LECTURE – sense of format and practise for the midterm – the old midterms
LINKAGE MAPPING
Genes on the same chromosomes are linked
Normally, without linkage, produce AaBb
AAbb parent had two chromosomes: Ab, Ab
And in aaBB, aB and aB are the two chromosomes
In F1 of linkage, it’s configuration will be Ab chromosome and an aB
chromosome
Parents were double homozygous, recombination would not thus have any
difference to the progeny without recombination
Mostly F1 will make parental gametes, which are Ab and aB
oBut with recombination in F1: AB and ab are the recombinant gametes
If linked, parental more common than recombinants
Unlike how under Mendelian ratios, the ratios are equal
THEN, what about trihybrid crossing?
Hypothesis and Extension
Linear order along chromosome – genes are
Recombination more likely if further apart
Can infer order of genes along chromosomes if look at many phenotypes and
many crosses
If Morgan is right, A+B+C = roughly A+C
Testing this…
Test cross because it unmasks the phenotypes
Mutants identified in lab
Ct and cv should have low proportion of recombinant phenotypes
Two inbred lines
oTest cross with vermilion eyes and cut wing edges
oProgeny produced
CAPITALS FOR THE PLUSES, LOWERCASE FOR THE NON-
PLUSES
oV+ct is double homozygote
oVct+is same
oF1 has Vct+and v+ct – these combinations most common because they are
parental types
oAnd parental types appeared in majority
oVermilion uncut is vct+
oRed cut is v+ct
oTwo different parental types should be half, half
oVermilion cut and red uncut is recombinant
First is double mutant
www.notesolution.com