Notatki Danniego part 5 Flashcards
The further the loci, …
the higher the chance of recombinance
-> More chance for a breakage between them
Recombination frequency represents
Recombination frequency equation:
Also: map of the chromosome:
distance between loci:
400 progeny, 44 recombinants
Recombinants/Total
=44/400 = 0.11 = 11%
- Build a map of the chromosome
1% is known as a centimorg
-> distance between loci - Now computers can recognise overlaps in fragments of DNA and stitch it together much quicker
3 point cross
P1 aa bb cc X ++ ++ ++
mutant wild type
f1: +a +b +c heterozygote wild types
Backcross with mutant stock
f2 Expect equal numbers of 8 possible combinations (2^3)
Expected 1:1:1:1:1:1:1:1
mutations Scute - bigger bristles on b ack of drosophila
Echinus - hairy eyes
Crossvainless - missing support structure in wing
P1: scsc ecec cvcv X ++ ++ ++
F1: sc+ ec+ cv+ triple hetorozygotes
F2: Expected equal combinations 1:1….
Actually:
sc ec cv 417 normal 430 sc ++ ++ 25 \++ ec ++ 29 \++ ++ cv 44 sc ec ++ 37 sc ++ cv ? \++ ec cv ?
Parentals ec cv = 417 + 29 \ in total 901
+ + = 430 + 25 /
Recombinants sc + = 25 \ in total 54
+ ec = 29 /
Map distance between scute x echinus = 54/982 = 5.5 map units
PRZEJRZEĆ BO COŚ MI TU NIE PASUJE
3 point cross question
Phenotype frequency:
A. x y z 501
B. + + + 487
C. x y + 15
D. + + z 10
E. x + + 16
F. + y z 9
total 1038
x and y : E + F = 25/1038 = 2,4
y and z : C + D = 25/1038 = 2.4
x and z : C+ D+ E+ F = 50/1038= 4,8
4.8 x y z 2.4 2.4
Rec frequency:
x-y= 0.024
y-z= 0.024
- 024^2= 0.000576
- 000576 x 1038 = 0.60
3 point cross: double crossovers
ZDJECIE TABELKI
- Double crossover unlikely in a short stretch of DNA
cv and c+ : C+ D+ G+ H= 93 93/1448= 6.4
c+ and ve : E+ F+ G+ H= 193 193/1448= 13.2
cv and ve : C+ D+ E+ F= 268 268/1448=18.5
6.4+13.2 =/ 18.5
figures don’t add up because you need to take into account the double crossover
-> G and H must be counted twice when computing map distances
cv and ve : C+ D+ E+ F+ 2(G+H)
45+ 40+ 89+ 94+ 2(3+5) = 284
284/1448= 19.6 map units
cv-c+ (6.4) + c+-ve (13.2)= 19.6
Coefficient of interference
Compares actual number of double crossovers with no. of expected igf each crossover is independent
- > is the number of recombinants over A-B-C equal to the product of A-B x B-C
- > if not there is CHIASMA INTERFERENCE
Recombination frequency cv-c+ = 0.064
Recombination frequency c+-ve = 0.132
Expected proportion of double recombinants =
= 0.064 x 01.132= 0.0084
0.0084 x 1448 = 12 double recombinants
Interference:
Expected 12 double recombinants
Observed only 8
Observed/Expected
8/12 = 0.66
COEFFICIENT OF INTERFERENCE=
= 1 - 0.66 = 0.33
- > females have a higher recombination rate
- > male drosophila never recombine
- > recombination hotspots
Drosophila in adolescence have ….
peculiar chromosomes in spit glands
- > need to make silk protein
- > 100s of copies of the gene
- > Endomytosis - replicate the genes within those cells
- > 1024 bands lined up
FISH -
Fluorescent In Situ Hybridisation
- > Where in genome in sequence for β-chain haemoglobin?
- > Sequence the protein
- > Infer the DNA sequence for protein
- > make mirror image of sequence
- > dye the image
- > finds opposite strand on DNA and binds whilst still dyed
- > Now you know where the gene for β-chain haemoglobin is
Drosophila larva salivary glands have huge chromosome - like structures made up of 1024 individual chromosomes lying close together as a single structure
-> …
-> Polytene chromosomes
These come from ENDOMITOSIS; the multiplying of copies of a gene leaded in large amount; in this case the one used in making the protein pupa
Positron effect variegation
- Gene is shifted by translocation close to the edge of some heterochromatin
- > DROSOPHILA WHITE EYE IS ON X ( potem zmnieszyć, jak już zrobimy diseases list) but can be translocated to chromosome 4
- > is usually recessive
- > normal allele loses part of its dominance after translocation
- > results in white and red patches - Cell division genes translocated to near active parts of DNA
- Causes out of control growth
- Cancer tumours
Somatic Cell Hybridization
Attacking a mouse and human cell together to make a hybrid
- Sendai virus
- > sticks to mouse cell
- > human cell floats past and also sticks to virus
- > virus tries to get in, cells open up = hybrid cell
- > Heterokaryote
Sorting the hybrid cells from normal cells
- use a poison that needs two enzymes (A and B) to break it down
- Mouse line is A- B+ (dies)
- Human line is A+ B- (dies)
Half of the hybrid cells are A+ B+ (survives)
-> half the hybrids are A- B- (dies) - > Human chromosomes are rejected
- > Irradiate hybrid lives and slit into tiny places
- > only small part is ejected
Haplotype
A set of DNA variations or polymorphisms that tend to be inherited together
Selective sweep
- The reduction or elimination of variation among the nucleotides in neighbouring DNA of a mutation as the result of recend and strong positive natural selection
- Example:
- > Drosophila mutation for insecticide resistance in the 40s/50s
- > Now every fly that is tested has a long (thousands of bp) section of DNA that is the same/unbroken
- > Recombination is rarely found in this section
Gene & Environment - linked together
siamese cat - mutant
coat colour due to ability to make melanin.
Siamese has black nose, feet, tail, white body
