Question

1. Maize plants homozygous for the recessive gene "variable sterile" (va) exhibit irregular distribution of chromosomes during meiosis. Yellowish-green seedlings are the results of another recessive gene called "virescent" (v). A third recessive called "glossy" (g) produces shinny leaves. All three genes are linked. Two homozygous plants were crossed and produced an all normal F1. When the F were testcrossed, progeny phenotypes appeared as follows. virescent /V) 60 virescent, glossy (V,g 7 48 glossy g) variable sterile, virescent, glossy (Va,v, gl variable sterile, virescent (Va V) 270 4 variable sterile (Va) variable sterile, glossy 40 62 235 wild-type Totad 726 i) What were the genotypes and phenotypes of the original parents? ii) Diagram the linkage relationship and calculate the map distances ii) Calculate how much interfernce is observed

Answer 1

1. i) As cross between both homozygous plants produced all normal (wild type) progeny in F1 & when F1 are testcrossed we observed recessive phenotype, then one plant must be homozygous for wild type & another plant homozygous for recessive alleles. Though it may possible that plants are homozygous for one or two wild type trait & homozygous recessive for other trait(s); for example, if one plant is shown recessive trait for virescent, the other plant must be recessive for glossy & variable sterile only. But here they will be true breeding due to the fact that when F1 are testcrossed we observed non-recombinant phenotypes as wild type & triple mutant.

So, genotype (phenotype) of the parents $\rightarrow$v+v+ va+va+ gl+gl+ (Wild type) & vv vava glgl (Virescent, variable sterile & glossy) ('+' indicate wild type allele)

ii) Non-recombinant phenotypes will be highest in number & double cross-over type will be least in number. So, wild type & triple recessive will be non-recombinant & glossy & variable sterile, virescent will be double cross-over type. Further, non-recombinant & double cross-over type will be differ in middle gene only. By comparing wild type with glossy (Or, triple recessive type with variable sterile, virescent) we find that glossy (gl) gene will be in middle. So, correct gene order will be variable sterile- glossy- virescent or virescent- glossy- variable sterile.

Now, with correct gene order position of cross-over (shown by "/") is shown in below table.

Gene Order (Phenotype)	Progeny Number	Cross-over Type
v / gl+ va+ (Virescent)	60	Single cross-over
v gl / va+ (Virescent, glossy)	48	Single cross-over
v+ / gl / va+ (Glossy)	7	Double cross-over
v gl va (Virescent, glossy, variable sterile)	270	Non-recombinant
v / gl+ / va (Virescent, variable sterile)	4	Double cross-over
v+ gl+ / va (Variable sterile)	40	Single cross-over
v+ / gl va (Glossy, variable sterile)	62	Single cross-over
v+ gl+ va+ (Wild type)	235	Non-recombinant
Total	726

Now, recombination frequency = (Number of recombinant progeny / Total number of progeny)

So, v-gl recombination frequency = [(60+7+4+62) / 726] x 100% = 18.32% (Up to 2 decimal)

So, gl-va recombination frequency = [(48+7+4+40) / 726] x 100% = 13.64% (Up to 2 decimal)

Now, 1% recombination frequency = 1 map units (m.u.)

So, v-gl map distance = 18.32 m.u

So, gl-va map distance = 13.64 m.u

The genetic map is drawn in below picture.

- 18.32 m.u- 31.96 m.u

iii) Now, expected number of double cross-over = (Probability of recombination between v & gl x Probability of recombination between gl & va) x Total number of progeny = 0.1832 x 0.1364 x 726 = 18.14 (Up to 2 decimal)

So, we get fewer than expected double cross-over due to interference (As observed double cross-over is 11).

Now, coefficient of coincidence = Number of observed double cross-over / Number of expected double cross-over = (7+4) / 18.14 = 0.61 (Up to 2 decimal)

Now, interference = 1 - coefficient of coincidence = 1 - 0.61 = 0.39