Question

Genetic analysis has shown that three recessive genes d (dwarf), e (enlarged trunk) and f (fine leaf) are all found on chromosome #1 of apple. When a plant that was heterozygous for each of these markers was testcrossed, the following 1000 progenies were obtained:

wild type- 130 (+++)

fine leaf- 19 (++f)

enlarged trunk- 2 (+e+)

enlarged trunk, fine leaf – 350 (+ef)

dwarf -362 (d++)

dwarf, fine leaf -1 (d+f)

dwarf, enlarged trunk -16 (de+)

dwarf, enlarged trunk, fine leaf- 120 (def)

a) Determine the central gene.

b) Calculate recombination frequencies between each of these three pairs of genes.

c) Draw a genetic map for the location of these 3 genes on chromosome #1 of apple. Show the map distances (cM or m.u.) between each loci.

d) Determine the interference among crossover events within the region of the chromosome containing the three genes.

Answer 1

Since all the mentioned genes are on the same chromosome, they will be not independently assorted unless there is recombination occurring. Recombination of genes occurs in the process of gamete production by meiosis. If no recombination occurs all the genes of the same chromosome will be inherited as a whole, resulting in a set of progeny showing only the phenotypes of their parents present. But as we can see in the progeny table, there are offsprings which show a new combination of traits different from their parents. Hence, it is evident that recombination between the three genes has occurred.

It is important to realize that the test cross is carried out with homozygous recessive individuals. Hence, the parental generation cross is actually : +++/def (phenotype: wild type) X def/def (phenotype: dwarf, enlarged trunk, fine leaf). (+ refers to the dominant allele of the respective recessive allele). But because of crossing over, there are 8 genotype inputs possible from the heterozygote: +e+,d++,de+,+ef,++f,d+f,def, and +++. Gametes can carry any of these genotypes. The progenies with genotype combination having most frequencies are generally the parental type and the rest of them are recombinants. Also, double cross over recombinants is fewer because the chance of occurring recombination twice is faint.

genotype	phenotype	progeny number	organized after finding the middle gene	cross over state
+++	wild type	130	+++	single cross over between d and f
++f	fine leaf	19	+f+	single cross over between e and f
+e+	enlarged trunk	2		Double cross over (DCO1)
+ef	enlarged trunk, fine leaf	350	+fe	parental type
d++	dwarf	362	d++	parental type
d+f	dwarf, fine leaf	1		Double cross over(DCO2)
de+	dwarf, enlarged trunk	16	d+e	cross over between e and f
def	dwarf, enlarged trunk, fine leaf	120	dfe	single cross over between d and f

(a) A double cross over event puts the middle gene in between two flanking genes from the other homologous chromosome. Hence, in DCO progenies the middle gene is the one that flanks two parental type genes. By comparing the genotype of the double cross over (DCO) recombinants (+e+ and d+f) to the parental genotypes (+ef and d++), we notice the gene for fine leaf ( i.e. f) is in the middle gene.

(b) recombination frequencies calculated as:

single cross over frequency between e and f (SCOef) = [(19+16) + (2+1)]/1000 = 0.038

single cross over frequency between d and f (SCOdf)=[(120+130)+(2+1)]/1000= 0.253

(c) The map distance in centimorgan represents the recombination frequency between genes. This distance is calculated by multiplying the recombination frequency among two genes multiplied by 100.

It has already established that the middle gene is f. The distances between d and f and f and e are as follows:

distance between d and f = SCOdf*100=25.3 cM

distance between f and e = SCOef*100=3.8 cM

Based on this following is the map:

(d) Interference is calculated as :

Interference = 1- coefficient of coincidence

coefficient of confidence = observed double-crossover recombinants/ expected double-crossover recombinants.

observed double-crossover recombinants = DCO1 + DCO2 = 2+ 1= 3

expected double-crossover recombinants = expected double-crossover frequency *1000 =  SCOdf*SCOef*1000=9.614

so, the coefficient of coincidence = 3/9.614 = 0.312

Hence, interference = 1- coefficient of coincidence= 1- 0.312 =0.688