Answer linkage

From Wikionchus

• The probability that two randomly chosen P. pacificus mutants fall on the same Chromosome is roughly 1/6. Yes, this relies on equal chromosome sizes and some other simplifying assumptions. And, yes, this was only a warm-up question!

• Now, lets get serious:

• If the two mutations are on different chromosomes, each mutation has probability 1/2 of being transmitted to a particular gamete, independently of what happens to the other locus. All combinations are equally likely with probability 1/2 * 1/2 = 1/4. If they are tightly linked they behave like a single factor: Both end up in the gamete in 50% of the cases.

genotype	tightly linked /		different chromosomes
no mutation	1/2		1/4
both mutations	1/2		1/4
egl, not dpy	0		1/4
dpy, not egl	0		1/4

• Assuming recessivity of both mutants, the following phenotype frequencies are expected when two randomly chosen gametes of the above type meet:

phenotype	tightly linked /	different chromosomes
no mutation	3/4	9/16
both mutations	1/4	1/16
egl, not dpy	0	3/16
dpy, not egl	0	3/16

• This is exactly what the monk's peas told us...