Example 4: Recombination (bis)

Pairing 4.2: darkgreen/blue type II cock x skyblue hen

Genetic formula:
Cock bl⁺ D/bl D⁺

Hen bl D⁺/bl D⁺

The gametes produced by the cock are:

85% 42.5% bl⁺ D

42.5% bl D⁺

15% 7.5% bl⁺ D⁺

7.5% bl D

The gametes produced by the hen are all:
100% bl
D⁺

Punnet square:

bl⁺ D bl D⁺ bl⁺ D⁺ bl D

42.5% 42.5% 7.5% 7.5%

bl
D⁺ 100% bl⁺ D/bl D⁺ bl D⁺/bl D⁺ bl⁺ D⁺/bl D⁺ bl D/bl D⁺

Solution:

42.5% darkgreen/blue type II
42.5% skyblue
7.5% lightgreen/blue
7.5% cobalt

The last two are the result of recombination (7.5% + 7.5% = 15% recombination).
Note These percentages are only indications. Due to chance the results of real pairings may
very well be different from these percentages, but the result of an infinite number of
pairings will. Secondly the recombination frequency is just an estimate and not exactly known.

As before you can interchange the cock and the hen.

What will be the result of a very similar pairing, where we use an darkgreen/blue type I instead of a type II?

Pairing 4.3: darkgreen/blue type I cock x skyblue hen

Genetic formula:
Cock bl⁺ D⁺/bl D

Hen bl D⁺/bl D⁺

The gametes produced by the cock are (15% recombination):

85% 42.5% bl⁺ D⁺

42.5% bl D

15% 7.5% bl⁺ D

7.5% bl D⁺

The gametes produced by the hen are all:
100% bl
D⁺

Punnet square:

bl⁺ D⁺ bl D bl⁺ D bl D⁺

42.5% 42.5% 7.5% 7.5%

bl
D⁺ 100% bl⁺ D⁺/bl D⁺ bl D/bl D⁺ bl⁺ D/bl D⁺ bl D⁺/bl D⁺

Solution:

42.5% lightgreen/blue
42.5% cobalt
7.5% darkgreen/blue type II
7.5% skyblue

The last two are the result of recombination (7.5% + 7.5% = 15% recombination).
Note These percentages are only indications. Due to chance the results of real pairings may
very well be different from these percentages, but the result of an infinite number of
pairings will. Secondly the recombination frequency is just an estimate and not exactly known.

As before you can interchange the cock and the hen.

You can see that pairing 4.2 and pairing 4.3 produce exactly the same offspring, but there is a big difference in the frequencies! When you want to breed as many cobalts as possible, pairing 4.3 is more appropriate (42.5% vs 7.5% cobalt). But when you only have skyblues and olivgreens that are not split blue, you will have to use pairings 4.1 and 4.2. Once you have cobalts, you can use them to breed more:

Pairing 4.4: cobalt cock x skyblue hen

Genetic formula:
Cock bl D⁺/bl D

Hen bl D⁺/bl D⁺

The gametes produced by the cock are:
Recombining bl D⁺ with bl D leads to bl D and bl D⁺.
These are the same gametes as before recombination:
Recombination has no effect and can be ignored.

50% bl D⁺

50% bl D

The gametes produced by the hen are all:
100% bl
D⁺

Punnet square:

bl D⁺ bl D

50% 50%

bl
D⁺ 100% bl D⁺/bl D⁺ bl D/bl D⁺

Solution:

50% skyblue
50% cobalt

As before you can interchange the cock and the hen.

Some of you might have seen that the only allele of the bl gene is bl, so the offspring will only have the bl allele and will be 'blue'. Then we only have to consider the D gene:
If we forget the bl gene, the pairing becomes darkgreen x lightgreen: D⁺/D x D⁺/D⁺
This could very well been an example in a previous part, when you write it out the result will be: 50% darkgreen + 50% lightgreen.
If we now add 'blue' (bl/bl) to the offspring in order to get the result of the pairing cobalt x skyblue, we get:
'blue' (bl/bl) & 'darkgreen' (D⁺/D) = cobalt (bl D⁺/bl D)
'blue' (bl/bl) & 'lightgreen' (D⁺/D⁺) = skyblue (bl D⁺/bl D⁺)
This is the same result as before for pairing 4.4.

Recombination can occure in both parents:
Pairing 4.5: darkgreen/blue type II cock x darkgreen/blue type II hen

Genetic formula:
Cock bl D⁺/bl⁺ D

Hen bl D⁺/bl⁺ D

The gametes produced by the cock are (15% recombination):

85% 42.5% bl D⁺

42.5% bl⁺ D

15% 7.5% bl⁺ D⁺

7.5% bl D

The gametes produced by the hen are (15% recombination):

85% 42.5% bl D⁺

42.5% bl⁺ D

15% 7.5% bl⁺ D⁺

7.5% bl D

Punnet square:

bl D⁺ bl⁺ D bl⁺ D⁺ bl D

42.5% 42.5% 7.5% 7.5%

bl D⁺ 42.5% bl D⁺/bl D⁺ bl⁺ D/bl D⁺ bl⁺ D⁺/bl D⁺ bl D/bl D⁺

bl⁺ D 42.5% bl D⁺/bl⁺ D bl⁺ D/bl⁺ D bl⁺ D⁺/bl⁺ D bl D/bl⁺ D

bl⁺ D⁺ 7.5% bl D⁺/bl⁺ D⁺ bl⁺ D/bl⁺ D⁺ bl⁺ D⁺/bl⁺ D⁺ bl D/bl⁺ D⁺

bl D 7.5% bl D⁺/bl D bl⁺ D/bl D bl⁺ D⁺/bl D bl D/bl D

Solution:

18.0625% skyblue (18.0625%)
36.1250% darkgreen/blue type II (18.0625% + 18.0625%)
6.3750% lightgreen/blue (3.1875% + 3.1875%) *
6.3750% cobalt (3.1875% + 3.1875%) *
18.0625% olivgreen (18.0625%)
6.3750% darkgreen (3.1875% + 3.1875%) *
6.3750% olivgreen/blue (3.1875% + 3.1875%) *
0.5625% lightgreen (0.5625%) **
1.1250% darkgreen/blue type I (0.5625% + 0.5625%) **
0.5625% mauve (0.5625%) **

No star indicates that no recombination has occured and corresponds to the
4 cells on the top left of the Punnet square

Two stars (**) indicate that in both parents recombination has occured,
these correspond to the 4 cells at the bottom right of the Punnet square.

A single star (*) means that recombination has occured in one of the parents
(cells on the top right or on the bottom left of the Punnet square).

Although only two genes are involved, the solution is already fairly complicated. As long as recombination occured only in one parent, the solution was managable.