Respuesta :
Answer:
Explanation:
a). Order of genes = wx-v-sh
b).
Distance between gene wx and v = 7 m.u.
Distance between gene v and sh= 30 m.u.
Distance between gene wx and sh= 33.64 m.u.
c).
Coefficient of Coincidence=0.81
Interference=0.19
Exaplanatiion:
Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.
Hence, the parental (non-recombinant) genotype is Wx Sh V / wx sh v.
1).
If single crossover occurs between Wx & Sh..
Normal combination: Wx Sh / wx sh
After crossover: Wx sh / wx Sh
Wx sh progeny= 1515+280 = 1795
wx Sh progeny = 1531+292 = 1823
Total this progeny = 3618
The recombination frequency between Wx & Sh = (number of recombinants/Total progeny) 100
RF = (3618/10756)100 = 33.64%
2).
If single crossover occurs between Sh & V..
Normal combination: Sh V / sh v
After crossover: Sh v / sh V
Sh v progeny= 94+1531 = 1625
sh V progeny = 87+1515 = 1602
Total this progeny = 3227
The recombination frequency between Sh & V = (number of recombinants/Total progeny) 100
RF = (3227/10756)100 = 30%
3).
If single crossover occurs between Wx & V..
Normal combination: Wx V / wx v
After crossover: Wx v / wx V
Wx v progeny= 94+280= 374
wx V progeny = 87+292 = 379
Total this progeny = 753
The recombination frequency between Wx & V = (number of recombinants/Total progeny) 100
RF = (753 / 10756)100 = 7%
Recombination frequency (%) = Distance between the genes (mu)
Wx-------7mu------V-----------30mu--------------Sh
Locus1(Wx)------Locus 2(V)-------Locus 3 (Sh)
Expected double crossover frequency = (RF between Wx & V) * (RF between V & Sh)
= 7% * 30% = 0.021
The observed double crossover frequency = 87+94 / 10756= 0.017
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0.017 / 0.021
= 0.81
Interference = 1-COC
= 1-0.81 = 0.19
Genetic distances and genes order are necessary to get CC and IC. a) The genes order is Wx---V----Sh. b) Distances between genes are Wx-V = 0.07 MU. V-Sh = 0.3 MU. Wx-Sh = 0.37 MU. c) CC= 0.8 / IC = 0.2.
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Available data:
→ Three linked genes ⇒ Wx, Sh, and V
→ A three-point testcross is performed ⇒ WxwxShshVv x wxwxshshvv
To solve this excersice, we need to know that
- 1% of recombination frequency = 1 map unit = 1 cM.
- The maximum recombination frequency is always 50%.
- The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant one.
- The recombination frequencies between two genes determine their distance in the chromosome, measured in map units.
By knowing the recombination frequencies, we can calculate distances between the three genes in the problem and figure the genes order out.
We have the number of descendants of each phenotype product of the tri-hybrid cross.
- wx sh V 87
- Wx Sh v 94
- Wx Sh V 3,479
- wx sh v 3,478
- Wx sh V 1,515
- wx Sh v 1,531
- wx Sh V 292
- Wx sh v 280
The total number of individuals is N = 10,756
a)
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:
Parental)
- Wx Sh V 3,479
- wx sh v 3,478
Simple recombinant)
- Wx sh V 1,515
- wx Sh v 1,531
- wx Sh V 292
- Wx sh v 280
Double recombinant)
- wx sh V 87
- Wx Sh v 94
Now, let us compare the parental gametes with the recombinant gametes
Parentals) WxShV wxshv
Double Recombinant) WxShv wxshV
They only change in the position of the alleles V/v.
This suggests that the position of the gene V is in the middle of the other two genes, Wx and Sh, because in a double recombinant only the central gene changes position in the chromatid.
So, the order of the genes is:
----Wx---- V -----Sh ----
Now we will call
- Region I to the area between Wx and V
- Region II to the area between V and Sh.
b)
Once established the order of the genes we can calculate genetic 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 Wx and V genes, and P2 to the recombination frequency between V and Sh.
- P1 = (R + DR) / N
- P2 = (R + DR)/ N
Where:
- R is the number of simple recombinants in each region,
- DR is the number of double recombinants in each region, and
- N is the total number of individuals.
So:
P1 = (R + DR) / N
P1 = (292 + 280 + 87 + 94) / 10,756
P1 = 756/10,756
P1 = 0.07
P2 = (R + DR)/ N
P2 = (1515 + 1531 + 87 + 94) / 10,756
P2 = 3227 / 10756
P2 = 0.3
Now, to calculate the recombination frequency between the two extreme genes, Wx and Sh, we can just perform addition or a sum:
P1 + P2= Pt
0.07 + 0.3 = Pt
0.37 = 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).
Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:
- GD1 = P1 x 100 = 0.07 x 100 = 7 MU
- GD2 = P2 x 100 = 0.3 x 100 = 30 MU
- GD3 = Pt x 100 = 0.37 x 100 = 37 MU
c)
To calculate the coefficient of coincidence, CC, we must use the next formula:
CC= observed double recombinant frequency/expected double recombinant frequency
Note:
- 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 = ((87 + 94)/10,756)/0.07x0.3
CC = (181/10756)/0.021
CC = 0.8
The coefficient of interference, I, is complementary with CC.
I = 1 - CC
I = 1 - 0.8
I = 0.2
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