleles donated by the male
on top and those from the female vertically) to figure out what
genotypes the offspring will have:

       b     b
    -----------
B | Bb   Bb
b | bb    bb

So, we get 2 offspring that are (Bb) and 2 that are (bb).  Put another
way, 50% are (Bb) and have brachydactyly.  The remaining 50% are (bb)
and are normal.


You can read more about Punnett squares, their use, and their
interpretation at Wikipedia:

http://en.wikipedia.org/wiki/Punnett_Square

I hope this answer was helpful.  Best of luck in your studies.  Feel
free to request clarification.

       -welte-ga

Very nice, I have another question coming at you that is a bit more involved.