Answer key, practice problems 2000--week 4

1. (a) When there are equal numbers of the four bases, the probability of any particular base occurring at a given position is 1/4. The probability finding a G base at a given position is therefore 1/4. The probability of finding GG is (1/4)(1/4), etc. Thus, the probability of finding GTCGAC is (1/4)6 or 1/4096. In other words, the SalI site will on average be found once every 4096 bp; the average fragment from a SalI digenst will be 4096 bp.
(b) The sequence GCGGCCGC will be found every 48 bp; the average fragment will be 48 bp = 65536 bp.
(c) When the base composition is skewed, the probability of finding a particular base at a given position is likewise skewed. In this instance, the A-T content = 60%, so the A content is 30%; the probability of finding an A base is 0.3. By the same token, p(T) = 0.3, p(C) = 0.2, and p(G) = 0.2. Thus, for a DNA molecule that is 60% A-T, the probability of finding the sequence GCGGCCGC is the product of the individual probabilities--

    p(GCGGCCGC) = (0.2)8 = 2.56 x 10-6. The average sized band will be the reciprocal of this, i.e., 390625 bp.

2. (a) The fragment lengths could be determined by gel electrophoresis of the digested DNA along with standards of known size, followed by staining for DNA with ethidium bromide and photographing of the gel under ultraviolet light.

(b) This being linear DNA, a single cut will yield two pieces, two cuts will give three pieces, and so on. Thus, XbaI (which gave 3 pieces) must have 2 sites in the molecule, and PstI must have a single site.
(c) First of all, we can see that for each digest, the fragment lengths add up to 25 kb – that is the length of the molecule.

Considering the XbaI digest first: there are three possible ways of explaining this result-- the 5 kb fragment is the middle fragment, the 7 kb fragment is the middle fragment, or the 13 kb fragment is the middle fragment (shown below as the three separate horizontal lines; the bars sticking up from the horizontal lines represent the possible locations of the XbaI sites). Now consider the PstI digest. It cuts only once in the molecule, and it makes an 8 kb fragment and a 17 kb fragment. Therefore, the PstI site has to be 8 kb from one end of the molecule (either the right end or the left end). The double digest resolves the ambiguities. After digesting with both enzymes, we get 4 fragments (as we should); when we compare the XbaI single digest with the double digest, we see that two of the resulting fragments (the 5 kb fragment and the 7 kb fragment) are the same between the two digests. However, instead of the 13 kb fragment, we now gt two fragments, 3 kb and 10 kb--i.e., PstI cuts within the 13 kb fragment produced by XbaI to give a 3 kb fragment and a 10 kb fragment.

Putting these bits of information together--Pst I cuts 8 kb from one end of the DNA molecule; this cut site is within the 13 kb fragment produced by XbaI, and is 3 kb from one end of that 13 kb fragment. The only XbaI map that allows these results is the third one (the 13 kb fragment is the middle fragment). The completed map is shown at the bottom. (If you don't understand this--take each of the three possible XbaI maps, and in each one, try placing the PstI site 8 kb from one end, and see if that gives the expected fragments; then try placing the PstI site 8 kb from the other end and see if that works; etc.)

3. Because a person who is heterozygous at the locus would "express" both alleles--i.e., both alleles would be manifested in an assay for the RFLP--the relation between the two alleles is best described as "co-dominant".

4.

5. Remember that the goal of the somatic cell hybrid approach is to get different cell lines to lose different human chromosomes at random. In the presence of aminopterin, the cells can grow only if they retain the TK gene and the HGPRT gene. If the human cells provided the functional HGPRT gene, the hybrid cells would not be able to lose the chromosome carrying that gene if the cells were grown in the presence of aminopterin.