Home

Course and Grading Info

Course Schedule

Homework Assignments

Exam Information

Genetics 301 Lecture #6
Spring 2003, January 30, 2003

In mammals, unlike Drosophila, the Y is male-determining. We discussed the mapping of the male-determining gene on the Y chromosome in humans. The study of chromosome abnormalities allowed this gene to be localized to the short arm of the Y chromosome.

The Chi-square test provides a method to test how well data fit a specific hypothesis.

Steps include: State hypothesis, Predict proportions and convert to expected numbers, Calculate (O-E)2/E for each class of progeny, Sum the values for each class, Determine P value from sum (for degrees of freedom = classes –1) from figure, Conclude if results fit model. The hypothesis is rejected if p < .05 (significant deviation) and strongly rejected if p < .01 (highly significant deviation). Problem 3.10 was done as an example.

There are many more genes than there are chromosomes (about 30,000 vs. 23 haploid number of chromosomes in humans). It is reasonable to expect some genes to be inherited together. Genes on different chromosomes, and genes far apart on the same chromosome typically show independent assortment. Genes near each other on the same chromosomes will tend to be inherited together (linkage), which can be exploited for genetic mapping.

Morgan’s lab first demonstrated linkage in Drosophila. The cross involved two genes on the X chromosome, w (white eyes) and M (miniature wings). About 35% of the progeny of the testcross of females heterozygous for both genes to homozygous recessive males are recombinant. This works whether the mutants are on the same chromosome (in cis or coupling) or on different chromosomes (in trans or repulsion). Drosophila males are unsual in that they do not show recombination.

Other pairs of genes show different percentages of recombinant progeny (e.g., only about 1.2% between yellow body (y) and white eyes (w)). This observation led to the idea that linkage data can be used to develop genetic maps. Sturtevant, a student in Morgan’s lab, came up with this idea. For genetic mapping, 1% recombinant progeny is termed 1 map unit (or one centimorgan, cM). Genes can be ordered, with the genes showing more map units between them being further apart on the chromosome. The map distance is calculated by summing the smaller intervals between two genes. For example, where the % recombinant progeny is: y-cv 13.3, y-r 7.5, r-cv 6.2, the correct calculation of the y-cv distance is 7.5 + 6.2 = 13.7.

Crossing over occurs at the 4-strand stage. A single crossover between two genes at meiosis results in 2 parental-type chromatids and two recombinant chromatids. Double crossovers result in the following proportions of recombinant chromatids: Two-stand double crossover gives 0% recombination. Four strand double crossover gives 100% recombination. (This is equally common as the two strand DCOs). Three stand double crossover gives 50% recombination, and is twice as common as either the 2 stand or 4 strand event. Overall the result is that for genes far from each other, there will be independent assortment (50% recombinant progeny).

<Previous Next>