1. What is the standard measuring unit for a genetic map?

Figure 5.4 shows the "definition" of the map unit, the centimorgan (cM). Two genes alpha and beta that are 1 cM apart will show a recombination frequency of 1%. Thinking of it in terms of a bunch of cells going through meiosis, we see that a single crossover will occur between alpha and beta in 2% of the cells, generating a collection of gametes in which 1% of them have the recombinant (non-parental) genotype.

2. What are the effects of a cross-over outside the region between two genes, or two cross-overs in the region between the genes?

Figure 5.5 shows that a cross-over outside the region between two genes does NOT produce gametes that are recombinant for these two genes.

Figure 5.6 shows that two crossovers, involving the same pair of chromatids, in the region between two genes also does NOT produce gametes that are recombinant for these two genes.

3. How does the measurement of recombination frequencies between pairs of genes allow us to start constructing a genetic map?

An example is presented in Figure 5.7. Three of the several thousand genes on the Drosophila X chromosome are "y", "rb", and "cv". The measured recombination frequency (percent non-parental gametes) for standard crosses involving alleles of genes "y" and "rb" is 7.5%. For cross "rb" x "cv", the measured recombination frequency is 6.2%, and for "y" x "cv", it is 13.3 %.

From these numbers, we can conclude that (1) Yes, these three genes all are on the same chromosome; and (2) The loci of these three genes must be such that gene rb is in the middle, as shown in part c of the figure.

By doing further crosses involving other genes, one can start to figure out the actual loci of all of the genes.

4. Over the length of a chromosome, what map distances (and recombination frequencies) do we see, and how does physical distance correspond with genetic map distance?

Figure 5.8 shows chromosome #10 of corn. The map "starts" at the telomere at the end of the short arm. At 18 cM we get to gene "rp1". We continue on, and eventually get to the other end, with gene "gln1" at position 173 cM. Genes that are significantly closer together than 50cM are "linked", whereas genes rp1 and gln1 are genetically unlinked.

Figure 5.15 shows chromosome 2 of Drosophila. The point of this figure is that crossing-over is, in general, much rarer in heterochromatin than in euchromatin. Two genes separated by a lot of heterochromatin may appear to be a lot closer genetically (cM distance between them) than they are physically (# base pairs distance between them).

Problem S-5: "Drawing the Results of Crossing-over".

For oogenesis or spermatogenesis going on in you, consider only chromosome #4, for which one of the chromosomal arms is significantly longer than the other. Assume that you are heterozygotic for three genes alpha, beta, and gamma that are on chromosome #4. Gene alpha is on the long arm, quite near the centromere. Gene beta is near the telomere on the long arm. Gene gamma is in the middle of the short arm. Make the two chromosomes #4 you got from your parents be two different colors. Use one side of one full sheet of plain paper to draw four accurate diagrams showing your chromosomes #4 in meiotic anaphase I for the following situations. Show accurate color-coding and allele designations in all four diagrams.

a. No crossing-over occurred during prophase I.

b. A single crossover occurred, half way out the long arm.

c. Two crossovers occurred, involving the same two chromatids, at locations about 1/3 and 2/3 of the way out the long arm.

d. Two crossovers occurred, involving all four chromatids, at locations about 1/3 and 2/3 of the way out the long arm.


Problem S-6: "Pea Chromosome #1 Map"

Consider five genes (alpha, beta, gamma, delta, epsilon) known to be on chromosome #1 in peas. You know the location of gene delta, and that gene beta is somewhere to the "left" of gene delta. You know nothing about the locations of genes alpha, gamma, or epsilon on this chromosome.
You want to map these genes on chromosome #1, so you do a bunch of appropriate crosses of homozygous parents, thus generating appropriate heterozygous F1s and then the F2s. Genetic analysis shows that the gametes produced by the F1s were as shown in the table below (from 250 cells going through meiosis, generating 1000 gametes, for each heterozygote).
(For example, "from cross #1" below gives the F1 gametes that resulted from the original parental cross AABB x aabb. That is, the "parental-type" gametes are AB and ab.)

from cross #1:    AB = 475 ...   Ab = 25 ... 
  aB = 25  ....   ab = 475

from cross #2:    AG = 425 ...   Ag = 75 ... 
  aG = 75 .....   ag = 425

from cross #3:    BG = 400 ...   Bg =100 ...   bG =100 ....  
bg = 400

from cross #4:    AD = 485 ...   Ad = 15 .....  aD = 15   .....
ad = 485

from cross #5:    BD = 490 ...   Bd = 10 ....   bD = 10  ..... 
bd = 490

from cross #6:    GE = 275 ...   Ge =225 ...   gE =225 .....   ge = 275

from cross #7:    AE = 250 ...   Ae =250 ...   aE =250 .....   ae = 250

Construct the genetic map of pea chromosome #1, showing the locations of all five genes, listing cM distances separating them to the extent possible.