Mendel's second law (of independent assortment) breaks down in one important way - when two genes lie close together on the same chromosome. In this case, alleles which were inherited from one parent tend to stick together when transmitted to the next generation because they are part of the same DNA molecule. However, the further apart two genes are along the molecule, the more likely it is that a meiotic crossover can occur between them and thus create a new combination of alleles. By measuring the frequency of recombinant chromosomes in the progeny, we can estimate the distance that separates the two genes and can make a genetic map.
Such maps are not made just to satisfy scientific curiosity, they have direct application both in science and in medicine.
The mathematics of linkage analysis can be frightening. However, the fundamentals of the subject are easy to grasp particularly if we start by studying experimental organisms in which we can set up those crosses which we need. Later on we must consider how to make the same type of measurements in humans whose "crosses" cannot be experimentally manipulated but where we must make deductions from families in which genetic diseases (or interesting variants) are present.
|A female Drosophila melanogaster laying an egg. Note her (wild type) red eye colour. From Microsoft Encarta Encyclopedia|
The first observation of data from a dihybrid cross which did not fit the predicted Mendelian 9:3:3:1 ratio were made by Bateson and Punnett investigating genes affecting flower pigmentation and seed shape in sweet peas. However, the first thorough investigation was conducted by Thomas Hunt Morgan using the fruitfly Drosophila melanogaster. To minimize the variables he considered a cross between fruitflies in which the females were heterozygous at two loci, the genes purple and vestigial (one affecting eye colour and the other wing shape) while the males were homozygous for each mutant gene. The mutant alleles are written pr and vg respectively and the "wild type" alleles are written pr+ and vg+.
"precross" pr+pr+ vg+vg+ x prpr vgvg / / / / backcross pr+pr vg+vg x prpr vgvg (females) (males) | | | V progeny pr+pr vg+vg, pr+pr vgvg, prpr vg+vg, prpr vgvg phenotype wild type red, vestigial purple, normal purple, vestigial expected ratio 1 : 1 : 1 : 1
The table shows the actual progeny observed:
|wild type||pr+, vg+||1339||709.75|
|red, vestigial||pr+, vg||151||709.75|
|purple, normal||pr, vg+||154||709.75|
|purple, vestigial||pr, vg||1195||709.75|
The reason is that the two genes under consideration are on the same chromosome as shown below in the diagram and the parental combinations of alleles remain together unless separated by a recombination event.
The closer together two genes are, the more likely they are to be inherited in the parental combination.
We can quantify this statement. We define the Recombination Fraction, theta.
q = (number of recombinant progeny) / (total number of progeny)
Theta must lie in the range from 0 to 0.5. A value of 0 means that the two genes are so close to each other that they never recombine, a value of 0.5 implies that the genes are unlinked. (The maximum value of 0.5 and not 1 comes from the fact that crossing over takes place after DNA replication and involves only one chromatid per chromosome of the pair, see diagram.) In this case, the recombination fraction between the purple and vestigial genes is (151+154) / (1339+151+154+1195) = 0.107
Because genes are organized along a linear structure, a DNA molecule, we could expect to create a linear map if we measure the frequencies of recombination between them. We make the recombination frequency the measure of the genetic distance between loci. The map unit corresponds to 1% recombination and is named a centiMorgan in honour of Thomas Hunt Morgan. So the purple and vestigial genes above are 10.7cM apart.
|gene 1||gene 2||recombination frequency|
The data in the question above come from seven separate crosses each involving only two genes. If a cross is set up involving more than one pair of genes it can confirm the approximately linear nature of the map and, by consideration of the relative frequencies of different classes of offspring it will also show directly the order of the genes.
Here is an example of a cross set up by Sturtevant between Drosphila females heterozygous for yellow, white and miniature and males hemizygous for the same three mutations. Notice that in this case the genetic notation has changed to an alternative form. When it is quite clear which gene we are referring to, it is acceptable simply to refer to the wild type allele as + instead of for instance, y+ or w+ etc.
|female gamete||recombination?||number of offspring||frequency|
|+ + +||parental||3501||0.664|
|y w m||3471|
|+ + m||between w and m||1754||0.329|
|y w +||1700|
|y + +||between y and w||28||0.0057|
|+ w m||32|
|y + m||double recombination||6||0.00086|
|+ w +||3|
The male gamete received by any progeny in the above cross is either an X chromosome bearing y w m or a Y chromosome, in either case the genotype of the female gamete is obvious. The order of the genes is as written but it could have been deduced by considering which is the rarest class of female gametes. That must have been due to double recombination and its frequency will be, at most, the product of the frequencies of the two single crossovers. In fact, because the presence of one crossover reduces the likelihood of any other in its neighbourhood, the observed frequency of double crossovers is less than this.
The map distances deduced from the above data are:
y to w, 100 * (28 + 32 + 6 + 3) / 10,495 = 0.66 cM w to m, 100 * (1,754 + 1,700 + 6 + 3) / 10,495 = 33cM y to m, 100 * (1,754 + 1,700 + 28 + 32) / 10,495 = 33.48 cM
The topics include
Read the relevant chapter in any basic (not specifically aimed at human) genetics textbook. One good possibility is Weaver and Hedrick, Genetics, 3rd edition, pub. W.C. Brown, ch. 5 pp104 - 113. See the end of the next lecture for the relevant material from the recommended course text books.
Unfortunately the virtual fly lab to which the question below refers went commercial so we can no longer do this activity. So imagine it!! and then look at the answer. I'll improve this question when I have time!