1. The B and N genes are 25 map units apart on a chromosome. Following a dihybri
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Question
1. The B and N genes are 25 map units apart on a chromosome. Following a dihybrid testcross, how many offspring of 250 would you expect to display a recombinant phenotype? (0.5 pt) a. b. If you performed a dihybrid testcross with a cis dihybrid, what would be the genotypes of both recombinant offspring phenotypic classes? (0.5 pt) 2. Listed below are the phenotypes of the offspring from a dihybrid testcross for pollen color (coded by the C gene) and texture (coded by the Tgene): Offspring phenotype orange, fuzzy pollen orange, smooth pollen yellow, fuzzy pollen yellow, smooth poller Number of offspring 125 26 32 a. Which of these phenotypic groups represent the parental type (nonrecombinant)? How do you know? (1 pt) b. What is the recombination frequency between these linked genes? (0.5pt) Draw a linkage map the the C and T genes based on this information and label the distance between them. (0.5 pt) c.Explanation / Answer
1)
a)given distance between two genes is 25mapunits
we know tht 0.01 recombinant freq=1map unit
=>25 mapunits=0.25 recombinant freq
recombinant freq=no of recombinant /total progeny
=>0.25=recombinants/250=62.5 recombinants
b)test cross is done between BbNn and bbnn
the gametes are BN,Bn,bN,bn from BbNn
and bn from bbnn
by cross we get BbNn,Bbnn,bbNn,bbnn
therefore recombinants are Bbnn and bbNn
2)
a)the parents are CCPP(yellow and smooth-dominant) and ccpp(orange and fuzzy-recessive)
the penotype ratio of f2 gen is 9:3:3:1
9-yellow and smooth(of this 1 is non recombinant-we can know this by the dihybrid cross)
1 is the non recombinant-orange and fuzzy
given yellow and smooth are 117 therefore the non recombinants are 1/9*117=32
and orange and fuzzy are 125 (non recombinants)
therefore total non recombinants are32+125=157
b)recombinant req=no of recombinants /total progeny*100
=> total progeny =125+26+32+117=300
recombinants =300-157=143
therefore recombinant freq=143/300*100=47.3%
c)here the recombinant fre=47.3%
1% recombinant freq=1 mapunit
=> distance between two genes is 47.3 mapunits
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