Transcript MONOHYBRID CROSS

IDENTIFICATION OF POLYMORPHIC ALLELES
14.04.09
Quiz
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If you are to prepare a %3 agarose gel,
what should be the amount of agarose
in the 75 ml 1xTAE buffer? Show your
calculations.
If you cross a heterozygous wildtype
with an ebony mutant what will be the
phenotypic ratio of your F1 generation?
Quiz

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
If you are to prepare a %3 agarose gel,
what should be the amount of agarose
in the 75 ml 1xTAE buffer? Show your
calculations.
In 75 ml for %1 percent: 0,75 gr
%3 percent: 2,25 gr
e
+
e
ee
+e
e
ee
+e
1:1 ebony: wildtype
Answer:
F1 generation
e
e
F2 generation
+
+e
+e
+
+e
+e
e
+
e
ee
+e
+
+e
++
Phenotype ratio3:1, wild-type:ebony
Genotype ratio 1:2:1, homozygot WT, heterozygot WT, homozygot ebony
Today’s experiments:
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1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
You will work as 4 groups
First prepare 2% agarose gel
Weigh 0,8 gr agarose into a flask
Put 40ml 1X TBE buffer
Dissolve the agarose completely by heating in the microwave
When boils, remove the flask from the microwave
Waits until it cools to 55C
Add 2.5µl EtBr and mix
Pour the melted agarose into gel apparatus
Let it to harden
Mix 5 µl of loading dye+ 5 µl of DNA sample, load on agarose gels
Run agarose gels at 150 V for 15 min.
Observe under UV light
Electropherosis:
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1.
2.
A method to seperate, identify and
purify DNA fragments
2 types of gels:
Agarose gels
Polyacrylamide gels
Agarose Gels:
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From sea weed
Cheap, non—toxic
Low resolving power, but a higher range
(200bp-50kb)
Run in horizantal configuration
In order to observe, EtBr is used
EtBr intercalates DNA and it flouresces under
UV light, so we can detect the location of the
DNA fragments on the gel
Agarose Gels:
Agarose gel electrophoresis unit
Agarose Gels:
Polyacrylamide gels:
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Highly toxic, synthetic chemicals
Prepared with acrylamide and bisacrylamide.
In the presence of free radicals, it polymerizes
into long chains
Polyacrylamide gels:
By changing acrylamide and bisacrylamide ratio, you can change the size
of the pores
Polyacrylamide gels:
High resolution power, but a shorter range
(5bp-500bp)
 Vertical configuration
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Polyacrylamide gels:
Polyacrylamide gels:
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In order to visualize DNA, silver staining
method can be used
Ag+ ions bind to (-)ly charged DNA,
Reduced to Ag which has a brown color
Polymorphisms:
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1.
2.
Common variation in DNA sequence
It is a kind of variation related to biodiversity,
genetic variation, and adaptation
Presence of more than one genetically distinct type
in a single population
Useful tools in genetic studies for linkage analysis,
prenatal diagnosis, criminal cases and paternity
tests
RFLP (restriction fragment length polymorhism)
VNTR (variable number of tandem repeats)
RFLP:
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Restriction enzymes can recognize and
cut specific DNA sequences
Ex: Msp I enzyme can recognize CCGG
Cfo I enzyme can recognize GCGC
EcoR I can recognize GAATTC
RFLP:
-/-
+/-
+/+
RFLP:
VNTR: (variable number of tandem repeats)
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Can be found on many chromosomes, and
often show variations in length between
individuals
Each variant acts as an inherited allele, can
be used for personal or parental identification
Their analysis is useful in genetics and
biology research, forensics, and DNA
fingerprinting.
Today’s experiments:


1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
You will work as 4 groups
First prepare 2% agarose gel
Weigh 0,8 gr agarose into a flask
Put 40ml 1X TBE buffer
Dissolve the agarose completely by heating in the microwave
When boils, remove the flask from the microwave
Waits until it cools to 55C
Add 2.5µl EtBr and mix
Pour the melted agarose into gel apparatus
Let it to harden
Mix 5 µl of loading dye+ 5 µl of DNA sample, load on agarose gels
Run agarose gels at 150 V for 15 min.
Observe under UV light
Today’s experiments:
You will make a paternity test
1) On each bench, you have 5 DNA samples:
mother, child and 3 father candidates
2) We will identify the father by checking 2
polymorhisms on different chromosomes
1. group RFLP (on chromosome 2)
2. group VNTR (on chromosome 5)
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Today’s experiments:
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1. group RFLP (on chromosome 2)
DNA fragment includes a Single nucleotide
polymorhism (G or T)
Msp I enzyme  CCGG
CCGG
600bp
CCTG
Digest with Msp I
400bp
200bp
600bp
Today’s experiments:
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marker
On agarose gel: load 3 µl 100bp marker
mother
child
father3
father1
father2
600
500
bp
400
300
bp
200
bp
100
-/-
+/-
+/+
+/-
-/-
Today’s experiments:
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2. group VNTR (on chromosome 5)
DNA fragment includes different number of (CA)
repeats
You dont need to load marker
Expected results:
800
700
600
RFLP:
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1. group:
mother
child
marker
father3
father1
father2
616bp
500
400bp
400
300
200bp
200
100
-/-
+/-
-/-
+/+
+/-
VNTR:
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2. group:
mother
child
marker
father3
father1
father2
800
4
700
600
3
500
400
2
300
200
1
100
2/4
1/2
2/6
1/2
1/3
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Monohybrid cross results
Ebony vs. wildtype
Hypothesis:
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Body color is an autosomal trait and
ebony type is recessive to the wild type”
To test the accuracy of our hypothesis,
we need to calculated to which extent
our observed results are departed from
the expected results
Count Ebony vs. Wildtype Drosophila
F1 generation
e
e
F2 generation
+
+e
+e
+
+e
+e
e
+
e
ee
+e
+
+e
++
Phenotype ratio3:1, wild-type:ebony
Genotype ratio 1:2:1, homozygote WT, heterozygote WT, homozygote ebony