GENE MAPPING

1. GENE MAPPING
PRESENTED BY:
NAME : RASHMI JALAN

2. WHAT IS GENOME MAPPING ?
Genome Mapping is used to identify and record the location of genes and
the distances between genes on a chromosome.
Genome mapping provided a critical starting point for the Human Genome
Project.
WHAT ARE GENETIC MAPS
AND GENE MAPPING?
Genetic Maps helps describes the spatial arrangement of genes on a
chromosome.
Genetic Mapping is the process of determining the order of and relative
distance between genes on a chromosome based on their pattern of
inheritance. It is also called as Linkage mapping.

3. WHO GAVE THE CONCEPT OF GENETIC
MAPPING?
ALFRED HENRY STURTEVANT,
a 19 years old, undergraduate
student of Morgan.
Created 1st genetic map of
a chromosome from
Drosophila melanogaster(a
fruit fly) in 1913.
Proposed that the frequency of
‘crossing over’ and ‘linkage’ between
two genes could help determine their
location on a chromosome.
It is possible to estimate
the distance between
the genes by finding
out how often various
characteristics are
inherited together.
He formulated that “proportion of crossover could be used as an index of
distance between any two factors”

4. TECNIQUES OF GENE MAPPING
Linkage is ∝ 1/recombination frequency.(Taking this into consideration)
Selection of variable traits
Initial steps in mapping are
assign the genes to a particular chromosome
establishment the proximity of traits to
one another

5. MAP UNIT
One “map unit” (or “Morgan”) in genome map distance is the
distance that produces a recombination frequency of 1%.
Therefore:
Map distance (in map units) = recombination frequency X 100
MAP DISTANCE = (Recombinant gametes) X 100
(Recombinant gametes) + (nonrecombinant gametes)

6. Types of Gene Mapping
LINKAGE ANALYSIS
GENE
ASSOCIATION
ANALYSIS
Through cross breeding
experiments. Includes two-
factor test cross and three-
factor test cross.
Includes pedigree
analysis in case of
humans.

7. Linkage Analysis with Two Point Test Cross
• The phenotype of the testcross progeny is determined by the
gametes from the heterozygous parent.
• Each phenotype in a testcross has a unique genotype (unlike in
the F2 of dihybrid cross)
A testcross lets us “count”
the number of
and non recombinant
gametes
• Cross an individual that is heterozygous for each gene with an
individual that is homozygous recessive for each gene.
So, to map the distance
between two genes

8. Example of Two Point Test Cross
Example: Tomato plants: Fruit shape & Texture genes:
A heterozygous round, heterozygous smooth plant
(RrPp) was crossed with a long, peachy (rrpp) plant. The
results are given in the table below:
Smooth round 39
Smooth long 463
Peachy round 451
Peachy long 47

9. Arrange the phenotypic
classes into pairs, with
each different
phenotype represented
in each pair.
Smooth Round
Peachy Long
Smooth Long
Peachy Round
Look at the numbers to
determine which class is
recombinant (lesser
numbers) and which is
nonrecombinant
(greater numbers)
Smooth round 39
Peachy long 47
Smooth long 463
Peachy round 451
STEP 1:
STEP 2:
}
}NONRECOMBINANT
}RECOMBINANT

10. Determine the linkage (cis
or trans) of the alleles in
the nonrecombinant
heterozygote parent.
In this particular cross, the
linkage is trans
Smooth round 39
Peachy long 47
Smooth long 463
Peachy round 451
STEP 3:
STEP 4:
Calculate the map
distance:
R – P gene distance
= 86/1000 X 100 =
8.6 m.u.
Smooth round 39
Peachy long 47
Smooth long 463
Peachy round 451
}RECOMBINANT
}NONRECOMBINANT
}RECOMBINANT
}NONRECOMBINANT
R--------------8.6 m.u.---------------------
Result :

11. THE DOUBLE CROSSOVER PROBLEM
 Double crossovers occur whenever two crossover events occur between two
genes.
 If this occurs, then the recombinant progeny will not be counted, because
each allele “goes back” to its original linkage
 For this reason, the map distance given by a 2-factor testcross often is too
low.
 2 factor test cross only give the relative distance between considered factors
but unable to point out their orders.
To solve this problem we perform 3- factors test cross.

12. 3- FACTOR TEST CROSS
Let us presume that there are three
genes A, B and C present on the
same chromosome.
 A 3 point test- cross is made,
which involves crossing of a tri-
hybrid ABC/abc (obtained from a
cross ABC/ABC X abc/abc) with
triple homozygous recessive
abc/abc.
 The progeny obtained will
represent the gametes formed by
the hybrid.
 Presuming A-B-C as the order of
genes,

13.  Hypothetical frequencies of
eight types of progenies of
above cross.
Linkage Gene Mapping Construction:
 Linkage maps are prepared
with the help of
recombination frequencies.

14. Let us consider an example from maize involving three endosperm
characters. These three characters are
 colored aleurone (C) versus colorless aleurone (c),
 full endosperm (Sh) versus shrunken endosperm (sh)
 non-waxy endosperm (Wx) versus waxy endosperm (wx).
The data presented by C.B. Hutchinson in 1922 are given in Fig.
8.18.
The three recombination values, i.e., C-Sh, Sh-Wx, C-Wx should be
worked out in order to find out the linear order of the three genes,
C, Sh and Wx. In the data presented, the progeny of parental types
are present in higher frequencies.
C and sh are present together in P1, therefore, the progeny showing
their separation would be recorded as recombination between C
and Sh. Similarly recombination between sh and Wx as well as between
C and Wx could be recorded.
The mathematical relationship among the recombination values of three genes may be utilized for
determining the gene order. From the values of X, Y and Z of the example in the fig 18.7., the
order of genes can be worked out.

15. In the example (Fig. 8-18), the recombination value
 C-Wx (21.7%) is nearly equal to recombination value of (C-sh) + (sh – Wx) = 3.5 + 18.4 =
21.9%.
 Therefore, sh should be located between C and Wx.
Another way of determining gene order is comparison of the allelic combination of paren-
tal and double crossover recombinant classes of progeny.
Out of the eight (4 pairs) phenotypic classes of progeny, one pair has the highest fre-
quency representing parental (non-recombinant) class; and one pair has the lowest
frequency representing double crossover recombinant class.
In the example (Fig. 8.18), the highest frequency progeny class develops from non-
recombinant gametes, C sh Wx and c Sh wx; and the lowest frequency progeny class
develops from double crossover recombinant gametes, C Sh Wx and c sh wx.
Comparison of allelic arrangements of non-recombinant gametes with double crossover
recombinant gametes [(C sh Wx and C Sh Wx) or (c Sh wx and c sh wx)] shows that Sh or
sh stands out as the changed locus indicating its position in the middle. Therefore, the
gene order will be C-sh-Wx.

16. So, the linkage map of
above example is:

17. COMMENTS AND GENE ASSOCIATION ANALYSIS
Linkage analysis is a tremendously important and powerful approach in medical
genetics because it is the only method that allows mapping of genes, including
disease genes, that are detectable only as phenotypic traits.
GENE ASSOCIATION ANALYSIS
Gene association analysis is a type of genetic mapping which stdies
Pedigree analysis provides ways for localizing genes on human chromosomes.

18. • Heritable diseases
• Cancer
• Used for gene therapy.
Identify genes
responsible for
diseases.
• Plants or Animals
• Disease resistance
• Meat or Milk Production.
Identify genes
responsible for traits.
Uses of Gene Mapping
“Precision health will enable us to give people individually targeted health information and
treatment that will allow them to live as healthy as possible as they move through live”

19. Limitations of Gene Mapping
The resolution of genetic
maps depends on the
number of crossovers that
have been scored.
Genes that have several tens of
kb apart may appear at the
same position on the genetic
map.
Genetic maps have
limited accuracy.
Presence of recombination
hotspots means that crossovers
are more likely to occur at same
position rather than others.
A map
generated
by a genetic
techniques is
rarely
sufficient for
directing
sequencing
phase of a
genome
project