1. IDENTIFICATION OF POLYMORPHIC ALLELES

2. Quiz
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?

3. Quiz
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

4. e + ee +e
1:1 ebony: wildtype

5. Answer: + e +e e + ee +e ++ F1 generation F2 generation
Phenotype ratio3:1, wild-type:ebony
Genotype ratio 1:2:1, homozygot WT, heterozygot WT, homozygot ebony

6. Today’s experiments: 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

7. Electropherosis:
A method to seperate, identify and purify DNA fragments
2 types of gels:
Agarose gels
Polyacrylamide gels

8. Agarose Gels: 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

9. Agarose Gels:
Agarose gel electrophoresis unit

10. Agarose Gels:

11. Polyacrylamide gels: Highly toxic, synthetic chemicals
Prepared with acrylamide and bisacrylamide.
In the presence of free radicals, it polymerizes into long chains

12. Polyacrylamide gels:
By changing acrylamide and bisacrylamide ratio, you can change the size
of the pores

13. Polyacrylamide gels: High resolution power, but a shorter range
(5bp-500bp)
Vertical configuration

14. Polyacrylamide gels:

15. Polyacrylamide gels:
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

16. Polymorphisms: 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)

17. RFLP:
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

18. RFLP:
-/ / /+

19. RFLP:

21. VNTR: (variable number of tandem repeats)
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.

23. Today’s experiments: 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

24. 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)

25. Today’s experiments: 1. group RFLP (on chromosome 2)
DNA fragment includes a Single nucleotide polymorhism (G or T)
Msp I enzyme  CCGG
CCGG
CCTG
600bp
Digest with Msp I
400bp
200bp
600bp

26. Today’s experiments: On agarose gel: load 3 µl 100bp marker father2
mother
child
father1
600
500 bp
400 bp
300
200 bp
100
-/ / / / /-

27. Today’s experiments: 2. group VNTR (on chromosome 5)
DNA fragment includes different number of (CA) repeats
You dont need to load marker

28. Expected results:

29. RFLP: 1. group: marker child father3 mother father1 father2 616bp
800
700
600
616bp
500
400bp
400
300
200bp
200
100
-/ / / / /-

30. VNTR: 2. group: marker mother child father3 father1 father2 4 3 2 1
800 4 3 2 1
700
600
500
400
300
200
100
2/ / / / /3

31. Monohybrid cross results
Ebony vs. wildtype

32. Hypothesis:
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

33. Count Ebony vs. Wildtype Drosophila
F1 generation + e +e e + ee +e ++
F2 generation
Phenotype ratio3:1, wild-type:ebony
Genotype ratio 1:2:1, homozygote WT, heterozygote WT, homozygote ebony