W9 Flashcards
how can a test cross be used and how is it used
- cross of heterozygote to homozygous recessive
- allows us to tell if linkage is present
- if the genotype ratio is the same, independent assortment has occured since all offsprings genotype in equal ratio
- if not 1:1:1:1, linkage has occurs since the ratio of phenotypes in offspring of testcrosss of double heterozygous is significant
- why does this work?
- homozygous parent not contribute to phenotype, therefore we see what other parent contributes
what is linkage
during meiosis, crossing over(chiasma) takes place and exchange of maternal and paternal chromatids takes place
FACTS:
- more genes exist than chromosomes
- several genes can exists on the same chrom.
- independent assortment occurs when genes are more than 50 MAP UNITS apart
NOTATION:
AB/ab cis arrangement
Ab/aB trans arrangement
the numerator and denominator come from a pair of homologues
- linkage explains why even tho we only expect non recombinant gametes, we also get recombinant gametes
why can trisomy on 13, 18,19 occur
- less number of genes, therefore can fit extra chromosome
- in other chromosomes, deleterious effects too great
describe crossing over
- can be chiasmata; multiple crosses
results:
1. non recombinants occur between sister chromatids
- more frequent
2. recombination; between non sister chromatids
- exchange between maternal and paternal chromatids
- 2 of 4 chromatids
- less frequent
THE GREATER THE MAP DISTANCE, THE MORE RECOMBINATION, UNTIL 50 units reached
THERE IS NO CROSSING OVER AND RECOMBINATION IN MALE DROSOPHOLIA AND LUCILIA
- thus, only produce two types of gametes
- it is impossible to get greater than 50% of recombinants
what is map distance
number of recombinants/ total number of offspring multiple by 100
- relative distance between gene loci measured by recombinant frequency
- 1 map unit (centimorgan)= 1% recombinations
what is interference and how can it be deduced
- crossover in one region influence crossover in another region
- frequency of double recombinants ie. MAP distance for DR1MAP distance for DR2total #offspring
- coefficient of coincidence=OBS/EXP= number of offspring with double recombination/frequency of double recombination
- interference = 1-CC
Posiitve Interference: observed DR less than expected
- crossover in one region inhibits crossover in adjacent region
Negative interference: observed DR more than expected dr
- CROSSOVER IN ONE region promotes crossover in adjacent region
- we often get positive interference especially if the genes are close because physically, one crossover makes a second difficult
- the further apart, the less likely interference to occur
- there are hot spots for interference, ie. peaks in graph
outline the method to deduce which allele in middle from a 3 point cross
- determine is it independent assortment or linkage
- linkage means greater than 50mu apart
- do we have a 1:1:1:1 ratio from a testcross - identify the non recombinant phenotypes (NR)
- identify the middle locus
A. find the DR, these are the ones with the two smallest number
B. long for the single difference between the double recombinant phenotypes and the non-recombinant phenotype
C. label as DR - rewrite out the correct genotypes for both parents
- in the correct order - assign gamete genotypes to all offspring phenotypic categories
- only mutations are mentioned in the name, wild type blah blah does not exist
- determine if the gene loci are in CIS or TRANS arrangement
cis: dominant trait for both loci are on the same chromosome
trans: dominant trait for both loci are on different chromosome
note: write in notation of linkage with the kine in the middle. above the line refers to coming from one chromosome whereas below the line refers to coming from a different chromosome - calculate map distances
- must be done for each region, there are two regions and these are differentiated by region 1 with SR1 and region 2 with SR2
eg. map distance for region 1=(number of single recombinants + number of double recombinants)/total number of individuals *100 - draw Map of chromosome
what is interference
crossover in one region influences crossover in another region
- interference = 1 - CC
method:
CC=(OBS#DRS)/(EXP#DRS)
1. frequency(DRs)= (map distance region 1/100) * (map distance region 2/100)=ANS
2. EXP #DRs = ANS * total number of individuals = BLAH
3. interference = 1 - (OBS#DRS)/(EXP#DRS) = 1- (OBS#DRS)/BLAH)
positive interference: OBS DR fewer than expected. the first crossing over event INHIBITS the chances of a 2nd crossing over events
what is the ratio for dominant epistasis
12:3:1
what is the ratio for recessive epistasis
9:3:4
how do we tell how many gene loci
if all these ratio add up to 16, there are 2 gene loci
- this was from the dihybrid cross
- furthermore, these gene loci are assorting independently
if there is one gene loci, then the ratio adds up to 4
