Please don\'t just draw the map. I need an explanation as to why you\'re drawing
ID: 301151 • Letter: P
Question
Please don't just draw the map. I need an explanation as to why you're drawing it the way you are. step by step or bad rating. thank you. please be detailed in all your answers.
1. Multiple two-point testcross experiments were performed and recombination frequencies between each gene pair were calculated. Draw a map of the genes in the correct order. Write the distance between each gene in map units.
C – D 50% RF
C – F 46% RF
B – C 25% RF
D – F 10% RF
D - B 30% RF
B – F 20% RF
2. The following testcross produces the progeny shown: AaBb × aabb à 10 AaBb, 40 Aabb, 40 aaBb, and 10 aabb. What is the percentage of recombination between the A and B loci? Were the genes in the AaBb parent in coupling or repulsion?
Explanation / Answer
1. No possible gene order satisfies the given set of conditions exactly. If we keep the gene order BFD or DFB, it satisfies for these three genes, however adding the 4th gene C does not fit from any side to satisfy CF and CD recombination frequency. Since CF is given 46% (odd number), so should be present at the extreme end of the gene order as rest of the gene pair frequencies are even numbered. With CD frequency 50%, it specifies the genes must be on different chromosomes in our case so satisfy the most appropriate gene order as DFBC,
2. Given; testcross, AaBb×aabb
Since the genotypes found most frequently are the parental genotypes, therefore,
Genotypes can be assigned as:
Aabb=40………………..Parental
aaBb=40………………..Parental
aabb=10………………..Recombinant
AaBb=10……………….Recombinant
With the help of above data, the percentage recombination between the A and B loci can be calculated as:
=Number of recombinant offspring/ total number of offspring×100
=10+10/10+10+40+40×100
=20/100×100
=20%
Since the recombinant frequency is between 0- 50%, the two genes are linked i.e on the same chromosome undergoing independent assortment, so we can infer that AaBb parent is in the coupling (cis) phase.
3. We can only suggest the genotypes from the given information. To be able to designate specific recombinants and nonrecombinants we need the genotype frequency which is not given in our case. So, the possible gametic contribution of the heterozygous parent is as follows:
ABC
Abc
aBC
Abc
aBBc
AbC
ABc
abC