Vampire Shugoshinator crosses pure-breeding plants with dominant phenotypes aeri
ID: 262309 • Letter: V
Question
Vampire Shugoshinator crosses pure-breeding plants with dominant phenotypes aerial roots (A), hairy leaves(H), and blue flowers (B) to a pure-breeding plant with recessive phenotypes stringy roots (a), smooth leaves (h),and white flowers (b). The F1 progeny are test-crossed to pure breeding stringy roots, smooth leaves, white flower plants. The test-cross progeny are shown in the table below.
Root type ,Leaf appearance, Flower color ,Number
Stringy hairy blue 60
aerial hairy white 47
stringy smooth white 270
aerial hairy blue 234
stringy smooth blue 46
aerial smooth white 61
a) Write the genotypes of the plants that gave rise to these progenies. Show the arrangement of the alleles on the chromosomes.
b) How many phenotypic classes do you expect to see in the cross you listed above?
c) What classes are missing and what do they represent? Explain fully!
d) Calculate the distances between these gene loci. Clearly show your work.
Distance between A and H _____________
Distance between H and B _____________
Distance between A and B _____________
e) Draw a genetic map of the three genes.
f) How many double crossover progeny are expected? ______ Clearly show your work.
Explanation / Answer
Answer:
Stringy hairy blue 60- (aHB) (Single cross over 1)
aerial hairy white 47- (AHb) (Single cross over 1)
stringy smooth white 270- (ahb) (non-recombinant)
aerial hairy blue 234 - (AHB) (non-recombinant)
stringy smooth blue 46 - (ahB) (Single cross over 1)
aerial smooth white 61 - (Ahb) (Single cross over 2)
a).
Homozygous dominant (AHB/AHB) x homozygous recessive (ahb/ahb)---Parents
AHB/ahb ------------------------------------------------F1
AHB / ahb x ahb/ahb -------------------------------------------------------------------------Test cross
b). 6
c). Double cross over progeny are missed. Those are aerial, smooth, blue and stringy, hairy, white phenotypes. It might be due very close linkage between the genes and very short distance between single cross over 1 & single cross over 2.
d).
Distance between A and H ____16.85 cM or mu_________
Distance between H and B __12.95 cM or mu___________
Distance between A and B ___29.81 cM or mu__________
e).
A----------16.85cM--------H-----------12.95cM--------------B
f). Expected double crossover progeny = 2.18 %
Explanation:
Parental (non-recombinant) genotypes is AHB/ahb
1).
If single crossover occurs between A& H.
Normal combination: AH/ah
After crossover: Ah/aH
Ah progeny= 61
aH progeny = 60
Total this progeny = 121
The recombination frequency between A&H = (number of recombinants/Total progeny) 100
RF = (121/718)100 = 16.85%
2).
If single crossover occurs between H & B..
Normal combination: HB/hb
After crossover: Hb/hB
Hb progeny= 47
hB progeny = 46
Total this progeny = 93
The recombination frequency between H&B= (number of recombinants/Total progeny) 100
RF = (93/718)100 = 12.95%
3).
If single crossover occurs between A & B..
Normal combination: AB/ab
After crossover: Ab/aB
Ab progeny= 47+61= 108
aB progeny = 46+60 = 106
Total this progeny = 214
The recombination frequency between A&B= (number of recombinants/Total progeny) 100
RF = (214/718)100 = 29.81%
Recombination frequency (%) = Distance between the genes (cM)
A----------16.85cM--------H-----------12.95cM--------------B
Expected double crossover frequency = (RF between A & H) * (RF between H & B)
= 0.1685% * 0.1295% = 0.0218 = 2.18%
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