The fly genome is fairly large and contains many hundreds of genes. Therefore, finding the particular gene or genes responsible for any given disease has always been a tricky task, quite literally like finding a needle in a haystack.

Traditionally, the search for a disease gene begins with linkage analysis. In this approach, the aim is to find out the rough location of the gene relative to another DNA sequence called a genetic marker, which has its position already known.

The principle of linkage analysis
All our chromosomes come in pairs, one inherited from our mother and one from our father. Each pair of chromosomes contains the same genes in the same order, but the sequences are not identical. This means it should be easy to find out whether a particular sequence comes from our mother or father. These sequence variants are called maternal and paternal alleles.

In the case of the disease gene, the alternative alleles will be the normal allele and the disease allele, and they can be distinguished by looking for occurrences of the disease in a family tree or pedigree. Genetic markers are DNA sequences that show polymorphism (variations in size or sequence) in the population. They are present in everyone and they can be typed (the allele can be identified) using techniques such as the polymerase chain reaction.

Linkage analysis: The top diagram shows paternal (blue) and maternal (red) chromosomes aligned in a germ cell, a cell that gives rise to eggs or sperm. Three DNA sequences are shown, labelled A, B and C. The capital letters represent the paternal alleles and the lower case letters represent the maternal alleles. The middle panel shows the physical process of recombination, which involves crossing over of DNA strands between the paired chromosomes. The bottom panel shows what happens when the crossover is resolved. The maternal and paternal alleles are mixed (recombined) and these mixed chromosomes are passed to the sperms or eggs. If A is the disease gene and B and C are genetic markers, recombination is likely to occur much more frequently between A and C than it is between A and B. This allows the disease gene to be mapped relative to the markers B and C.

This ability to determine the parental origin of a DNA sequence allows us to show whether recombination has taken place. Recombination occurs in germ cells ë the cells that make eggs and sperm. In these cells, the maternal and paternal chromosomes pair up and exchange parts. After recombination, the chromosomes contain a mixture of maternal and paternal alleles. These mixed up chromosomes are placed in our eggs or sperm and passed to our children (see Figure).

As recombination occurs more or less at random, if there is a large distance between two DNA sequences on a chromosome, there is a good chance that recombination will occur between them and the maternal and paternal alleles will be mixed up (see A and C in the figure). In contrast, if two DNA sequences are very close together, they will recombine only rarely. The maternal and paternal alleles will tend to stay together (see A and B in the Figure).

Disease genes are mapped by measuring recombination against a panel of different markers spread over the entire genome. In most cases, recombination will occur frequently, indicating that the disease gene and marker are far apart. Some markers however, due to their proximity, will tend not to recombine with the disease gene and these are said to be linked to it.

Ideally, close markers are identified that flank the disease gene and define a candidate region of the genome between 1 and 5 million bp in length. The gene responsible for the disease lies somewhere in this region.

CLASS PRACTICAL

In class we were given the F1 progeny of a mating between homozygous parents for three traits:

As we see from the Punnett Square above the F1 will be segregated into two classes: ALL the males are triple mutant, as you observed in class, and all the females are wildtype for all three traits.

Next, you took F1 males and mated them to F1 sisters;

IF there was complete linkage, i.e. all three genes were next to each other on the X chromosome then you would predict the following F2 progeny from these parents:

However, we know that the genes are not next to each other so there will be some crossing-over between the X chromosomes within the mother (the father has only a single X chromosome by definition so there is no crossing-over) resulting in the following possible gametes:

Scoring the resulting F2 generation may give something like the following data:

We know from fact that the parental classes will be in the majority because it is more likely that there will be no crossing-over as compared to crossing-over. We see that this is the case with 300 and 280 flies in those classes. The next rule is that the two lowest classes WILL be the double-recombinants, because the probability to two cross-overs is always lower that a single cross-over. And these are the two classes with 15 and 10 flies.

We now can order the genes for the first time; instead of what we have been writing, wmf, the actual order from the data suggest it to be, fwm, with the white eyed gene being located between the other two.

To determine the actual diatance between the genes we use the following formula:

Genetic Distance (cM) = (number of recombinants for that class pair)/(Total progeny) x 100

So the distance between the forked gene locus and the white eyed locus is:

(80 + 85)/(1000) x 100 = 16.5 cM

The distance between the miniture gene locus and the white eyed locus is:

(120 + 110)/(1000) x 100 = 23.0 cM

Thus we can conclude so far that the forked gene is closer to the white eyed gene than is the miniture gene locus.