Expert: Walter Hintz - 7/9/2005

Question
How can I make a punnet chart showing the crossing of (P1)2 jackalopes with long ears  resulting in all (F1)jackalopes with ALL Short ears?

In my thinking.... if couldn't happen.  Is this correct?  Or am I wrong?  Please explain.


Answer
Hi Durnerin;
 The way the question is worded you cannot show a punnett square where ALL have short ears. You could get results however where all have short ears assuming that long ears are dominant and that the parents are heterozygous(and that there are such animals)
Example:
 Let L equal long ears and l equal short ears
   Parents Ll    x    Ll

    L             l

L  LL           Ll


l  Ll          ll

   The odds of all with short ears are 1   in   4

There is another consideration though again involving the wording of the question.  In Mendelian genetics a P1 generation implies Homozygous parents  (  LL x ll)  in this case there would be no short ears in F1