1. The recognition sequence is GG(A/T)CC. For positions 1,2,4 and 5 in this sequence
only 1 base out of four will lead to cutting. For positions 3 in the sequence, two bases out
of 4 will lead to cutting. Therefore, the odds of having this exact sequence in a random
DNA molecule will be:
1/4 x1/4 x 1/2 x 1/4 x 1/4 = 1/512; or it will cut once every 512 base pairs.
The haploid genome of the fish is (3.8 x 106); expecting a cut every 512 nucleotides
would produce (3.8 x 106)/512 or 7.4 x 105 fragments. Ans: (c).
2. Inserting a foreign DNA fragment into the BglI site would lead to a disruption of the
reading frame in the kan gene. The selection would therefore be for cells that were
tetracycline resistant and kanamycin sensitive. Ans: (d).
3. In this diagram, a black dot is indicative of the presence of a particular allele at one of
these gene loci. During meiosis, we would expect the X and Y chromosome to segregate
from one another at random. If a gene were located on either the X or the Y, then it
would be expected to be in approximately ½ of the sperm- i.e. about ½ of the sperm
would be expected to carry an X chromosome, and about ½ a Y. Therefore, gene locus C
would be expected to carry a gene located on a sex chromosome. Ans: (c).
4. In this instance, if an individual were homozygous for particular allele on an
autosome, we would expect for it to be located on both chromosome homologs and to
segregate to all gamete cells in the male. Therefore, gene locus A would be expected to
carry the same allele, and be found in all cells. Gene loci B and D would also be
autosomal, but they are heterozygous for two different alleles. Is there any evidence of
linkage between the B and D genes?
5. In the family pedigee shown, the unaffected mother is homozygous for a 8 kb
fragment linked to the wild type allele found at the disease gene locus (where the probe
will hybridize in the DNA in this figure is indicated in red:
+
+
The father, however, is heterozygous for a restriction fragment length polymorphism:
there is an additional site in the DNA in this region on one chromosome homolog, cutting
the 5 kb piece into two pieces. In addition, examination of the family pedigree shows
that the disease gene is segregating with the polymorphism indicated by the two smaller
fragments, as shown below:
+
D
This polymorphism can be seen throughout the
genome, and in actuality there may be
numerous different RFLP “alleles”, in
populations, so the answer to question 5 is
incorrect. Ans: (b) false.
6. As noted above, analysis of the family pedigree indicates the disease gene locus in this
family is associated with the 3 + 2 kb fragment pattern, and the 5 kb pattern is associated
with the wild type allele. Ans: (d). How do you explain the pattern for individual II-9?
7. deoxyadenosine triphosphate is the nucleotide tri-phosphate normally incorporated
during DNA synthesis. Consequently, we would expect ALL the sequencing reactions to
contain this dNTP. Ans: (e)
8. Reading 5’ to 3’ on the strand being synthesized, the sequence would be:
5’- GATACTGTTACC-3’ Consequently, the TEMPLATE STAND WOULD READ:
3’- CTATGACAATGG-3’
Ans: (a).
To derive the pathway, the compound that will support the growth of the most number of
mutants would be the last compound in the pathway. Following that logic:
ED FA CB Z
Similarly, a mutant early in the pathway would be able to survive on any compound
synthesized downstream of the enzymatic block. Following that logic:
ED FA CB Z
5
2
6
4
3
1
9. B is the immediate precursor to Z. Ans: (e).
10. A mutation in gene 2 would block the pathway to F, leading to a potential buildup of
precursor D. Ans: (c).