Dr Maninder tetrad analysisssssssssss.pptx

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1. Tetra d analysis 1

2. I NTRODUCTION 2 Certain species of lower Eukaryotes such as algae and fungi which spend most of their lifecycle in haploid state (Only one set of chromosome) , have also been used in mapping studies. Fungi can also reproduce sexually by the fusion of two haploid cells to create a diploid zygote (2n).The diploid zygote can then proceed through meiosis to produce four haploid cells or spores known as Tetrad . In some, meiosis is followed by mitosis to form 8 spores known as Octad .

3. O RGANISM USED FOR THE STUDY OF THE TETRAD ANALYSIS 3 Lower eukaryotes such as algae and fungi are used for the study of tetrad analysis . Especially in fungi, Sachharomyces cerevisiae , Coprinus lagopus, Chlamydomonas reinhardtii (Tetrads) and Neurospora crassa ,Aspergillus nidulans (Octads) are used extensively.

4. Tetrad analysis n 4 n 2n interpha s e Tetrads in ascus m i to s is Octads in ascus Each of the product of meiosis can be cultures separately to study their phenotypes theryby inferring their genotypes.

5. T YPES OF TETRAD Two types of tetrad :- Ordered tetrad:- when some species of fungi produce a very tight ascus that prevents spores from randomly moving around. This can create a linear tetrad also called ordered tetrad. Ex . Neurospora crassa . Unordered tetrad:- when the ascus provides enough space for the tetrads or octads to randomly mix together, this type of tetrad arrangement is called unordered tetrad. Ex. Sachharomyces cerevisiae.

6. A NALYSIS OF ORDERED TETRAD The product of meiosis are contained in an ordered array of spores. Each mature ascus contains eight ascospores in four pairs,each pair represnting one of the products of meiosis. The ordered arrangement of meiotic product makes it possible to map each gene with respect to its centromere; i.e. to determine the recombination frequency between a gene and its centromere. Two cases are possible depending on whether or not there is a crossover between the gene and its centromere. 1.first division segregation (FDS) 2.second division segregation(SDS) 8

7. 1.FDS – in the absence of crossing over between a gene and its centromere, the alleles of the gene must separate in the first meiotic division , this separation is called FDS or Since the two types of centromeres or alleles of a gene A and a segregate to different nuclear areas after first meiotic division, the segregation is called First Division Segregation (FDS). 2. SDS(I) :- cross over occur between the gene and its centromere. 9 A a A a A a A a a a A A A a A 2 a 2 2 2 Meiosis I Meiosis II Mitosis

8. 2. SDS(II) :- cross over occur . 10 A a A a A a a A A a a A A a A 4 2 2 Unit of map distance is centimorgan ( cM ) Map distance = ½ No.of SDS Asci Total no. of Asci Meiosis I Meiosis II Mitosis X 100

9. How to find the Map distance Assume, 60 percent of a sample of asci have second - Division segregation pattern for alleles A and a. This means 60 percent of the cells Undergoes meiosis, had a crossover between alleles A gene and its centromere . Here, in each cell in which crossing over takes place, two of the chromatids are recombinant and two are non-recombinant. S0, the frequency of crossing-over of 60 percent corresponds to a recombination frequency of 30 percent. To determine the map distance, in case of SDS asci , only half of the spores are actually the product of crossover. Therefore, the map distance is calculated as Map distance = ½ No.of SDS Asci Total no. of Asci X 100

10. A NALYSIS OF UNORDERED TETRAD 10 Unordered tetrad analysis can be used to map genes in dihybrid crosses.This analysis can determine if two genes are linked or assort independently. Three pattern of segregation are possible in the tetrad when two pairs of alleles are segregating.  1.Parental ditype:- 4:4 2.Non- parental ditype:- 4:4 3.Tetratype:- 2:4:2

11. A B a b A B A b B a A B A B 4 4 a b A b A b B a 4 4 a B I Parental ditype (PD) II Non Parantal Ditype (NPD) Meiosis I Meiosis II Mitosis Meiosis I Meiosis II Mitosis a b 11 a b a b A B

12. A B a b A b a B a b A B A a b B a b 2 2 A B A b a B a b III Tetratype (TT) Meiosis I Meiosis II Mitosis A B 4 Total progeny Map distance= Recombinant progeny X 100 Recombinant progeny = NPD+1/2 TT 12

13. Interrupted Mating

14. Transfer of Genetic Material in Bacteria The three main mechanisms by which bacteria acquire new DNA are transformation, conjugation, and transduction. Transformation involves acquisition of DNA from the environment. Conjugation involves acquisition of DNA directly from another bacterium. Transduction involves acquisition of bacterial DNA via a bacteriophage intermediate.

15. Conjugation Conjugation is a mating process involving bacteria. It involves transfer of genetic information from one bacterial cell to another, and requires physical contact between the two bacteria involved. The contact between the cells is via a protein tube called an F or sex pilus . Basic conjugation involves two strains of bacteria: F+ and F-. The difference between these two strains is the presence of a Fertility factor (or F factor) in the F+ cells. Genetic transfer in conjugation is from an F+ cell to an F- cell, and the genetic material transferred is the F factor itself.

16. Process:- Basic conjugation occurs between an F+ cell and an F- cell. The difference between these two types of cells is the presence or absence of the F (fertility) factor, which is a circular DNA molecule independent of the bacterial chromosome (the larger circular molecule. The F+ cell initiates conjugation by extending an F pilus toward the F- cell. Among the genes present on the F factor are the genes encoding the proteins required for pilus construction. The F pilus, when finished, temporarily connects the two cells. On strand of the F factor is nicked, and begins unwinding from the other strand. The nicked strand begins to transfer through the F pilus to the F- cell. As it does so, this strand begins to be replicated, as does circular strand remaining behind in the F+ cell. Eventually, the nicked strand completely passes through to the recipient cell, and is completely replicated. This process produces a new F factor in the recipient cell. The pilus is broken, severing the connection between the two cells. Since both cells now contain an F factor, both cells are F+. The new F+ cell (which was the F- cell, can now initiate conjugation with another F- cell.

17. Hfr strain E. coli strain discovered as Hfr (high frequency of recombination) Hfr strain transfers chromosomal DNA to F - strains. This transfer begins at the origin of transfer The amount of DNA transferred depends on the time of conjugation

18. Mapping Genes on Bacterial Chromosomes Bacteria, since they are usually haploid, cannot have their chromosomes mapped by the same techniques as eukaryotes  They can, however, be mapped by using Hfr bacterial conjugation.

19. Interrupted mating The length of time a mating occurs, the more DNA is transferred. The Hfr DNA is transferred in a linear manner. By mating for different times, you can get DNA of several sizes, and determine the order of the genes, and how far apart they are (minutes).

20. For example, imagine that an F- cell has mutant alleles of two genes,  a  and  b  (the F- would therefore be  a -,  b -). If this cell undergoes conjugation with an Hfr cell that is  a +,  b + (in other words, wild type), the F- cell should undergo gene conversion to  a +,  b + when both of those genes have been transferred by conjugation. By determining how long it takes the  b  gene to transfer after the  a  gene has transferred, it is possible to get a relative idea of how far apart the two genes are on a chromosome. The experiment would be done this way:  a +,  b + Hfr cells would be mixed with  a -, b - F- cells. The time of mixing would be designated 'time zero'. At regular intervals, a small amount of the mixture would be removed and conjugation would be disrupted using a blender (the shear force of the blender would cause any pili to break). These bacteria would then be tested for gene conversion (for example, if the mutations rendered the F- bacteria auxotrophic, the bacteria could be tested by growing them on minimal medium, or minimal medium supplemented with the necessary nutrient required because of one or the other mutation). If the  a  gene was converted to wild type at 8 minutes after time zero, and the  b  gene was converted to wild type at 19 minutes after time zero, then the distance between the two genes would be '11 minutes' (because that was the difference in time required to transfer the  b  gene compared to the  a  gene). Bacterial map distances are always expressed in minutes, because of this technique.

21. Mapping via Interrupted Mating

22. Mapping of the E. coli chromosome This technique was utilized to map all genes of E. coli chromosome 100 minutes long (how long it takes to transfer over the entire chromosome)

23. Mapping procedure Genetic distance is determined by comparing their times of entry during an interrupted mating experiment Therefore these two genes are approximately 9 minutes apart along the E. coli chromosome