BIOL 2000 Lecture Notes - Centimorgan, Wild Type, Chromosome

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8 Feb 2013
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Chapter 4a-3
Recombinants
Aa X Bb x aabb Cis-dihybrid
o AB 150 Parental
o Ab 60 Recombinants
o aB 50 Recombinants
o ab 155 Parental
frequency of recombinant products = map distance
60+50 / 415 = 26.5%
The genes curly win and ebony are are 20 map units apart on chromosome 3.
A wild type fly is crossed by a fly homozygous recessive for mutation at both
genes, If the F1 is test crossed, what frewuecny of wild type winged and
ebony-eyed flies will be seen in the progeny?
o Map distance = frequency of recombinants
20% recombinants
80% parental
o + + / + + X e cy / e cy [ plus is wildtype and recessive is small letters]
F1 ++/ e cy x ecy/ ecy
Cis-di-hybrid
++ Parental 40%
+ cy Recombinant 10%
e + recombinant 10%
e cy Parental 20%
TTG and ttg are the same distance as they code for the same gene, so they are
in the same spot on the chromosome
Clicker:
In tomoato R and W are 15 cm apart. Gene R controls fruit (R=red r=orange)
and W controls flower color (W=yellow w=white). If it is cis-dihybrid for R and
W is test crossed. In test cross progeny, what is the frequency of orange fruit
and white plant?
o 42.5%
15 cM = 15% recombinants
RW / rw
RW 42.5% P
Rw 7.5% R
rW 7.5% R
rw 42.5% P
If there was no test cross, but an F2 generation. What is the probability of the
rrww?
o Self cross RW/rw
Two rw must fuse.
.425 x .425
= .18062
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