Week 4_lecture_4

Lecture 4 A H. Sturtevant, Morgan's undergraduate researcher, took the concept of distance between two genes being proportional to the number of cross over progeny and designed a mapping function. Sturtevant's mapping function stated that 1% crossing over was equal to one map unit.  Sturtevant was the first to create and publish a map of an organism. He was able to create a map of the X chromosome in Drosophila by using linkage groups. Linkage associations are made between genes on a chromosomes, ultimately a chromosome can be classified as one linkage group. Below is an illustration of Sturtevant's X chromosome. The genes on this map are placed relative to one another. Normal, dominant eye color in Drosophila is brick red. White and vermillion are eye color mutations and are recessive. Yellow is a recessive body color. Miniature and rudimentary are recessive wing mutations. Zero indicates the end of the chromosome. INCLUDEPICTURE "file:///d:/Fall2001/sturdmap.jpg" \* MERGEFORMATINET Two point crosses involve two genes: Examine the following cross, you will be expected to use the correct notation on your exam. As in earlier example, "Y" indicates the Y chromosome. P: wild-type female (y+w+//++) x yellow bodied, white-eyed male (yw/Y) F1: all wild-type Then cross the F1 females (y+w+//yw)(test cross or back cross) x yellow, white males (yw/Y) F2: 342 y+w+; 337 yw; 32 y+w; 35 yw+ (total of 746). When analyzing the F2 data the progeny which represent parental phenotypes should be identified first. These individuals are the most numerous. In this example y+w+ and yw are parentals. The recombinants should be identified next. These are the least numerous. In this example y+w and yw+ are recombinants. In the above example the female and male progeny are not differentiated. This is a clue that they have the same phenotype or genotype and are not sex-linked. In this example, y+w+//yw females and y+w+//Y make up the y+w+ parental class and have the same genotypes and phenotypes relative to yellow and white. The yw//yw females and the yw//Y males make up the mutant parental class. Recall that 1% recombination is equal to 1 map unit. To calculate map units take the sum of the recombinants and divide them by the total number of progeny. For this example: (32 + 35)/746 = 9% or 9 map units. This value is very close to the actual map distance. A map unit is a relative term for a relative distance. A map unit represents different amounts of DNA in different species. The distance represented by a map unit differs if the genes are near the centromere, telomere, etc. The cross over classes came from a cross over event between the yellow and white markers as shown below. If no crossing over had occurred, only parental phenotypes would be present in the F2. INCLUDEPICTURE "file:///d:/Fall2001/strdxovr.jpg" \* MERGEFORMATINET Recall that as the distance between two markers increases, the probability of a cross over event occurring between these two markers increases. Cross over events are relatively random and occur on average once per arm. What if a cross over occurred at the location below? INCLUDEPICTURE "file:///d:/Fall2001/norcombs.jpg" \* MERGEFORMATINET No recombinants would occur in the F2 because no genetic evidence would indicate that crossing over at this site occurred. What happens when a double cross over event occurs? INCLUDEPICTURE "file:///d:/Fall2001/dblxovr.jpg" \* MERGEFORMATINET In this case all of the progeny would have the parental phenotype. Only odd numbered cross overs can be detected between two genes. Looking back at the map, rudimentary is located 57.6 map units from yellow. Is rudimentary linked to yellow? Because these markers are far away from each other, a high rate of recombination would occur between them. These two markers would independently assort, following Mendel's law. Anytime two markers are 50 map units apart they are considered to be unlinked. 50% recombination, which is equal to 50 map units, is like tossing a coin. The outcome of these two genes assorting together or apart is random. Sturtevant would have found a 9:3:3:1 ratio is he had investigated the inheritance of these two markers. Miniature wing is linked to yellow and rudimentary is linked to mini. In essence, these linkages link rudimentary to yellow, but the 50 map units between these genes allows them to independently assort. If enough data is generated for genes on a particular chromosome, one chromosome can be shown to represent one linkage group. If not enough data is present, one chromosome can be broken up into several smaller linkage groups.