Answer and Explanation:
We have the number of descendants of each phenotype product of the tri-hybrid cross.
The total number, N, of individuals is 230.
<em>In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the thrid not, or that the three genes are linked</em>. In this example, in particular, the three genes are linked on the same chromosome.
Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the genotypes of the parental gametes with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:
<em>Parental</em>)
<em>Double recombinant</em>)
Comparing them we will realize that between
s u TU (parental)
s u tu (double recombinant)
and
S U tu (Parental)
S U TU (double recombinant)
They only change in the position of the alleles TU/tu. This suggests that the position of the gene TU is in the middle of the other two genes, S and U, because in a double recombinant only the central gene changes position in the chromatid.
So, the order of the genes is:
---- S ---- TU -----U ----
In a scheme it would be like:
Chromosome 1:
---s---TU---u--- (Parental chromatid)
---s---tu---u--- (Double Recombinant chromatid)
Chromosome 2
---S---tu---U--- (Parental chromatid)
---S---TU---U--- (Double Recombinant chromatid)
Now we will call Region I to the area between S and TU and Region II to the area between TU and U.
Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between S and TU genes, and P2 to the recombination frequency between TU and U.
P1 = (R + DR) / N
P2 = (R + DR)/ N
Where: R is the number of recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals. So:
P1 = (21+17+4+2)/230
P1 = 44/230
P1 = 0.191
P2 = (21+13+4+2)/230
P1 = 40/230
P1 = 0.174
Now, to calculate the recombination frequency between the two extreme genes, S and U, we can just perform addition or a sum:
P1 + P2= Pt
0.191 + 0.174 = Pt
0.365=Pt
The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).
The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product. Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:
GD1= P1 x 100 = 0.191 x 100 = 19.1 MU
GD2= P2 x 100 = 0.174 x 100 = 17.4 MU
GD3=Pt x 100 = 0.365 x 100 = 36.5 MU
To calculate the coefficient of coincidence, CC, we must use the next formula:
CC= observed double recombinant frequency/expected double recombinant frequency
<em>Note: </em>
- observed double recombinant frequency=total number of observed double recombinant individuals/total number of individuals
- expected double recombinant frequency: recombination frequency in region I x recombination frequency in region II.
CC= ((2 + 4)/230)/0.174x0.191
CC=(6/230)/0.0332
CC=0.7857
The coefficient of interference, I, is complementary with CC.
I = 1 - CC
I = 1 - 0.7857
I = 0.2143
