1. Answer:
1. A cDNA library made from kangaroo brain might not be very useful. It is not likely that dystrophin will be expressed to a reasonable degree in this tissue and so its cDNA is unlikely to be very easy to find in this source.
2. This is unlikely to be useful because it will contain either no or defective dystrophin cDNAs.
3. This should be useful. Is it a full length library? Anyway, it should contain dystrophin cDNAs. (Would you expect all the clones to be identical to each other? How will you screen the library?)
4. This should contain genomic clones from the DMD gene. How will you screen the library?
5. This ought to contain the kangaroo DMD gene. However, because DMD is a very large gene and because YACs are very prone to rearrangement, you should not place too much reliance on it. It would be better to find / make a BAC library or to confirm your results using kangaroo genomic DNA directly.
6. This appears to be your only source of dystrophin mRNA. You can either make a cDNA library from it or you could make cDNA (using reverse transcriptase) and then attempt to amplify kangaroo dystrophin sequences by PCR using primers derived from the human sequence.

1. • single copy sequence - almost all genes are contained in the single copy sequence component of the genome
   • moderately repeated sequence - some moderately repeated sequences may be present within genes and may be transcribed but they are usually within introns and hence are not present in the mature mRNA unless they are in the 5' or 3'UTRs. (Similarly, many mRNAs contain the highly repeated Alu element in a 3'UTR.)
   • minisatellite sequence - Again, this may be found in introns but very few minisatellites are found in mature mRNAs.
   • satellite sequence - Absolutely not! Great blocks of tandem repeats have no place in a normal mRNA!
2. • single copy sequence - Yes
   • moderately repeated sequence - See the previous answer, many of these sequences lie within genes but in noncoding DNA. Others are found between genes. Of course, there are just a few genes which are themselved repeated often enough in the genome to enter this class of DNA - the histones for instance and the rRNAs.
   • minisatellite sequence - Yes, sometimes.
   • satellite sequence - No, the blocks of satellite DNA are usually too big to be accommodated within a gene without destroying its function.

This section has remained in this file for historical reasons. If you are curious to know to what question this answer refers then click here

1. 
• If dominant, then the chance that III₆ will have affected children is close to zero (just the chance of a new mutation).
• If recessive then obligate carriers are:
  • I₂,
  • II₁, II₅, II₆, II₈,
  • III₁, III₂, III₄, III₅, III₆, III₇,
  • IV₁
• The following are at risk of being carriers:
  • III₁₀, III₁₁, III₁₂ all with chance 50%
2. 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

This section is here for historical reasons. If you want to see the questions to which these are the answers then look here

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

   
   The allele b is recessive to B but dominant to b^l. This illustrates the point that in a multiple allele system it may not be obvious which alleles are present.

3. 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 daughter 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

4. 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.

1. The protein is at its isoelectric point, (IEP). Ionisation depends on pH. At the IEP positive and negative charges balance out.

1. Two bands
2.  
   

   Isozymes a - d are products of the PGM1 locus which shows genetically determined polymorphism. Isozymes a and c are products of the PGM11 allele and isozymes b and d are products of the PGM12 allele. Isozymes e, f and g are products of the invariant PGM2 gene. Isozymes c, d, f and g are secondary isozymes which result from post-translational modification of a, b and e.

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 TAL LBA D

   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.

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

   gene 1	gene 2	recombination frequency
   yellow (y)	white (w)	0.010
   yellow	vermilion (v)	0.322
   white	vermilion	0.297
   vermilion	miniature (m)	0.030
   white	miniature	0.337
   white	rudimentary (r)	0.452
   vermilion	rudimentary	0.269

   

   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:

Summary of Results

From this I calculate the distance between the two genes to be:
100 * (627 + 592 + 604 + 622) / 9857 = 24.8 cM

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 too much I think.