Class Notes (1,100,000)
CA (650,000)
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Human Biology (800)
HMB265H1 (300)
Stephen Wright (60)
Lecture
Department
Human BiologyCourse Code
HMB265H1Professor
Stephen WrightThis preview shows half of the first page. to view the full 3 pages of the document.
LECTURE ā sense of format and practise for the midterm ā the old midterms
LINKAGE MAPPING
Genes on the same chromosomes are linked
īNormally, without linkage, produce AaBb
īAAbb parent had two chromosomes: Ab, Ab
īAnd in aaBB, aB and aB are the two chromosomes
īIn F1 of linkage, itās configuration will be Ab chromosome and an aB
chromosome
īParents were double homozygous, recombination would not thus have any
difference to the progeny without recombination
īMostly F1 will make parental gametes, which are Ab and aB
oBut with recombination in F1: AB and ab are the recombinant gametes
īIf linked, parental more common than recombinants
īUnlike how under Mendelian ratios, the ratios are equal
THEN, what about trihybrid crossing?
Hypothesis and Extension
īLinear order along chromosome ā genes are
īRecombination more likely if further apart
īCan infer order of genes along chromosomes if look at many phenotypes and
many crosses
īIf Morgan is right, A+B+C = roughly A+C
Testing thisā¦
īTest cross because it unmasks the phenotypes
īMutants identified in lab
īCt and cv should have low proportion of recombinant phenotypes
īTwo inbred lines
oTest cross with vermilion eyes and cut wing edges
oProgeny produced
īCAPITALS FOR THE PLUSES, LOWERCASE FOR THE NON-
PLUSES
oV+ct is double homozygote
oVct+is same
oF1 has Vct+and v+ct ā 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
īFirst is double mutant
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