SAQ answers

Lecture 1

1. The trailing strand would be unable to be replicated all the way to the end (where would the RNA primer be placed to prime the Okazki fragment synthesis?) and would shorten every cell generation. The leading strand would be replicated to the end of the trailing strand and so would also shorten (but one cell division cycle behind the trailing strand. Eventually the chromosome would be lost. It has been suggested that the syndrome progeria which leads to premature aging, (the affected children die of diseases of old age in their teens) is caused by a mutation in the gene coding for the enzyme telomerase. The similar syndrome, Werner's syndrome has been proved to be caused by mutations in a DNA helicase gene (involved in DNA unwinding).
2. Four times that amount, or 12 picograms. (Diploid cells with replicated chromosomes)

Lecture 2

1. a) 46 chromosomes are present in a premeiotic germ cell. When meiosis begins each has replicated and when it condenses in division can be seen to be composed of two chromatids.
   b) Each human gamete contains 23 chromosomes, each at that point composed of a single chromatid.
   c) The number of chromosomes changed as a result of the first meiotic division in which homologous chromosomes go to opposite poles of the spindle.
2. 

• If dominant, then the chance that III₆ will have affected children is zero.
• If recessive then obligate carriers are:
• I₂,
• II₁, II₅, II₆, II₈,
• III₁, III₂, III₄, III₅, III₆, III₇,
• IV₁

• III₁₀, III₁₁, III₁₂ all with chance 50%

1. Several genes are involved in determining cat coat colour but only two are segregating in this cross, the genes Black (alleles B and b^l) and Agouti (alleles A and a).
   The initial cross is:

          BB AA    x    blbl aa
      brown tabby     cinnamon solid
                   |
                   |
                   V
F1              Bbl Aa
             brown tabby
These cats are intercrossed to give:

F2     BB AA            |
       Bbl AA  (x2)     |
       BB Aa  (x2)      |  9 brown tabby
       Bbl Aa   (x4)    |
       blbl AA       |
       blbl Aa  (x2) |  3 cinnamon tabby
       BB aa                 |
       Bbl aa   (x2)         |  3 black
       blbl aa       |  1 cinnamon

Lecture 3

1. FHC is genetically heterogeneous. It is also a disease with reduced penetrance. The high frequency of sporadic cases might be caused by a high mutation frequency and, given the number of possible gene targets this could be so. However, as the causative mutations are being found, most can in fact be traced back to subclinically affected parents so this is a case of either low expressivity (minor symptoms) or reduced penetrance. It is not known what is the phenotype of a homozygote, might it be lethal? Nor is it known what is the result of simultaneous heterozygosity for more than one of the relevant genes.
2. The tabby gene is irrelevant here. The first cross was of the form:

          blbl     x     BB
         cinnamon  |     black
                   |
                   V
                  blB
                 black

The second cross was:
          blbl     x     Bb
         cinnamon  |     black
                   |
                   V
                blB  blb
             black  chocolate
                 1  :  1

Lecture 4

1. The pedigree looks like this and the woman must be a carrier.
1. The chance that any son will be affected is ½. The chance that the first child will be an affected son is ¼, (half a chance of being a son, half a chance of being affected).
2. Similarly the chance of any daughter being affected is also ½ (the daughter must receive one affected X chromosome from her father), the chance of the first child being an affected daugter is thus ¼.
3. Again, the chance that any daugter will be a carrier is ½ and so the chance that the first child is a carrier daughter is ¼

If the woman had had an affected brother rather than an affected father, then her chance of being a carrier goes from 1 to ½. All the odds in the question thus reduce by a factor of 2 to 1/8

2. Perhaps the husband is an XX male. His karyotype should be examined more closely. DNA tests for the presence of Y chromosome DNA and for the gene SRY could be carried out.

Lecture 5

1. Answer:
1. A whole genome cosmid library should contain clones covering the desired region. How will you screen it to find the correct few clones out of the several hundred thousand clones in the library?
2. This is unlikely to be useful.
3. This may be useful. Is alpha-1-antitrypsin expressed in the embryo? If not the cDNA library will not contain any relevant clones.
4. This will certainly contain alpha-1-antitrypsin cDNA clones at a good frequency. Again, how will you screen it? Is it in an expression vector to enable you to screen with antibodies?
5. Wrong chromosome - so not useful. It would be nice to have access to a chromosome 14 library though.
6. This might be useful, but I would be reluctant to place too much reliance on any DNA which has at some stage in its evolution passed through a YAC. These clones are very prone to internal deletions and rearrangements. It might be a good source of DNA fragments to probe the first cosmid library to obtain cosmids from the area.
2. Begin with:
1. A Search of the Genome Database (GDB)
2. A search of EST databases such as Unigene

Follow up any clones listed. Sometimes it will be necessary to contact authors but, since the advent of gridded libraries it is often possible to obtain named clones from public depositories (for instance in the UK the Human Genome Resource Centre (on the same research campus as the Sanger Centre).

3. When the gene (in this case - genes) has been identified, the DNA from members of a family can be studied in order to find the mutation(s) which is (are) causing a problem in that family. Once this is done all family members can, if they wish, be screened to confirm whether or not they are carrying a mutation. Advice can then be given on the likelihood of transmitting the disease to their offspring. It will also be possible to offer antenatal diagnosis of genetic disease by chorionic villus sampling or by amniocentesis, followed by PCR amplification of the mutation site and then by a mutation detection technique (possibly DNA sequencing).

Lecture 6

1. Steeling ourselves to ignore the questionable message expressed by this gene:

WEE WET WET ARE NOT ALL BAD

point mutation

WET WET WET ARE COT ALL BAD

point mutation

WEW ETW ETA REN OTA LLB AD

deletion of one nucleotide leading to frameshift

WET WET WET ART TEN OTA LLB AD

insertion of two nucleotides leading to frameshift

WET WET ARE NOT ALL BAD

deletion of a trinucleotide

WET WET WEG OOD MOR NIN GTA REN OTA LLB AD

insertion (of a repetitive element)

WET WET WET WET WET WET ARE NOT ALL BAD

trinucleotide expansion


2. The mutation frequency of the X linked genes is much greater than that of the autosomal gene because the X linked genes can undergo unequal cross over at meiosis. Depending on the position of the crossover this can generate chromosomes on which there is only one gene, either the red sensitive gene or the green sensitive as shown in the figure. (In reality there are often multiple genes present for the green sensitive opsin, but this does not significantly affect the argument.) Unequal crossover can also generate chromosomes which contain more than two genes or even hybrid genes. For the full story turn as usual to the protanopia and deuteranopia references in OMIM.

Lecture 7

1. The boy may have a deletion on his X chromosome covering the three genes XG, KAL1 and STS causing a "contiguous gene syndrome". You might ask "Is there any family history of this combination of symptoms?" paying particular attention to the male relatives of his mother. You should also ask for a detailed karyotype examination, can any deletion be observed in the terminal region of Xp in the boy? What about his mother's X chromosomes, do both appear normal or is there a deletion on one of them? If there is no family history and the mother's X does not appear to be deleted then perhaps the boy has inherited a new mutation. You could look at the human map for that region and find polymorphic DNA markers. Has the boy failed to inherit any alleles from his mother for markers near the two genes?
2. On the contrary. Any reproductive problems in heterozygotes arise because of recombination within the region of the inversion. If the chromosome pair is homozygous for the inversion, it will be able to pair without forming an inversion loop and there will be no production of unbalanced gametes as a consequence of crossovers within the loop. (It has been suggested that inversions of this type can help along the process of speciation by making infertile the hybrids between species which differ by an inversion.)
3. The genes whose absence causes Turner's syndrome must be located on Xp. (Logically, since males with only one X chromosome do not have Turner's syndrome, the genes must also be on the Y chromosome). The posession of three copies of Xq is not lethal because the iXq is X inactivated.

Lecture 8

1. These are the data from an experiment by Sturtevant (Morgan's graduate student).

The distances are given in cM from the furthest left gene, yellow, by adding together consecutive intervals. Note that the interval between widely spaced markers (e.g. w to r, 45.2cM) is less than the sum of the intervals within it (56.6cM). This is because in cross involving widely spaced genes there is opportunity for double recombination - which puts the genes back into the parental combination and thus reduces the apparent recombination frequency. This also has the effect that some distances (for instance from y to r), by being the sum of several smaller intervals, amount to greater than 50cM on the map even though when we measure any distance in a cross it cannot exceed 50cM.

2. Enter the Virtual Fly Lab here or here. Click on the "design a cross between two flies" button.

• Cross one. Mate white miniature males to wild type females. All the offspring should be wild type in appearance (you will get about 10,000 in total).
• Cross two. Mate the female offspring of cross one to the male parent. (You are performing a typical test cross or backcross.) Again you will get about 10,000 progeny.
• I got back:

Lecture 9

1. a) A centiMorgan is a unit of recombination distance. It is equivalent to 1% recombination between two loci.
   b) A Lod score of 3 is enough to convince that two loci are linked. It will be associated with a recombination fraction which is the estimate of the distance between the two loci. If Z is greater than 3 then it is more than 1000 times more likely that the loci are linked than that they are segregating randomly.
   c) Yes. Z_max is 3.4. So the two genes are linked at a genetic distance of 30cM (which is not very close).
2. a) RFLP = Restriction Fragment Length Polymorphism. An RFLP will be most useful for family studies when it is likely to be informative. This will be the case when it has many common alleles.
   b) The relevant part of the pedigree is repeated here.
   
   Allele 2 of the RFLP appears to be linked to the disease mutation in the main part of the pedigree. I have filled in with red to show the presence of this chromosome. Individual III₅does indeed carry the chromosome. At first sight it may appear that her partner, III₆may contain the same chromosome because he also has allele 2 at the RFLP locus. However, his allele 2 comes from his father. The only way that he could carry the disease allele would be if a recombination event had transfered the mutation onto a chromosome marked with allele 1 in creating individual II₅ and a second recombination had then transfered the mutation onto a chromosome marked with allele 3. The chance of this happening is 1% x 1% = 1 in 10,000. Low enough not to worry him I think.

Lecture 10

1. Assume that both populations are in Hardy Weinberg equilibrium.
   Let the frequency of the M allele, f(M), = p and f(N) = q.
   a) The frequency of heterozygotes, f(MN)_Somerset is expected to be 2pq.
   Therefore f(MN) = 2 x 0.49 x 0.51 = 0.4998
   b) Similarly f(NN)_Kyoto = q² = 0.65² = 0.4225
2. First estimate p and q for the infants.
   p (=f(allele 1)) = ((2 x 44) + 45)/(2 x 100) [2 allele ones per homozygote, one per heterozygote out of 200 alleles in total (100 individuals each with 2 alleles)]