Tetrad analysis by rk

gene mapping

1. Tetrad analysis
RanjeetKumar Taram
MSc.IIsem(Microbiology)
S.o.S. in Life Science
Pt. RavishankarShuklaUniversity, Raipur
1

2. CONTENTS
Introduction
Organisms used and Advantages
Types of tetrad
Analysis of tetrads
Problems
References
2

3. INTRODUCTION
Certain species of lower Eukaryotes such as algae
and fungi which spend most of their lifecycle in
haploid state, 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.
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4. Tetrad analysis
n
n
2n
interphase
Tetrads in
ascus
mitosis
Octadsin
ascus
4

5. ORGANISM USEDFORTHESTUDYOF
THETETRADANALYSIS
Lower eukaryotes such as algae and fungi are
used for the study of tetrad analysis .
Especially in fungi, Sachharomycescerevisiae,
Coprinuslagopus, Chlamydomonasreinhardtii
(Tetrads) and
Neurosporacrassa,Aspergillusnidulans
(Octads) are used extensively.
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6. ADVANTAGES
Haploid in nature.
Produce very large numbers of progeny.
The life cycles of these organisms tend to be short.
Potential for analyzing all of the products from each
meiotic division.
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7. TYPESOFTETRAD
Two types of tetrad :-
Ordered tetrad:-when some species of fungi
produce a very tight ascusthat prevents spores
from randomly moving around. This can create a
linear tetrad also called ordered tetrad.
Ex . Neurosporacrassa.
Unordered tetrad:-when the ascusprovides
enough space for the tetrads or octadsto
randomly mix together, this type of tetrad
arrangement is called unordered tetrad.
Ex. Sachharomycescerevisiae.
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8. ANALYSISOFORDERED TETRAD
The product of meiosis are contained in an ordered array of
spores. Each mature ascuscontains eight ascosporesin
four pairs,eachpair represntingone 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 locus and its centromere.
1.first division segregation (FDS)
2.secound division segregation(SDS)
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9. 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.
2. SDS(I):-cross over occur between the gene and its centromere.
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A
a
a
A
A
a
A
a
4
4
a
A
A
a
A
a
A
A
a
a
a
a
A
A
A
A
a
a
2
2
2
2
Meiosis I Meiosis II Mitosis
Meiosis I
Meiosis II Mitosis

10. 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
Map distance= Recombinant progeny
Unit of map distance is centimorgan(cM)
Total progeny
X 100
Recombinant progeny = ½ No of SDS progeny
Meiosis I Meiosis II
Mitosis

11. 11
Problems:-
We have given the following data, calculate gene distance between
gene and centromere.
aAAaaA
aAAaaA
aAaAAa
aAaAAa
AaaAaA
AaaAaA
AaAaAa
AaAaAa
34485463
B b b B B b
B b b B B b
B b B b b B
B b B b b B
b B B b B b
b B B b B b
b B b B b B
b B b B b B
12310815121824
Map distance= Recombinant progeny
Total progeny
X 100

12. ANALYSISOFUNORDERED TETRAD
Unordered tetrad analysis can be used to map
genes in dihybridcrosses.Thisanalysis 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
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13. AB
ab
AB
Ab
Ba
AB
AB 4
4
ab
Ab
Ab
Ba
4
4Ba
13
I Parental ditype(PD)
II Non ParantalDitype(NPD)
Meiosis I
Meiosis II
Mitosis
Meiosis I Meiosis II
Mitosis
ab
ab
ab
AB

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AB
ab
Ab
aB
ab
AB
Ab
a
B
ab
2
2AB
Ab
aB
ab
III Tetratype(TT)
Meiosis I Meiosis II
Mitosis
AB
4
Map distance= Recombinant progeny
Total progeny
X 100
Recombinant progeny = NPD+1/2 TT

15. 15
Problems:-
We have given the following data, calculate gene distance between two genes.
AB AB ab ab 1750
AB Ab aB ab 220
AB aB Ab ab 20
Ab aB Ab aB 10
Lys-His-Lys+His-Lys-His-Lys+His+Lys-His-Lys-His+Lys-His-
Lys-His-Lys+His-Lys-His-Lys+His+Lys-His-Lys-His+Lys-His-
Lys-His-Lys+His-Lys-His+Lys+ His-Lys+His+Lys+His-Lys+His+
Lys-His-Lys+His-Lys-His+Lys+ His-Lys+His+Lys+His-Lys+His+
Lys+His+Lys-His+Lys+His+Lys-His+Lys+His+Lys+His-Lys+His-
Lys+His+Lys-His+Lys+His+Lys-His+Lys+His+Lys+His-Lys+His-
Lys+His+Lys-His+Lys+ His-Lys-His-Lys-His-Lys-His+Lys-His+
Lys+His+Lys-His+Lys+ His-Lys-His-Lys-His-Lys-His+Lys-His+
136 2 23 93 121 1 24
Auxotroph +
Prototroph –
OR
Wild type +
Mutant –

16. REFERENCE
16
Griffiths, A. J. F., et al. "Introduction to genetic
analysis, p 351–392." (2015)
https://ocw.mit.edu/courses/biology/7-03-genetics-
fall-2004/lecture-notes/lecture8.pdf
https://www.ncbi.nih.gov/books/NBK21972/

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