How to Figure Genotype from a Pedigree
Step 1 – What color is your rabbit? Look up the base genotype on the chart!
Take for example our senior JW doe Black Magic…
Black Magic is a black, from the chart aaB-C-D-E-
Step 2 – What color rabbits are in the pedigree? Look up their base genotypes.
Black Magic’s sire is a blue, aaB-C-ddE-
His sire is a pointed white, aaB-ch-D-E-
His dam is a blue, aaB-C-ddE-
Black Magic’s dam is also a black aaB-C-D-E-
Her sire is a Siamese sable, A-B-chl-D-E-
Her dam is a chestnut, A-B-C-D-E-
Step 3 – Organize the information in a chart (not strictly necessary, but I can’t follow without some organization!)
Take for example our senior JW doe Black Magic…
Black Magic is a black, from the chart aaB-C-D-E-
Step 2 – What color rabbits are in the pedigree? Look up their base genotypes.
Black Magic’s sire is a blue, aaB-C-ddE-
His sire is a pointed white, aaB-ch-D-E-
His dam is a blue, aaB-C-ddE-
Black Magic’s dam is also a black aaB-C-D-E-
Her sire is a Siamese sable, A-B-chl-D-E-
Her dam is a chestnut, A-B-C-D-E-
Step 3 – Organize the information in a chart (not strictly necessary, but I can’t follow without some organization!)
Step 4 – look for the double recessive alleles. Note: one had to come from each parent AND that is the only allele that can be passed from that rabbit to its offspring. Fill those in!
Step 5 – look at the blanks left in the top line of your pedigree. Write in the POSSIBLE alleles that might hide in those blanks (note, not every allele can be hidden, only ones that are the same or recessive to the allele showing!)
Step 6 – Repeat for the next line of the pedigree – but note that the blanks here can ONLY use the alleles that were listed (either in the genotype or as possibles) from the parents! I underline and call ‘likely’ the ones that were in an actual genotype in the above line, and note the rest as possible. Repeat for each line.
So based on the pedigree, Black Magic IS carrying dilute and MAY be carrying either shaded or pointed (note can’t carry both). It is possible that she is carrying chocolate, tort, chinchilla or REW.
How to figure Genotype from offspring
Information about offspring can add a lot of information about the alleles a rabbit is carrying. After all, the offspring had to get those alleles from one of the parents! I’ll use our Jersey wooly buck Frost as an example here. We got Frost before we decided that we were going to breed rabbits – so we didn’t pay extra to get his pedigree. A mistake on our part – ALWAYS get the pedigree unless the rabbit is fixed. You never know.
Step 1 – look up the base genotype on the chart!
Frost is an opal – A-B-C-ddE-
Step 2 – Look up the genotype of the rabbit you bred this rabbit to on the chart.
We bred Frost to Chestnut (who is a chestnut) – A-B-C-D-E-
Step 3 – Add any information from the pedigrees that you can.
We don’t have any additional information on Frost. But we know Chestnut’s sire was a REW, making her A-B-CcD-E-
Step 4 – What colors are the kits? Look up the kits’ base genotypes on the chart.
Chestnut and Frost have two kits together – Cameron is a squirrel – A-B-chd-ddE- and Kristine is a chinchilla A-B-chd-D-E-
Step 5 – organize all this information into a chart!
Step 1 – look up the base genotype on the chart!
Frost is an opal – A-B-C-ddE-
Step 2 – Look up the genotype of the rabbit you bred this rabbit to on the chart.
We bred Frost to Chestnut (who is a chestnut) – A-B-C-D-E-
Step 3 – Add any information from the pedigrees that you can.
We don’t have any additional information on Frost. But we know Chestnut’s sire was a REW, making her A-B-CcD-E-
Step 4 – What colors are the kits? Look up the kits’ base genotypes on the chart.
Chestnut and Frost have two kits together – Cameron is a squirrel – A-B-chd-ddE- and Kristine is a chinchilla A-B-chd-D-E-
Step 5 – organize all this information into a chart!
Step 6 – Look for the double recessives. A double recessive parent must give this allele to his/her offspring. A double recessive offspring must have gotten one allele from EACH parent. Here, Cameron is a dilute so he must have gotten a dilute allele from EACH parent. Since Frost is also a dilute, we know that all his offspring must be dilutes or dilute carriers, therefore Kristine is a dilute carrier. Step 7 – A bit tricky… use your knowledge of dominant and recessive traits and the fact that the alleles have to come one from each parent to fill in the blanks. For example, both of the kits show the chd allele and the only blank on the parents’ C series gene is Frost – so he is Cchd. We now know that the kits got their chd allele from Frost, so they must have gotten their other C series allele from their mother (C or c). The C can’t be hidden, so the kits must both be chdc.
The information we’ve gained from this breeding is that (1) Frost is a chinchilla carrier, (2) both Chestnut and Kristine are dilute carriers and (3) both Cameron and Kristine are REW carriers. This is important information that we now have about Frost and Chestnut that you can’t tell from their pedigrees!
We’ve also bred Frost to a Black Magic (aa) – resulting in self kits (aa), so we know he is a self carrier (because those kits had to get the self allele from BOTH Black Magic and Frost). This increases the probability that Cameron and Kristine are also self carriers, information which doesn’t otherwise appear on their pedigrees! I use that cross below as an example of planning a breeding (before we knew what he was carrying!).
Sometimes information about multiple generations of offspring allow you to backtrack to a grandparent's genotype. Black Magic's son Cooro (also black) went on to produce a (martenized) pointed offspring - Prussia. Given that Prussia's mother (Serenity) had to be carrying either REW or chinchilla (not pointed), we know that the pointed gene had to come from Cooro -- and that Cooro got it from Black Magic.
We’ve also bred Frost to a Black Magic (aa) – resulting in self kits (aa), so we know he is a self carrier (because those kits had to get the self allele from BOTH Black Magic and Frost). This increases the probability that Cameron and Kristine are also self carriers, information which doesn’t otherwise appear on their pedigrees! I use that cross below as an example of planning a breeding (before we knew what he was carrying!).
Sometimes information about multiple generations of offspring allow you to backtrack to a grandparent's genotype. Black Magic's son Cooro (also black) went on to produce a (martenized) pointed offspring - Prussia. Given that Prussia's mother (Serenity) had to be carrying either REW or chinchilla (not pointed), we know that the pointed gene had to come from Cooro -- and that Cooro got it from Black Magic.
Test crossing is a term geneticists use to describe breeding an animal (or plant) with an unknown genotype against one with a known genotype in order to determine the unknown genotype. The best animals for test crosses are those with a completely recessive genotype. For rabbits the absolute best animal to use for test crossing would be a REW that had two lilac torts for parents – which would be genotype aabbccddee (unless you are testing for Vienna carriers, in which case you need something with vv). Test crossing is really good at unmasking those hidden recessives, but if you don’t get kits showing the recessive traits, that doesn’t prove that they aren’t there – it’s always possible that the random flip of a coin gives you all heads!
Planning Breeding from Color Genetics
The number one reason for knowing the color genotype of your rabbits before breeding them is to avoid producing unshowable colors. As most breeders will tell you, body type is far more important than color in planning your breeding – EXCEPT in avoiding those non-showable colors. The second most important reason is to tip the odds in favor of producing colors that you like (or know will sell).
I’m going to use our Jersey woolies Frost and Black Magic as an example to show how to predict kit colors from genotype. We did this cross BEFORE we did the one above between Frost and Chestnut and before I learned very much about color genetics.
Frost is an opal buck A-B-C-ddE-. We don’t have his pedigree, so we don’t know what other traits he is carrying.
Black Magic is a black doe. From her pedigree, we know she is carrying dilute and MAY be carrying either shaded or pointed (but not both). It is possible that she is carrying chocolate, tort, chinchilla or REW. So she has genotype aaB-C-DdE-
A figure called a Punnett square (after its inventor) is used to determine the probabilities of certain offspring for each gene. In a Punnett square, you write the genotype of one parent across the top and the genotype of the other parent vertically on the sides of the square. Some people combine the genes into giant squares, but I find it much simpler to deal with the genes one at a time and then combine the math. I’ll start with the “D” gene in this example because we have the best information there.
I’m going to use our Jersey woolies Frost and Black Magic as an example to show how to predict kit colors from genotype. We did this cross BEFORE we did the one above between Frost and Chestnut and before I learned very much about color genetics.
Frost is an opal buck A-B-C-ddE-. We don’t have his pedigree, so we don’t know what other traits he is carrying.
Black Magic is a black doe. From her pedigree, we know she is carrying dilute and MAY be carrying either shaded or pointed (but not both). It is possible that she is carrying chocolate, tort, chinchilla or REW. So she has genotype aaB-C-DdE-
A figure called a Punnett square (after its inventor) is used to determine the probabilities of certain offspring for each gene. In a Punnett square, you write the genotype of one parent across the top and the genotype of the other parent vertically on the sides of the square. Some people combine the genes into giant squares, but I find it much simpler to deal with the genes one at a time and then combine the math. I’ll start with the “D” gene in this example because we have the best information there.
Combining the B and D series…
Add the “A” series.
So - Chestnut (AaB-Dd) and opal (AaB-dd) are the most likely kits. Black (aaB-Dd) and Blue (aaB-dd) OR Black otter (ataB-Dd) and blue otter (ataB-dd) are likely. Chocolate/lilac/chocolate agouti/lilac agouti/chocolate otter/lilac otter are unlikely, but possible.
Add the “C” series.
Add the “C” series.
This didn’t change the ‘likely’ kits but did add a whole lot of unlikely but possible combinations…
Chestnut (AaB-C-Dd) and opal (AaB-C-dd) are the most likely kits. Black (aaB-C-Dd) and Blue (aaB-C-dd) OR Black otter (ataB-C-Dd) and blue otter (ataB-C-dd) are likely. Chocolate series, lilac series, chinchillas, shadeds, pointeds, REW are all possible but unlikely.
And last but not least, add the E series…
Chestnut (AaB-C-Dd) and opal (AaB-C-dd) are the most likely kits. Black (aaB-C-Dd) and Blue (aaB-C-dd) OR Black otter (ataB-C-Dd) and blue otter (ataB-C-dd) are likely. Chocolate series, lilac series, chinchillas, shadeds, pointeds, REW are all possible but unlikely.
And last but not least, add the E series…
Again, this doesn’t change the ‘likely’ kits, but does add to the list of possibles (the downside of those unknowns!). Chestnut (AaB-C-DdE-) and opal (AaB-C-ddE) are still the most likely kits. Black (aaB-C-DdE-) and Blue (aaB-C-ddE-) OR Black otter (ataB-C-DdE-) and blue otter (ataB-C-ddE-) are likely. We add the red/orange/fawn/creams and the torts and foxes and sallanders to the ever-growing list of possibles.
So what did we actually get? I know you are dying to know!
So what did we actually get? I know you are dying to know!
Hey – they all look like selfs! How did that happen? Chestnut and opal were supposed to be the most likely! Well, probability is like flipping a coin – sometimes you get all heads! (BTW they are all bucks too – 3 selfs from this cross is just as likely as 3 boys out of 3 – should be 50-50) Seeing selfs let us know that Frost was definitely carrying the self allele – so those otter kits we were kind of hoping for weren’t even really possible. And we did get about half dilute.
But wait! Life was going to throw us another curve! Here’s how the kits looked after a couple weeks.
But wait! Life was going to throw us another curve! Here’s how the kits looked after a couple weeks.
What happened to that second blue kit?! He doesn’t look like a blue anymore?! Thanks to another breeding with Frost, we now know Frost IS carrying the chd gene – making the chinchillas, squirrels etc that we had tossed into the possible but unlikely category much more possible. Terrance slowly shifted back to a true blue color, with darker eyes and retaining a little shimmery agouti patch on his nose - we concluded he is a self squirrel (self blue chin). In any case, given what we know now, we won’t be doing this cross again! Rule 2 – chins (including chin carriers like Frost) should be matched with agoutis, not selfs! [Rule 1 is not to cross shadeds into agoutis, which we also likely did here not realizing that Black Magic could be a shaded carrier!]
Working with multiple genes at once...and unknown information!
Classic genetics texts generally have you calculating the various ways the genes from each parent can group together to be passed to the offspring -- and then writing these gene combinations across the top and down the side of a much larger Punnett square. I want to show you what I think is a much easier way to combine multiple small Punnett squares using a simple pie chart...it also makes it much easier to deal with those unknowns...
In our second year of breeding Jersey woolies, Jillian fell in love with tan pattern. So she bought a very nice typed black otter doe (Personality) and bred her to her favorite baby buck, Cameron - a squirrel.
In the exercise above, Cameron was one of the kits we figured out the genotype for -- A-B-chdcddE-. The symbol - means I don't know what his second gene is -- it could be another A (one from each parent is possible), it could be a (which we know his father Frost is hiding) or it could even be at (we don't know for sure what his mom, Chestnut, is hiding). We did a pedigree analysis for Personality and figured out her genotype at-B-C(chd/c)D-Ee. This symbol (chd/c) means I don't know whether her second gene is chd or c - but I know it has to be one of those two genes based on her parents (and it can not be chl or ch).
So genetically speaking, Cameron x Personality = A-B-chdcddE- x at-B-C(chd/c)D-Ee.
Classic genetics has you figure out all the possible combinations of genes that each parent can give - 1 gene from each pair...
Cameron's possible combinations are: ABchddE, ABcdE, -BchddE, -BcdE, ABchdd-, ABcd-, -Bchdd-, -Bcd- A-chddE, A-cdE, --chddE, --cdE, A-chdd-, A-cd-, --chdd-, --cd-
Personality's are: atBCDE, atBCDe, atBC-E, atBC-e, atB(chd/c)DE, atB(chd/c)De, atB(chd/c)-E, atB(chd/c)-e, at-CDE, at-CDe, at-C-E, at-C-e, at-(chd/c)DE, at-(chd/c)De, at-(chd/c)-E, at-(chd/c)-e,
Now, the classic method would have us write all 16 of Frost's possible combinations across the top of a chart and all 16 of Personality's down the side, then copy across and down to fill out 256 squares! Many of which will have lots of dashes which don't tell us much.
Instead, let's take this one gene at a time...
A-gene
In our second year of breeding Jersey woolies, Jillian fell in love with tan pattern. So she bought a very nice typed black otter doe (Personality) and bred her to her favorite baby buck, Cameron - a squirrel.
In the exercise above, Cameron was one of the kits we figured out the genotype for -- A-B-chdcddE-. The symbol - means I don't know what his second gene is -- it could be another A (one from each parent is possible), it could be a (which we know his father Frost is hiding) or it could even be at (we don't know for sure what his mom, Chestnut, is hiding). We did a pedigree analysis for Personality and figured out her genotype at-B-C(chd/c)D-Ee. This symbol (chd/c) means I don't know whether her second gene is chd or c - but I know it has to be one of those two genes based on her parents (and it can not be chl or ch).
So genetically speaking, Cameron x Personality = A-B-chdcddE- x at-B-C(chd/c)D-Ee.
Classic genetics has you figure out all the possible combinations of genes that each parent can give - 1 gene from each pair...
Cameron's possible combinations are: ABchddE, ABcdE, -BchddE, -BcdE, ABchdd-, ABcd-, -Bchdd-, -Bcd- A-chddE, A-cdE, --chddE, --cdE, A-chdd-, A-cd-, --chdd-, --cd-
Personality's are: atBCDE, atBCDe, atBC-E, atBC-e, atB(chd/c)DE, atB(chd/c)De, atB(chd/c)-E, atB(chd/c)-e, at-CDE, at-CDe, at-C-E, at-C-e, at-(chd/c)DE, at-(chd/c)De, at-(chd/c)-E, at-(chd/c)-e,
Now, the classic method would have us write all 16 of Frost's possible combinations across the top of a chart and all 16 of Personality's down the side, then copy across and down to fill out 256 squares! Many of which will have lots of dashes which don't tell us much.
Instead, let's take this one gene at a time...
A-gene
We can see right away that half the kits get an agouti gene from Frost and will be agouti...but how do we deal with the blanks on the other half? Let's try writing in the options instead...
Well, that doesn't have any blanks, but it still looks confusing! What it means is that IF Frost is carrying A, then his A genes will dominate and all the kits will be agouti. If both Frost AND Personality are hiding a, the the upper right box will be ata (25% tan kits) and the lower right will be aa (25% self kits). Otherwise, both the upper and lower right boxes will be at- and those kits will be tan pattern.
But that's a lot of 'if's' what are the actual odds? - AND it still looks really confusing. So let's try something we are all a little more used to interpreting. Let's change a Punnett square into a pie chart! Just drop the parents boxes (we really care only about kits at this point) and round the corners a bit...
But that's a lot of 'if's' what are the actual odds? - AND it still looks really confusing. So let's try something we are all a little more used to interpreting. Let's change a Punnett square into a pie chart! Just drop the parents boxes (we really care only about kits at this point) and round the corners a bit...
Each of the Punnett square boxes is now a quarter of the pie. And each pie piece still has the same gene symbols floating in it. Now we need to figure out what each of those boxes really means in terms of phenotype...box one is easy -- one genotype = agouti.
Box 2 is also pretty easy -- it doesn't matter what gene Personality gives the kit, that agouti gene from Frost is still dominant.
The third box gets a little trickier...Maybe Frost is hiding an agouti gene that will dominate. But if he is hiding at or a, then Personality's at gene will show. In the absence of any other knowledge, let's consider those possibilities equally likely -- and just split the pie piece evenly. Now in reality, Frost hidden gene is either A or it isn't. If we figure that out later, we are going to want to go back and adjust. This method isn't giving us the exact ratio of kit colors (that quarter of the pie is goingto be all agouti if that's what Frost is hiding - but it is telling us how likely we are to get certain colors from this cross - taking into account our current lack of information.
Now for that last quarter...a little trickier as both parents have multiple options... Let's deal with the chance that Frost is hiding agouti first...exactly as above...
Getting there...let's look at that last piece...this is either going to give tan kits (if either or both parents are hiding at) or self (only if BOTH parents are hiding a). Since BOTH parents have to be hiding a to get a self, only one-fourth of this pie-piece is self.
Rearrange the slices a little and - voila!
Please note what this chart does and doesn't say. It does NOT say that if you have 10 kits in your litter you should expect 1 self, 3 tans and 6 agoutis. It DOES say that if you do this cross, not knowing what genes the parents carry, you have a 3% chance of getting some self kits (25% if both parents happen to carry self averaged against the 0% if they don't), a 30% chance of getting tan kits and a 66% chance of getting agoutis (averaging 50% if Frost doesn't carry agouti against 100% if he does). IF you want agoutis and tans, but hate selfs, this tells you this is a pretty safe cross (low odds of getting self).
Now let's add another gene to the mix -- I always like to do the E-gene second, as it works together with the A-gene to define pattern.
Personality is Ee. Cameron is E-. Remember, the dash can stand for any of the E-alleles which are equal to or recessive to E -- so the dash can be E, Ej or e. First build the Punnett square...
Personality is Ee. Cameron is E-. Remember, the dash can stand for any of the E-alleles which are equal to or recessive to E -- so the dash can be E, Ej or e. First build the Punnett square...
You can see that three of the four boxes contain an E. So at least 75% chance of kits being an E- genotype (regular agouti, otter or self). The box on the lower right takes a little more thinking, as Cameron's hidden gene will determine the kit's pattern. Ee is normal, Eje is harlequin, and ee is non-extension. Now, we could just assign those all equal probability...but knowing that harlequin/tri-color is very rare in Jersey woolies, it doesn't seem right to assume we have an equal chance of getting that gene - even though we don't know for sure. Looking back at Cameron's pedigree, we can see that his mother is a chestnut -- and there are no non-extension or harlequin/tri in his background on that side. His sire, on the other hand, was unpedigreed, so a much bigger mystery. So I think it is most likely that Cameron is hiding an E-allele, possible, but less likely that he is hiding e, and unlikely, but just slightly possible that he is hiding Ej. So I'm going to divide the Punnett square into a pie chart like this...
That's roughly an 85% chance of Full extension, 10% chance of non-extension (possible) and a 5% chance of harlequin (not likely, but not impossible). Note that this is the odds given our lack of knowledge -- in actuality Cameron is carrying either E, Ej or e - not all three. So the 85% is an average of the likelihood that he is carrying E (which gives 100% full extension) with the probability he is carrying one of the other genes (which gives only 75% full extension). Likewise, the 5% is an average of the small possibility he is carrying harlequin (which would give 25% chance of harlequin kits) with the probability he is carrying one of the other genes (which gives NO harlequin kits).
To combine the information we have for the A-gene and the E-gene, we have to apply the percentages we found for the E-gene to EACH SLICE of the A-gene pie chart... that is, we have to subdivide the 'agouti slice' (representing A-) into agouti (A-E-), 'red' (A-ee) and harlequin (A-Eje), then subdivide the 'tan slice' (representing at-) into tan (at-E-), torted otters (at-ee), and harlequin (at-Eje), then the 'self slice' (representing aa) into self (aaE-), tort (aaee) and harlequin (aaEje).
If you like math, you can simply multiply the percentages -- 66% agouti (A-) x 85% full extension (E-) = 56% agouti (A-E-). repeat for each combination.
If you don't, you can just divide each slice of the pie chart by eye...
To combine the information we have for the A-gene and the E-gene, we have to apply the percentages we found for the E-gene to EACH SLICE of the A-gene pie chart... that is, we have to subdivide the 'agouti slice' (representing A-) into agouti (A-E-), 'red' (A-ee) and harlequin (A-Eje), then subdivide the 'tan slice' (representing at-) into tan (at-E-), torted otters (at-ee), and harlequin (at-Eje), then the 'self slice' (representing aa) into self (aaE-), tort (aaee) and harlequin (aaEje).
If you like math, you can simply multiply the percentages -- 66% agouti (A-) x 85% full extension (E-) = 56% agouti (A-E-). repeat for each combination.
If you don't, you can just divide each slice of the pie chart by eye...
Let's repeat for the B- gene. Both Cameron and Personality are B-, but neither has any chocolate in their background.
As with E, at least 75% odds favor B- (3 of 4 boxes) and we just need to focus on the lower right corner. This box will be genotype bb ONLY IF BOTH parents are hiding a b gene. Chocolate is fairly rare in Jersey woolies (neither chocolate agouti, nor chocolate otters are accepted) so it seems pretty unlikely (though not impossible) that this box is going to be bb. So I estimate the odds that both parents are carrying chocolate at 5% (that's probably high). 95% B- and 5% bb. Now, I could go ahead and subdivide the pie chart above to include this right now, but I like working with B and D together to get the 4 base colors, so I'm going to do that first, and then get back to the pie chart.
Cameron is squirrel, which is a blue-based color, genotype dd. Personality is black, with blue in her background. She could be either DD or Dd.
Cameron is squirrel, which is a blue-based color, genotype dd. Personality is black, with blue in her background. She could be either DD or Dd.
In the upper two boxes, Personality's D is the dominant gene, so those are not dilute color (black or chocolate). The lower two boxes, color is determined by Personality's hidden gene - either full color or dilute. Given that dilute is a fairly common gene (and Personality even has it in her pedigree), I assign a 50-50 chance to which gene she is hiding - D or d. That causes EACH of the bottom boxes to be split 50-50 between Dd and dd genotypes. Overall, that's 75% chance for Dd (top two boxes plus half of each bottom box) and 25% (half of 2 bottom boxes) dd. Remember, the real odds for kits is either 100% full color (if Personality is hiding D) or 50% (if she is hiding d) - the 75-25 is an average of those two possibilities (given that we don't know which she really carries).
I like to take a quick look at the breakdown of the 4 base colors before combining with the pie chart... in this case 5% of the full color shift to chocolate - 75% x 5% = 4% and 5% of the dilute shift to lilac - 25% x 5% = 1%. So I the odds are 71% black-based, 4% chocolate-based, 24% blue-based and 1% lilac based. If you don't like the math, you can just combine the two pie charts by taking a sliver off each piece of the 75-25 pie.
Now we need to divide each piece of the pattern pie (from the A and E genes) into smaller parts based on the B and D genes. You can do these one-by-one or all at once.
I like to take a quick look at the breakdown of the 4 base colors before combining with the pie chart... in this case 5% of the full color shift to chocolate - 75% x 5% = 4% and 5% of the dilute shift to lilac - 25% x 5% = 1%. So I the odds are 71% black-based, 4% chocolate-based, 24% blue-based and 1% lilac based. If you don't like the math, you can just combine the two pie charts by taking a sliver off each piece of the 75-25 pie.
Now we need to divide each piece of the pattern pie (from the A and E genes) into smaller parts based on the B and D genes. You can do these one-by-one or all at once.
Some of the pie slices are getting really small now -- if they get too small to see, I just mark them as <1%. I also didn't bother to subdivide groups that are completely unshowable (like torted otters) or unshowable in this breed (as reds and harlequins aren't showable in Jersey woolies).
Whew! Just one 'primary' gene left to go - the C-gene. Cameron's genotype is chdc. Personality's first gene is C, and her second is either chd or c. Try that Punnett square on your own...
You should get 2 squares with Personality's C gene as the dominant. You should get 1 square with Cameron's chd gene as the dominant. And you should get one square that depends on Personality's hidden gene - chdcc or cc. So 50% are a full color genotype (C-), 25-50% (average 33%) are chd- and 0-25% (average 12%) are REW (cc).
The C-gene also needs to be added to the pie chart to get the full picture of possible kit colors. This means that each piece of the pie needs to be divided into 3 pieces -- half the piece for the normal color, one-third for the chin-version of the color, and the remaining little slice for REW. Which brings us to a small dilemma -- because REW is epistatic and can hide all the other genes, we are going to have a ton of little tiny slices that all say REW. These should be rearranged to fall next to each other so that we have one 12% slice that just says REW. To be honest, I usually skip ahead, figure out REW first and take that slice out of the pie FIRST before splitting it all up (i.e., before I even do the A gene). Another way to do this is to draw the pie for the C-gene first, and then use the chart above with the other 4 genes to split up the C-gene pie. But that's also hard to do if your pie has as many pieces as this one does. The other way, which is what I'm going to use here, is to just take REW out of the biggest piece, add a little bit to one side, and 'smoosh' all the other colors just bit to accommodate. That works pretty well visually when REW is a small percentage (as here, only 12%).
Whew! Just one 'primary' gene left to go - the C-gene. Cameron's genotype is chdc. Personality's first gene is C, and her second is either chd or c. Try that Punnett square on your own...
You should get 2 squares with Personality's C gene as the dominant. You should get 1 square with Cameron's chd gene as the dominant. And you should get one square that depends on Personality's hidden gene - chdcc or cc. So 50% are a full color genotype (C-), 25-50% (average 33%) are chd- and 0-25% (average 12%) are REW (cc).
The C-gene also needs to be added to the pie chart to get the full picture of possible kit colors. This means that each piece of the pie needs to be divided into 3 pieces -- half the piece for the normal color, one-third for the chin-version of the color, and the remaining little slice for REW. Which brings us to a small dilemma -- because REW is epistatic and can hide all the other genes, we are going to have a ton of little tiny slices that all say REW. These should be rearranged to fall next to each other so that we have one 12% slice that just says REW. To be honest, I usually skip ahead, figure out REW first and take that slice out of the pie FIRST before splitting it all up (i.e., before I even do the A gene). Another way to do this is to draw the pie for the C-gene first, and then use the chart above with the other 4 genes to split up the C-gene pie. But that's also hard to do if your pie has as many pieces as this one does. The other way, which is what I'm going to use here, is to just take REW out of the biggest piece, add a little bit to one side, and 'smoosh' all the other colors just bit to accommodate. That works pretty well visually when REW is a small percentage (as here, only 12%).
Now, obviously in one litter, you aren't going to get all of these colors. Some slices are 'either or' meaning that if you get one color, the other isn't really possible. So think of this as a dart board -- throw a dart for each kit -- you are more likely to have darts land in the big sections, less likely to have darts land in the small sections. But it is entirely possible to have all your darts land in one little section (beating the odds).
Curious what we got? 2 chins, 2 black otters, 1 opal and 1 blue ermine.
Curious what we got? 2 chins, 2 black otters, 1 opal and 1 blue ermine.