In a three-point cross between a flowering plant heterozygous at the following l
ID: 6806 • Letter: I
Question
In a three-point cross between a flowering plant heterozygous at the following loci LzlzGlglSusu and a homozygous recessive flowering plant (lzlzglglsusu) you obtain the gene combinations of the progeny frequencies below. You know the genes are all on the 3rd chromosome of the plants so there must be a fair amount of recombination taking place. Using these data, determine the gene in the middle, map the chromosomes in order and give distances between them. Now calculate the interference and coefficient of coincidence on the above data set and tell me what it meansPhenotype of Testcross Genotype of Gametes Number of Progeny
Wildtype Lz Gl Su 286
Sugary Lz Gl su 4
Glossy, sugary Lz gl su 40
Lazy, glossy, sugary lz gl su 272
Lazy lz Gl Su 33
Lazy, sugary lz Gl su 44
Glossy Lz gl Su 59
Lazy, glossy lz gl Su 2
In a three-point cross between a flowering plant heterozygous at the following loci LzlzGlglSusu and a homozygous recessive flowering plant (lzlzglglsusu) you obtain the gene combinations of the progeny frequencies below. You know the genes are all on the 3rd chromosome of the plants so there must be a fair amount of recombination taking place. Using these data, determine the gene in the middle, map the chromosomes in order and give distances between them. Now calculate the interference and coefficient of coincidence on the above data set and tell me what it means
Phenotype of Testcross Genotype of Gametes Number of Progeny
Wildtype Lz Gl Su 286
Sugary Lz Gl su 4
Glossy, sugary Lz gl su 40
Lazy, glossy, sugary lz gl su 272
Lazy lz Gl Su 33
Lazy, sugary lz Gl su 44
Glossy Lz gl Su 59
Lazy, glossy lz gl Su 2
Explanation / Answer
Given data is
Phenotype of Testcross Genotype of Gametes Number of Progeny
Wild type Lz Gl Su 286
Sugary Lz Gl su 4
Glossy, sugary Lz gl su 40
Lazy, glossy, sugary lz gl su 272
Lazy lz Gl Su 33
Lazy, sugary lz Gl su 44
Glossy Lz gl Su 59
Lazy, glossy lz gl Su 2
Total progeny = 740
Recombinant frequency between Lz and Gl = 40 + 33+ 44+ 59 = 176 / 740 = 0.24
Recombinant frequency between Gl and Su = 4 + 44+ 59+ 2 = 109 / 740 = 0.14
Recombinant frequency between Lz and Su = 4+ 40+ 33+ 2 = 79 / 740 = 0.10
From above frequencies the map distances are
Lz----------Su---------------Gl
ß--0.10--àß----0.14---à
ß--------0.24-------------à
coefficient of coincidence = actual double recombinant frequency / expected double recombinant frequency.
Actual double recombinant frequency = 44 + 59 = 103
Expected double recombinant frequency = 176 / 740 * 109 / 740 = 0.035 = 25.9 per 740.
coefficient of coincidence = 103 / 25.9
= 3.97
Interference = coefficient of coincidence – 1
= 3.97 – 1 = 2.97
From above calculations it shows that there is a strong cross over between the genes during meosis.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.