Class Notes (1,100,000)
CA (650,000)
UTSG (50,000)
HMB265H1 (300)
Lecture

Notes taken during lecture


Department
Human Biology
Course Code
HMB265H1
Professor
Stephen Wright

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LECTURE โ€“ sense of format and practise for the midterm โ€“ 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, itโ€™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 โ€“ 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 thisโ€ฆ
๎€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 โ€“ 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
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