SOME EXAMPLES OF BREEDING CALCULATIONS

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This section assumes basic knowledge of how genes are passed from parent to kitten and an understanding of the effect of dominant and recessive genes. It also requires understanding of the use of Punnet Squares (P-Squares), so is meant to be read after the 'How to work out the results of a breeding' section.

This section uses abbreviations for the different genes, so is designed to be used in conjunction with the 'Overview of the different genes used' section.

To make these examples easier to understand, it is assumed that no genes are carried except those stated (i.e. if it says 'Black', the assumption is that no chocolate/cinnamon alleles are carried unless otherwise stated).

The examples here are grouped into three sections, graded according to the complexity of the calculations, and getting increasingly more difficult. These examples are designed to be read (and understood) in the order written:-

Basic Examples
Solid Black (carrying Dilute) with a Solid Blue
Black Mackerel Tabby (carrying Chocolate) with a Chocolate Mackerel Tabby
Black Ticked Tabby (carrying Dilute and Mackerel Tabby) with a Blue Mackerel Tabby
Intermediate Examples
Chocolate Tabby Point (carrying Dilute, Cinnamon and Classic Tabby) with a Fawn Classic Tabby Mink
Cinnamon Classic Tabby (carrying Non-Agouti and Dilute) with a Lilac Point (masking Mackerel Tabby and carrying Cinnamon and Classic Tabby)
Natural Ticked Tabby Mink (masking Spotted Tabby and carrying Non-Spotted and Classic Tabby) with a Sable Classic Tabby Burmese
Advanced Examples
Black Ticked Tabby and White Harlequin (carrying Cinnamon and Classic Tabby) with a Chocolate Mackerel Tabby and White Van (carrying Cinnamon and Classic Tabby)
White (masking Red and Black Mackerel Tabby and carrying Non-White, Classic Tabby, Dilute, Non-Agouti and Chocolate) with a Lilac Tortie (masking Classic Tabby)
Red Point (masking Black and Classic Tabby and carrying Dilute and Cinnamon) with a Fawn Tortie Classic Tabby (carrying Non-Agouti and Points)

Basic Examples:-
The female's parentage is not included, because it has no bearing on these breedings.

Example 1:
Male: Solid Black (Sire: Solid Black, Dam: Solid Blue)
Female: Solid Blue

Because the male's Dam is Blue, we know that he must carry Dilute.

Male: aa BB Dd
Female: aa BB dd

Because both cats are aa and BB, there is no need to include this in the P-Square - it is guaranteed.
(See note above about assumptions regarding the carrying of non-stated genes)

(Male across the top, female down the left)

  D d
d Dd dd

Because both the female's genes are both the same, hers only need input once.

½ Dd, ½ dd, and we know that both parents are aa and BB, so:
½ aa BB Dd, ½ aa BB dd, meaning:
½ the kittens would be Solid Black, carrying Dilute
½ the kittens would be Solid Blue

Example 2:
Male: Black Mackerel Tabby (Sire: Chocolate Mackerel Tabby, Dam: Black Mackerel Tabby)
Female: Chocolate Mackerel Tabby

Because the male's Sire is Chocolate, we know that he must carry Chocolate.

Male: Bb AA tt
Female: bb AA tt

Because both cats are AA and tt, there is no need to include this in the P-Square - it is guaranteed.
(See note above about assumptions regarding the carrying of non-stated genes)

(Male across the top, female down the left)

  B b
b Bb bb

Because both the female's genes are both the same, hers only need input once.

½ Bb, ½ bb, and we know that both parents are AA tt, so:
½ Bb AA tt, ½ bb AA tt, meaning:
½ the kittens would be Black Mackerel Tabby, carrying Chocolate
½ the kittens would be Chocolate Mackerel Tabby

Example 3:
Male: Black Ticked Tabby (Sire: Blue Ticked Tabby, Dam: Black Mackerel Tabby)
Female: Blue Mackerel Tabby

Because the male's Sire is Blue, and his Dam is Mackerel Tabby, we know that he must carry Dilute and Mackerel Tabby.

Male: BB Dd AA Tat
Female: BB dd AA tt

Because both cats are BB and AA, there is no need to include this in the P-Square - it is guaranteed.
(See note above about assumptions regarding the carrying of non-stated genes)

(Male across the top, female down the left)

  D Ta D t d Ta d t
d t Dd Tat Dd tt dd Tat dd tt

(Because the female's genes are both the same, hers only need input once)

¼ Dd Tat, ¼ Dd tt, ¼ dd Tat, ¼ dd tt, and we know that both parents are BB AA, so:
¼ BB Dd AA Tat, ¼ BB Dd AA tt, ¼ BB dd AA Tat, ¼ BB dd AA tt, meaning:
¼ the kittens would be Black Ticked Tabby, carrying Dilute and Mackerel Tabby
¼ the kittens would be Black Mackerel Tabby, carrying Dilute
¼ the kittens would be Blue Ticked Tabby, carrying Mackerel Tabby
¼ the kittens would be Blue Mackerel Tabby

Are you still following? If not, go back and read through this again; you may also want to read the 'How to work out the results of a breeding section'. If you still don't understand, post a message on the help forum with your problem and someone should be able to help you understand. DO NOT read any further if you don't understand - this only gets more complicated.

Intermediate Examples:-
The female's parentage is only included where it has a bearing on the results of the breeding.

After each example, only read on if you understand that example.

Example 1:
Male: Chocolate Tabby Point (Sire: Blue Classic Tabby, Dam: Cinnamon Classic Tabby)
Female: Fawn Classic Tabby Mink

Because the male's Sire is a Blue Classic Tabby, and his Dam is a Cinnamon Classic Tabby, we know that he must carry Dilute, Cinnamon and Classic Tabby (note that both his parents must have been carrying Point).

Male: bb1 Dd AA tbtb cscs
Female: b1b1 dd AA tbtb cbcs

Because both cats are AA and tbtb, there is no need to include this in the P-Square - it is guaranteed.
(See note above about assumptions regarding the carrying of non-stated genes)

(Male across the top, female down the left)

  b D cs b d cs b1 D cs b1 d cs
b1 d cb bb1 Dd cbcs bb1 dd cbcs b1b1 Dd Cbcs b1b1 dd cbcs
b1 d cs bb1 Dd cscs bb1 dd cscs b1b1 Dd cscs b1b1 dd cscs

1/8 bb1 Dd cbcs
1/8 bb1 dd cbcs
1/8 b1b1 Dd cbcs
1/8 b1b1 dd cbcs
1/8 bb1 Dd cscs
1/8 bb1 dd cscs
1/8 b1b1 Dd cscs
1/8 b1b1 dd cscs

We know that both parents are AA and tbtb (meaning that all the kittens will be), so the kittens would be:
1/8 Chocolate Classic Tabby Mink, carrying Cinnamon and Dilute
1/8 Lilac Classic Tabby Mink, carrying Cinnamon
1/8 Cinnamon Classic Tabby Mink, carrying Dilute
1/8 Fawn Classic Tabby Mink
1/8 Chocolate Tabby Point, masking Classic Tabby and carrying Cinnamon and Dilute
1/8 Lilac Tabby Point, masking Classic Tabby and carrying Cinnamon and Dilute
1/8 Cinnamon Tabby Point, masking Classic Tabby and carrying Dilute
1/8 Fawn Tabby Point, masking Classic Tabby

Example 2:
Male: Cinnamon Classic Tabby (Sire: Solid Black, Dam: Lilac Classic Tabby)
Female: Lilac Point (Sire: Fawn Mackerel Tabby, Dam: Chocolate Classic Tabby. One of her past kittens with a Classic Tabby male was a Mackerel Tabby)

Because the male's Sire is a Solid, and his Dam is Lilac, we know that he must carry Dilute and Non-agouti.

Because the female's Sire is Fawn, and her dam is a Classic Tabby, we know that she must carry Cinnamon and Classic Tabby (Note that both her parents must have carried Point and Non-Agouti). Also, because she produced a Mackerel Tabby kitten in a mating with a Classic Tabby, we know that she must mask Mackerel Tabby.

Male: b1b1 Dd Aa tbtb
Female: bb1 dd aa tbtb

Because both cats are tbtb, there is no need to include this in the P-Square - it is guaranteed.
(See note above about assumptions regarding the carrying of non-stated genes)

(Male across the top, female down the left)

  b1 D A b1 D a b1 d A b1 d a
b d a bb1 Dd Aa bb1 Dd aa bb1 dd Aa bb1 dd aa
b1 d a b1b1 Dd Aa b1b1 Dd aa b1b1 dd Aa b1b1 dd aa

1/8 bb1 Dd Aa
1/8 bb1 Dd aa
1/8 bb1 dd Aa
1/8 bb1 dd aa
1/8 b1b1 Dd Aa
1/8 b1b1 Dd aa

1/8 b1b1 dd Aa
1/8 b1b1 dd aa

We know that both parents are tbtb (meaning that all the kittens will be), so the kittens would be:
1/8 Chocolate Classic Tabby, carrying Cinnamon, Dilute and Non-Agouti
1/8 Solid Chocolate, carrying Cinnamon and Dilute
1/8 Lilac Classic Tabby, carrying Cinnamon and Non-Agouti
1/8 Solid Lilac, carrying Cinnamon
1/8 Cinnamon Classic Tabby, carrying Dilute and Non-Agouti
1/8 Solid Cinnamon, carrying Dilute
1/8 Fawn Classic Tabby, carrying Non-Agouti
1/8 Solid Fawn

All of these kittens would carry the pointed gene.

Example 3:
Male: Natural Ticked Tabby Mink (Sire: Sable Classic Tabby Burmese, Dam: Seal Ticked Tabby Point. One of his previous kittens with a Classic Tabby female was Spotted Tabby)
Female: Sable Classic Tabby Burmese

Because the male's Sire is a Classic Tabby, we know that he must carry Classic Tabby and Non-Spotted. Also, because he produced a Spotted Tabby kitten in a mating with a Classic Tabby, we know that he must mask Spotted Tabby.

Male: BB Tatb Spsp cbcs
Female: BB tbtb spsp cbcb

Because both cats are BB, there is no need to include this in the P-Square - it is guaranteed.
(See note above about assumptions regarding the carrying of non-stated genes)

(Female across the top, Male down the left. This is because the male has too many changes to fit across the top)

  tb sp cb
Ta Sp cb Tatb Spsp cbcb
Ta Sp cs Tatb Spsp cbcs
Ta sp cb Tatb spsp cbcb
Ta sp cs Tatb spsp cbcs
tb Sp cb tbtb Spsp cbcb
tb Sp cs tbtb Spsp cbcs
tb sp cb tbtb spsp cbcb
tb sp cs tbtb spsp cbcs

1/8 Tatb Spsp cbcb
1/8 Tatb Spsp cbcs
1/8 Tatb spsp cbcb
1/8 Tatb spsp cbcs
1/8 tbtb Spsp cbcb
1/8 tbtb Spsp cbcs
1/8 tbtb spsp cbcb
1/8 tbtb spsp cbcs

We know that both parents are BB (meaning that all the kittens will be), so the kittens would be:

GENOTYPE:
1/8 Sable Ticked Tabby Burmese, masking Spotted Tabby, carrying Classic Tabby
1/8 Natural Ticked Tabby Mink, masking Spotted Tabby, carrying Classic Tabby
1/8 Sable Ticked Tabby Burmese, carrying Classic Tabby
1/8 Natural Ticked Tabby Mink, carrying Classic Tabby
1/8 Sable Spotted Tabby Burmese, masking Classic Tabby
1/8 Natural Spotted Tabby Mink, masking Classic Tabby
1/8 Sable Classic Tabby Burmese
1/8 Natural Classic Tabby Mink
PHENOTYPE:
1/4 Sable Ticked Tabby Burmese
1/4 Natural Ticked Tabby Mink
1/8 Sable Spotted Tabby Burmese
1/8 Natural Spotted Tabby Mink
1/8 Sable Classic Tabby Burmese
1/8 Natural Classic Tabby Mink

Still following? As before, if you're not, go back and read through this section again. You may find the 'How to work out the results of a breeding section' is of some help, although we are now mostly beyond that level. If you still don't understand, post a message on the help forum with your problem and someone should be able to help you understand. DO NOT read any further if you don't understand - this only gets more complicated.

Advanced Examples:-
In these examples, there are too many genotypes to write them all, so I will write the phenotypes and simply make a note as to the different genotypes within each phenotype. I will also change from fractions to percentages for these problems, because percentages are easier to use in more complex numbers.

After each example, only read on if you understand that example.

Example 1:
Male: Black Ticked Tabby and White Harlequin (Sire: Black Ticked Tabby and White Bicolour, Dam: Cinnamon Classic Tabby and White Harlequin)
Female: Chocolate Classic Tabby and White Bicolour (Sire: Chocolate Classic Tabby and White Van, Dam: Cinnamon Mackerel Tabby)

Because the male's Dam is a Cinnamon Classic Tabby, we know that he must carry Cinnamon and Classic Tabby. Because the female's Dam is Cinnamon, we know that she must also carry Cinnamon, and because her sire was a Van and her dam had no White Spot, we know that she must be a S1s.

Male: Bb1 Tatb SS1
Female: bb1 tbtb S1s

(Female across the top, Male down the left. This is because the male has more changes than the female, so will fit better this way around)

  b tb S1 b tb s b1 tb S1 b1 tb s
B Ta S Bb Tatb SS1 Bb Tatb Ss Bb1 Tatb SS1 Bb1 Tatb Ss
B Ta S1 Bb Tatb S1S1 Bb Tatb S1s Bb1 Tatb S1S1 Bb1 Tatb S1s
B tb S Bb tbtb SS1 Bb tbtb Ss Bb1 tbtb SS1 Bb1 tbtb Ss
B tb S1 Bb tbtb S1S1 Bb tbtb S1s Bb1 tbtb S1S1 Bb1 tbtb S1s
b1 Ta S bb1 Tatb SS1 bb1 Tatb Ss b1b1 Tatb SS1 b1b1 Tatb Ss
b1 Ta S1 bb1 Tatb S1S1 bb1 Tatb S1s b1b1 Tatb S1S1 b1b1 Tatb S1s
b1 tb S bb1 tbtb SS1 bb1 tbtb Ss b1b1 tbtb SS1 b1b1 tbtb Ss
b1 tb S1 bb1 tbtb S1S1 bb1 tbtb S1s b1b1 tbtb S1S1 b1b1 tbtb S1s

PHENOTYPE:
12.5% (4/32) Black Ticked Tabby and White Bicolour (¼ of these carry Chocolate and Classic Tabby and are S1s, ¼ carry Cinnamon and Classic Tabby and are S1s, ¼ carry Chocolate and Classic Tabby and are Ss, ¼ carry Cinnamon and Classic Tabby and are Ss)
(IN THIS CASE, ROWS 1 & 2, COLUMNS 2 & 4)
12.5% (4/32) Black Classic Tabby and White Bicolour (¼ of these carry Chocolate and are S1s, ¼ carry Cinnamon and are S1s, ¼ carry Chocolate and are Ss, ¼ carry Cinnamon and are Ss)
(IN THIS CASE, ROWS 3 & 4, COLUMNS 2 & 4)
6.5% (2/32) Black Ticked Tabby and White Van (½ of these carry Chocolate and Classic Tabby, ½ carry Cinnamon and Classic Tabby)
(IN THIS CASE ROW 1, COLUMNS 1 & 3)
6.5% (2/32) Black Ticked Tabby and White Harlequin (½ of these carry Chocolate and Classic Tabby, ½ carry Cinnamon and Classic Tabby)
(IN THIS CASE ROW 2, COLUMNS 1 & 3)
6.5% (2/32) Black Classic Tabby and White Van (½ of these carry Chocolate, ½ carry Cinnamon)
(IN THIS CASE ROW 3, COLUMNS 1 & 3)
6.5% (2/32) Black Classic Tabby and White Harlequin (½ of these carry Chocolate, ½ carry Cinnamon)
(IN THIS CASE ROW 4, COLUMNS 1 & 3)
6.5% (2/32) Chocolate Ticked Tabby and White Bicolour (½ of these carry Cinnamon and Classic Tabby and are S1s, ½ carry Cinnamon and Classic Tabby and are Ss)
(IN THIS CASE ROWS 5 & 6, COLUMN 2)
6.5% (2/32) Cinnamon Ticked Tabby and White Bicolour (½ of these carry Classic Tabby and are S1s, ½ carry Classic Tabby and are Ss)
(IN THIS CASE ROWS 5 & 6, COLUMN 4)
6.5% (2/32) Chocolate Classic Tabby and White Bicolour (½ of these carry Cinnamon and are S1s, ½ carry Cinnamon and are Ss)
(IN THIS CASE ROWS 7 & 8, COLUMN 2)
6.5% (2/32) Cinnamon Classic Tabby and White Bicolour (½ of these are S1s, ½ are Ss)
(IN THIS CASE ROWS 7 & 8, COLUMN 4)
3.125% (1/32) Chocolate Ticked Tabby and White Van (carrying Cinnamon and Classic Tabby)
(IN THIS CASE ROW 5, COLUMN 1)
3.125% (1/32) Chocolate Ticked Tabby and White Harlequin (carrying Cinnamon and Classic Tabby)
(IN THIS CASE ROW 6, COLUMN 1)
3.125% (1/32) Cinnamon Ticked Tabby and White Van (carrying Classic Tabby)
(IN THIS CASE ROW 5, COLUMN 3)
3.125% (1/32) Cinnamon Ticked Tabby and White Harlequin (carrying Classic Tabby)
(IN THIS CASE ROW 6, COLUMN 3)
3.125% (1/32) Chocolate Classic Tabby and White Van (carrying Cinnamon)
(IN THIS CASE ROW 7, COLUMN 1)
3.125% (1/32) Chocolate Classic Tabby and White Harlequin (carrying Cinnamon)
(IN THIS CASE ROW 8, COLUMN 1)
3.125% (1/32) Cinnamon Classic Tabby and White Van
(IN THIS CASE ROW 7, COLUMN 3)
3.125% (1/32) Cinnamon Classic Tabby and White Harlequin
(IN THIS CASE ROW 8, COLUMN 3)

The percentage tells you the statistical percentage of kittens from these two cats that will have that Phenotype. The fractions tell you how many kittens out of every 32 produced by these cats, would have that Phenotype.

I know it's confusing using percentages and then saying "½ of these would be…" but it's probably easier to see how these relate to the P-Square, in this way.

You may need to read this a couple of times, comparing it back to the original genotypes of the parents. Once you are sure that you understand how I got these results, move on to the next example.

Example 2:
Male: White (Sire: Black Ticked Tabby, Dam: White. Past kittens have included a Solid Red male from a mating with a Solid Black female; a Black Mackerel Tabby from a mating with a Chocolate Classic Tabby; a Chocolate Mackerel Tabby from a mating with a Solid Fawn, a Solid Blue and a Classic Tabby)
Female: Lilac Tortie (Sire: Chocolate Classic Tabby, Dam: Blue Tortie Classic Tabby)

Because the male is white, we can't tell anything about his colour genotype just by looking at him (except, of course, that he is white!). However, because his Sire was a Black Ticked Tabby, we know that he is heterozygous for Dominant White. We also know that he must mask Red (he produced a Red kitten), Black (he produced a Black kitten from a mating with a Chocolate) and Mackerel Tabby (he produced a Mackerel Tabby from a mating with a Classic Tabby). He must carry Classic Tabby (he has produced a Classic Tabby kitten) and Chocolate or Cinnamon (he produced a Chocolate kitten); and he must mask or carry Dilute (he has produced Dilute kittens) and Non-agouti (he has produced Solid kittens). However, we know that it must be Chocolate, not Cinnamon that he carries, and that he carries, rather than masks dilute and Non-agouti (he produced a Chocolate Mackerel Tabby kitten from a mating with a Solid Fawn).

Because the female's parents are both Classic Tabby we know that she must mask Classic Tabby (and we still maintain the assumption that no genes are carried unless mentioned).

Male: Bb xoy Dd ttb Aa Ww
Female: bb xox dd tbtb aa ww

(Female across the top, Male down the left. This is because the male has too many changes to fit across the top)

Row No.   b xo d tb a w b x d tb a w
1 B xo D t A W Bb xoxo Dd ttb Aa Ww Bb xox Dd ttb Aa Ww
2 B xo D t A w Bb xoxo Dd ttb Aa ww Bb xox Dd ttb Aa ww
3 B xo D t a W Bb xoxo Dd ttb aa Ww Bb xox Dd ttb aa Ww
4 B xo D t a w Bb xoxo Dd ttb aa ww Bb xox Dd ttb aa ww
5 B xo D tb A W Bb xoxo Dd tbtb Aa Ww Bb xox Dd tbtb Aa Ww
6 B xo D tb A w Bb xoxo Dd tbtb Aa ww Bb xox Dd tbtb Aa ww
7 B xo D tb a W Bb xoxo Dd tbtb aa Ww Bb xox Dd tbtb aa Ww
8 B xo D tb A W Bb xoxo Dd tbtb aa ww Bb xox Dd tbtb aa ww
9 B xo d t A W Bb xoxo dd ttb Aa Ww Bb xox dd ttb Aa Ww
10 B xo d t A w Bb xoxo dd ttb Aa ww Bb xox dd ttb Aa ww
11 B xo d t a W Bb xoxo dd ttb aa Ww Bb xox dd ttb aa Ww
12 B xo d t a w Bb xoxo dd ttb aa ww Bb xox dd ttb aa ww
13 B xo d tb A W Bb xoxo dd tbtb Aa Ww Bb xox dd tbtb Aa Ww
14 B xo d tb A w Bb xoxo dd tbtb Aa ww Bb xox dd tbtb Aa ww
15 B xo d tb a W Bb xoxo dd tbtb aa Ww Bb xox dd tbtb aa Ww
16 B xo d tb A W Bb xoxo dd tbtb aa ww Bb xox dd tbtb aa ww
17 B y D t A W Bb xoy Dd ttb Aa Ww Bb xy Dd ttb Aa Ww
18 B y D t A w Bb xoy Dd ttb Aa ww Bb xy Dd ttb Aa ww
19 B y D t a W Bb xoy Dd ttb aa Ww Bb xy Dd ttb aa Ww
20 B y D t a w Bb xoy Dd ttb aa ww Bb xy Dd ttb aa ww
21 B y D tb A W Bb xoy Dd tbtb Aa Ww Bb xy Dd tbtb Aa Ww
22 B y D tb A w Bb xoy Dd tbtb Aa ww Bb xy Dd tbtb Aa ww
23 B y D tb a W Bb xoy Dd tbtb aa Ww Bb xy Dd tbtb aa Ww
24 B y D tb A W Bb xoy Dd tbtb aa ww Bb xy Dd tbtb aa ww
25 B y d t A W Bb xoy dd ttb Aa Ww Bb xy dd ttb Aa Ww
26 B y d t A w Bb xoy dd ttb Aa ww Bb xy dd ttb Aa ww
27 B y d t a W Bb xoy dd ttb aa Ww Bb xy dd ttb aa Ww
28 B y d t a w Bb xoy dd ttb aa ww Bb xy dd ttb aa ww
29 B y d tb A W Bb xoy dd tbtb Aa Ww Bb xy dd tbtb Aa Ww
30 B y d tb A w Bb xoy dd tbtb Aa ww Bb xy dd tbtb Aa ww
31 B y d tb a W Bb xoy dd tbtb aa Ww Bb xy dd tbtb aa Ww
32 B y d tb A W Bb xoy dd tbtb aa ww Bb xy dd tbtb aa ww
33 b xo D t A W bb xoxo Dd ttb Aa Ww bb xox Dd ttb Aa Ww
34 b xo D t A w bb xoxo Dd ttb Aa ww bb xox Dd ttb Aa ww
35 b xo D t a W bb xoxo Dd ttb aa Ww bb xox Dd ttb aa Ww
36 b xo D t a w bb xoxo Dd ttb aa ww bb xox Dd ttb aa ww
37 b xo D tb A W bb xoxo Dd tbtb Aa Ww bb xox Dd tbtb Aa Ww
38 b xo D tb A w bb xoxo Dd tbtb Aa ww bb xox Dd tbtb Aa ww
39 b xo D tb a W bb xoxo Dd tbtb aa Ww bb xox Dd tbtb aa Ww
40 b xo D tb A W bb xoxo Dd tbtb aa ww bb xox Dd tbtb aa ww
41 b xo d t A W bb xoxo dd ttb Aa Ww bb xox dd ttb Aa Ww
42 b xo d t A w bb xoxo dd ttb Aa ww bb xox dd ttb Aa ww
43 b xo d t a W bb xoxo dd ttb aa Ww bb xox dd ttb aa Ww
44 b xo d t a w bb xoxo dd ttb aa ww bb xox dd ttb aa ww
45 b xo d tb A W bb xoxo dd tbtb Aa Ww bb xox dd tbtb Aa Ww
46 b xo d tb A w bb xoxo dd tbtb Aa ww bb xox dd tbtb Aa ww
47 b xo d tb a W bb xoxo dd tbtb aa Ww bb xox dd tbtb aa Ww
48 b xo d tb A W bb xoxo dd tbtb aa ww bb xox dd tbtb aa ww
49 b y D t A W bb xoy Dd ttb Aa Ww bb xy Dd ttb Aa Ww
50 b y D t A w bb xoy Dd ttb Aa ww bb xy Dd ttb Aa ww
51 b y D t a W bb xoy Dd ttb aa Ww bb xy Dd ttb aa Ww
52 b y D t a w bb xoy Dd ttb aa ww bb xy Dd ttb aa ww
53 b y D tb A W bb xoy Dd tbtb Aa Ww bb xy Dd tbtb Aa Ww
54 b y D tb A w bb xoy Dd tbtb Aa ww bb xy Dd tbtb Aa ww
55 b y D tb a W bb xoy Dd tbtb aa Ww bb xy Dd tbtb aa Ww
56 b y D tb A W bb xoy Dd tbtb aa ww bb xy Dd tbtb aa ww
57 b y d t A W bb xoy dd ttb Aa Ww bb xy dd ttb Aa Ww
58 b y d t A w bb xoy dd ttb Aa ww bb xy dd ttb Aa ww
59 b y d t a W bb xoy dd ttb aa Ww bb xy dd ttb aa Ww
60 b y d t a w bb xoy dd ttb aa ww bb xy dd ttb aa ww
61 b y d tb A W bb xoy dd tbtb Aa Ww bb xy dd tbtb Aa Ww
62 b y d tb A w bb xoy dd tbtb Aa ww bb xy dd tbtb Aa ww
63 b y d tb a W bb xoy dd tbtb aa Ww bb xy dd tbtb aa Ww
64 b y d tb A w bb xoy dd tbtb aa ww bb xy dd tbtb aa ww

Female - White - 32 (Uneven Numbered Rows 1 - 15 & 33 - 47)
Male - White - 32 (Uneven Numbered Rows 17 - 31 & 49 - 63)
Female - Red - 16 (Column 1, Even Numbered Rows 2 - 16 & 34 - 48)
Male - Red - 16 (Column 1, Even Numbered Rows 18 - 32 & 50 - 64)
Female - Black Tortie Mackerel Tabby - 1 (Column 2, Row 2)
Female - Black Tortie Classic Tabby - 1 (Column 2, Row 6)
Female - Black Tortie - 2 (Column 2, Rows 4 & 8)
Female - Blue Tortie Mackerel Tabby - 1 (Column 2, Row 10)
Female - Blue Tortie Classic Tabby - 1 (Column 2, Row 14)
Female - Blue Tortie - 2 (Column 2, Rows 12 & 16)
Male - Black Mackerel Tabby - 1 (Column 2, Row 18)
Male - Black Classic Tabby - 1 (Column 2, Row 22)
Male - Black - 2 (Column 2, Rows 20 & 24)
Male - Blue Mackerel Tabby - 1 (Column 2, Row 26)
Male - Blue Classic Tabby - 1 (Column 2, Row 30)
Male - Blue - 2 (Column 2, Rows 28 & 32)
Female - Chocolate Tortie Mackerel Tabby - 1 (Column 2, Row 34)
Female - Chocolate Tortie Classic Tabby - 1 (Column 2, Row 38)
Female - Chocolate Tortie - 2 (Column 2, Rows 36 & 40)
Female - Lilac Tortie Mackerel Tabby - 1 (Column 2, Row 42)
Female - Lilac Tortie Classic Tabby - 1 (Column 2, Row 46)
Female - Lilac Tortie - 2 (Column 2, Rows 44 & 48)
Male - Chocolate Mackerel Tabby - 1 (Column 2, Row 50)
Male - Chocolate Classic Tabby - 1 (Column 2, Row 54)
Male - Chocolate - 2 (Column 2, Rows 52 & 56)
Male - Lilac Mackerel Tabby - 1 (Column 2, Row 58)
Male - Lilac Classic Tabby - 1 (Column 2, Row 62)
Male - Lilac - 2 (Column 2, Rows 60 & 64)

This means:
50% (64/128) White (½ male, ½ female)
25% (32/128) Red (½ male, ½ female)
12.5% (16/128) Assorted Torties (¼ Black, ¼ Blue, ¼ Chocolate, ¼ Lilac - ¼ Mackerel Tabby, ¼ Classic Tabby, ½ Solid)
The remainder various colours of Non-Red, Non-White males

As you can see, with this many options, these calculations become very unwieldy and difficult to work out. While the P-Square is fantastic for calculating the full results of a litter, sometimes we are only interested in what proportion of a litter will be a colouring that we require. I will now explain a method of working this out, without using a P-Square.

Example 3:
Male: Red Point (Sire: Blue Classic Tabby, Dam: Chocolate Tortie Classic Tabby Mink. Past kittens have included a Cinnamon and a Seal point in a mating with a Chocolate)
Female: Fawn Tortie Classic Tabby (Sire: Red Classic Tabby, Dam: Blue Point)

Because the male is Red, we can't tell much about his genotype just by looking at him. However, because his Sire was Blue and his Dam Chocolate, we know that he must carry Dilute and Chocolate or Cinnamon, and both parents are Classic Tabby so he must mask Classic. We also know that he must mask Black (he produced a Seal point when mated with a Chocolate) and carry Cinnamon (he produced a Cinnamon kitten).
Because the female's Dam is a Solid Point, we know that she must carry point and Non-Agouti (Note that both parents must have been carrying Cinnamon).

Male: Bb1 xoy Dd aa tbtb cscs
Female: b1b1 xox dd Aa tbtb Ccs

For this method of calculation it is important to know what colour you require - I am working to find what proportion of the litter will be Fawn Tortie Point (b1b1 xox dd aa cscs), and will go through each gene in turn.

We know that Cinnamon is recessive, so a cat has to have two Cinnamon alleles to show Cinnamon. You can see that the female will always pass on a Cinnamon allele to her kittens, but the male will only pass on a Cinnamon to 50% of his - the other 50% will receive a Black. This means that 50% of the kittens will be (genotypically) Black (Bb1), and 50% (genotypically) Cinnamon (b1b1). As I am trying to find out about Fawns, it is only the 50% Cinnamon that I am interested in (1 in every 2 kittens).

Because we are interested in Torties, we are only interested in the females - 50% of the Cinnamons or 25% of all of the litters (1 in every 4 kittens).

In this case, the females will always receive a Red gene from their father, but 50% will also receive a Red gene from their mother. This means that 50% of the females will receive 2 Red genes and therefore be Red (xoxo), and 50% will receive 1 Red gene and be Tortie (xox). I am only interested in the latter, so that leaves 12.5% of the litters that interests me - the 12.5% that are Cinnamon, Female and Tortie (1 in every 8 kittens).

Of these, we can see that all of the kittens would receive a dilute allele from their mother, but only 50% would receive one from their father. This gives us 6.25% of the litters who are Cinnamon (Dd) Torties and 6.25% who are Fawn (dd) Torties - the ones that I am interested in (1 in every 16 kittens).

While all of the kittens will receive a Non-Agouti gene from their father, 50% of them will receive an Agouti gene from their mother and therefore be Tabby (Aa). The other 50% will receive a Non-Agouti from both parents, making them Solid (aa), so 3.125% of the litters will be Solid Fawn Torties (1 in every 32 kittens).

In this case I am not interested in the tabbies, so I will ignore the Classic Tabby gene and move straight on to the Albino gene. As you can see, although all of the Solid Fawn Torties will receive a 'cs' from their Dad, 50% of them will receive a 'C' allele from their mother, with only 50% receiving a 'cs' from her. This means that just 1.5625% of my litters will display the colour that I want. This is only 1 in every 64 kittens produced by this mating, so I am very unlikely to get the results that I want from this match.

Remember, though, that although this mating is unlikely to produce the desired result straight away, every kitten will carry Cinnamon, Dilute, Non-Agouti and Point, making them good breeding cats for the next generation. If I got one cat out of this mating, which was a tortie female or red male, I'd definitely keep it to breed the next generation (this particularly, because it is the only one of the desired genes that cannot be 'carried').

Hopefully you have followed all of this, and now have a full understanding of how to calculate the results of any mating you may do. However, if you need to clarify anything, visit the Help forum and explain clearly where you are struggling, and someone will be sure to help!

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