Establishing linkage between his, ade, leu and lys genes in Saccharomyces cervisiae through random spore analysis

Lauren Lior-Liechtenstein

Lab Section A

Yeast Meiotic Mapping

Abstract

Genes are located within chromosomes, which can either come in haploid or diploid form. When certain genes are on the same chromosome, they are considered linked, and this can then be used to create a linkage map which shows the genes in their relative distances to each other. In haploid Saccharomyces cerevisiae this can be observed by carrying out random spore analysis and a subsequent chi-squared test, and comparing the recombinant frequency and the chi-squared value against a critical chi-squared value to determine if the gene pair is linked or not. The gene pairs his/leu, his/lys and leu/lys were found to be linked, while his/ade, ade/leu and ade/lys were either not linked or too far apart on the same chromosome to behave like it was linked. This tool is useful in identifying the linked genes on a chromosome and their roles in development in higher organisms.

Introduction

Genes are contained in chromosomes, which can sometimes hold thousands of genes. Those on the same chromosome would be considered linked because of the fact that inheritance occurs through genetic crosses. Independent assortment deals with the issue of combinations of chromosomes during meiosis, which will be referred to as recombination throughout this report. The aim of this experiment is to figure out whether the Saccharomyces cerevisiae genes his, ade, leu and lys are linked, or if they are on different chromosomes. We can figure this out by calculating the percentage or recombinants versus the percentage of parentals and performing the chi-squared test with the tabulated the results. With this knowledge in hand, a method of genetic mapping, where linked genes were placed in a linear fashion along a chromosome according to their recombination frequency. With the chance of crossing over at any two random points on a chromosome being 50%, the higher the recombination frequency, the more further apart the genes are. This kind of map is called a linkage map, and is based on recombination frequency. In order to produce this map in diploid organisms, a test cross must be performed with heterozygous and homozygous parents, so that the F1 genotypes can be determined.

S. cerevisiae can exist in either haploid or diploid form, and usually occurs in the later when grown normally. However, when grown in a special sporulation medium or starvation conditions, they will undergo meiosis and subsequently, sporulation. During this, four spores will group into four, and become packaged into a sac known as the ascus. Random spore analysis allows us to determine the map distances between certain genes, and is effective in that we can analyse a great number of spores.

In performing the chi-squared thest, the nullifying hypothesis used in this case will be that the genes are not ...

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In performing the chi-squared thest, the nullifying hypothesis used in this case will be that the genes are not linked, and this would be indicated by a chi-squared number lower than 3.841 and a recombination frequency higher than 50%. In order to see which genes are linked, however, we must consider the opposite and require the chi-squared number to be higher than 3.841 and the recombination frequency lower than 50%, which would reject the null hypothesis and increase the probability that the genes are linked.

Materials and methods

To establish which genes were linked, random spore analysis was performed, and this involved several steps, the first of which being to treat the sporulated cells with glusuclase, a snail enzyme, which breaks down the wall of the ascus. The four spores are then separated from each other through agitation, after which it is then plated on YEPD (Yeast Extract, Peptone, Dextrose) a rich medium, for both haploids and diploids to grow. The haploids can be distinguished by a red adenine marker (ade2) and are then transferred to a YEPD master plate. These are then plated onto mediums which are complete in terms of nutrients, except for one, such as -ade, -leu, -lys and –his.

Having done this, the recombination frequency can then be calculated for each gene pair using the number of recombinants and the following formula:

RF=(R/(R+P))x100 where R equals the number of recombinants and P equals the number of parentals. Using the number of parentals and recombinants, we can also carry out the chi-squared test to give a number which would then be compared to the critical value, in this case 3.841 to see if the genes were possibly linked. The two class equation for chi-squared is as follows:

X2=(P-E)2/E + (R-E)2/E where R=recombinant, P=parental and E=expected value (total spores/2).

The haploid strains being used for this experiment are:

CH560 a ura3-52 X CH859 a ade2 his4-519 leu2 lys2

Results

(Table 1)

Table 1: Shows the tabulated data from the S. cerevisiae spores, and which are parental and recombinant.

(Table 2)

Table 2: The table shows the results of the chi-squared test, together with the amount of parentals and reocombinants observed from the S. cerevisiae culture. It shows the results for gene pairs his, ade, leu and lys. From the table we can tell that the gene pairs which reject the null hypothesis are his/leu, his/lys and leu/lys, while the other gene pairs have accepted the null hypothesis, the significance of which will be explained in the discussion section.

Discussion

Table 2 (above) shows that there are three gene pairs which are possibly linked: his/leu, his/lys and leu/lys because they reject the null hypothesis. This is an example of A and B being linked, B and C being linked, and so therefore A and C must be linked as well. By looking at data in the last column on table 2, we can form a linkage map with the three genes.

(Fig I)

_____________22.72cM_____________ ______________25cM________________

| | | |

leu_____________________________his_________________________________lys

|_____________________________29.54cM_________________________________|

Fig I: This is a linkage map showing the three genes leu, lys and his on the same chromosome. We can see that they are on the same chromosome, and oculd possibly be found in the order above, or with lys and leu at different ends, but wither way, his would be in the middle, because the greatest RF is between lys and leu, so they are the furthest apart.

With this in mind, it is also important to note that three gene pairs failed the chi-squared test, which would indicate a high probability that they are not linked, and on different chromosomes, or just so far apart on the same chromosome that they behave as though they were not. Looking at my data in table 2, the gene pairs which failed the chi-squared display equal amounts of parentals and recombinants, which would indicate there is a greater chance of them not being linked because the chance of recombination is 50%.

For this experiment to produce conclusive data, it would have to be repeated several times, so that an emergent pattern cam be identified, where there is a consistent number of parental vs. recombinant spores with which to carry out the chi-squared test.

This experiment is a useful tool in identifying the links certain genes have with each other, and whether this linkage has any effect on development. If applied to humans, it could be useful in understanding how our genome works, and the interactions between different genes.

Bibliography

Howard, John, et al. 2004. Manual For the Laboratory, Harvard Summer School, Cambridge, MA.
Campbell, Neil A. and Reece, Jane B. 2002 Biology: 6th edition. San Francisco: Benjamin Cummings, 2002.

Yeast complementation analysis

(Table 1)

Table 1: This shows the results of the complementation analysis performed on the S. cerevisiae where plates 1-6 are known genes:

αhis1
αhis2
αhis3
αhis4
αhis5
αhis6

and 7-12 are unknown αhis “mutants”. To find the unknown type, we simply have to look at where there is no growth, which would indicate that they are of the same gene. The genotypes for 7-12 are as follows: