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A world of Genetics vs. Real World Experience

7K views 22 replies 16 participants last post by  scott furbeck 
#1 · (Edited)
We all know the stats from the genetics school book on how color is inherited Yellow x Black (yellow carrier) = 50% Black-%50 Yellow, Brown x Black (brown carrier)= 50%-50%. Black(yellow carrier)x Black (yellow carrier) = 75% black, 25% yellow. Simple right? Same with the Chocolates. We'll as I've been watching a few litters this year, toying with the idea of trying a new color, but not yet insane enough to eliminate a black parent completely ;). I've found they are nowhere close to schoolbook. So I thought it might be fun to get some real world stats based on color breeding from personal experience, see if we can get them to balance out to 50%.

So here goes this year so far

Yellow x black ( Y carrier) = 8 black 3 yellow
Chocolate x black (coco carrier) = 5 black 1 chocolate
Black(Y carrier) x black (y-carrier) = 7 black

So far all signs pointing to me staying with black ;)
 
#14 ·
This.
What most people do not understand is that the estimates using punnett squares (and lab color is not a typical Punnett square) are for each pup, not the entire litter.
It is exactly like lottery tickets, no matter how many you buy each ticket has the same odds as the others. Lets say a specific breeding combination has a 75% chance of being black and a 25% chance of being yellow. Each dog from that breeding has a 75% chance of being black and that is different then saying 75% of the litter will be black and 25% will be yellow. Over time and with a lage enough sample size you would see something close to a 75/25 mix, but not necessarily per litter.
Lastly, chromosomal segregation in conception is a crapshoot and completely random. Punnett squares are just mathematic estimates.
 
#3 ·
Plenty of parents out there with 9 girls because they weren't gonna give up until they got a boy (or vice-versa).

There's a reason geneticists work with fruit flies....
 
#4 ·
It is also not a typical Punnett square type situation. The percentages are the likelihood of each puppy in a litter being a particular color.... not a percentage of each litter. It sounds the same but it is definitely not.

Oh, and I think you should go play some money on black in a good game of roulette! ;)
 
#5 ·
I agree to go bet on black...

Two litters we have had of mixed colors
Black (chocolate carrier) X Chocolate 4 chocolate 4 black...better part of that was 6 females and 2 males (also theoretically a 50/50 chance each pup is male or female)
Black (yellow carrier) X Yellow 1 yellow 6 males
Black (chocolate carrier) X Chocolate Due Christmas Day we will see what we get

Two litters same mating we have bought pups out of
Black (yellow carrier) X Black (yellow carrier)
Mating 1
4 yellow and 2 black
Mating 2
6 black
 
#6 · (Edited)
Litter 1 – Black (chocolate factored) X Black (chocolate factored) = 13 pups, 9 black, 4 chocolate
Repeat litter 2 - 6 pups, 5 black, 1 chocolate
Repeat litter 3 - 13 pups, 5 black, 8 chocolate

My conclusion….
Color = Unpredictable
Accepting deposits expecting a large litter = small litter
Turning down deposits expecting small litter = large litter
Receiving more male deposits = more females
Receiving more female deposits = more males
Interest for black = chocolate
Interset for chocolate = black
 
#17 ·
Litter 1 – Black (chocolate factored) X Black (chocolate factored) = 13 pups, 9 black, 4 chocolate
Repeat litter 2 - 6 pups, 5 black, 1 chocolate
Repeat litter 3 - 13 pups, 5 black, 8 chocolate

My conclusion….
Color = Unpredictable
Accepting deposits expecting a large litter = small litter
Turning down deposits expecting small litter = large litter
Receiving more male deposits = more females
Receiving more female deposits = more males
Interest for black = chocolate
Interset for chocolate = black
This is so right. I thought about seeking a grant from the federal govt. to end the drought, all I have to do is schedule a company to come out and wash our windows, guaranteed rain.

John
 
#7 ·
Litter 1) Chocolate stud x Black bitch (chocolate factored), 9 chocolate, 1 black

Liter 2) Black stud (chocolate factored) x Chocolate bitch (from Litter 1) 1 chocolate, 9 black
 
#8 ·
How about this: Same male dog
litter 1: Black Male (Yellow factored) x black female (yellow Factored) = 9 pups all black
litter 2: Black Male (Yellow factored) x black female (yellow Factored) = 9 pups all black
litter 3: Black Male (Yellow factored) x chocolate female (yellow Factored) = 6 pups all black
litter 4: Black Male (Yellow factored) x yellow female (yellow Factored) = 10 pups all black
litter 5: Black Male (Yellow factored) x chocolate female (yellow Factored) = 8 pups all black
litter 6: Black Male (Yellow factored) x black female (yellow Factored) = 11 pups all black, all litters were born in a 6 year span with 6 different females !
 
#9 ·
Was the male tested for the yellow gene?
 
#10 · (Edited)
Hmmm I'm envisioning a new retriever game, guessing litter color, so far it's appearing Black is the best bet, then a Yellow, then chocolate. If we give odds on this that 9 chocolate:1 black would be quite a pay-out;)

I also have experience with a black male tested yellow carrier, shoot he was out of the yellow female, only ever produced one still born yellow in multiple breedings to different yellow females. I guess the black ones just swam faster ;)

Anyone dare to put odds on a tri-factor breeding?
 
#12 ·
My black trifactored male has sired four litters.......
All to yellow females.
Seven yellow three black
Seven yellow four black
Six yellow, one chocolate two black
Six yellow four black.....
 
#16 ·
Hmm not a statt-i-tion, but flip a quarter 5 times get heads, even though it's 50-50 every-time seems like tails becomes more likely to come up unless somethings wrong with the quarter.
 
#18 ·
You can predict the probability of getting say 5 Heads in a row or 5 Tails in a row, even though the probability of an H or T on a flip is independent from the binomial. Here's an example...I hope I did make any boboos...

Flip a coin, 50% chance of H, 50% chance of tails (assuming fair coin).

Flip the coin 2 times, possible outcomes:

HH, probability of (.5)(.5)=.25
TT, probability of (.5)(.5)=.25
HT, probability of (.5)(.5)=.25
TH, probability of (.5)(.5)=.25

By analogy, if it is a litter of 2 puppies with 50% chance of black vs. yellow, 25% of the time you will get all yellows; 25% of the time it is all black; 50% of the time it is 1 yellow, 1 black.

Say you flip the coin 3 times, possible outcomes:

HHH, probability (.5)(.5)(.5)=.125/ all yellows
HHT, (.5)(.5)(.5)=.125 /2 Y, 1 B
HTT, (.5)(.5)(.5)=.125 /1 Y, 2 B
HTH, (.5)(.5)(.5)=.125 /2 Y, 1 B
HHT, (.5)(.5)(.5)=.125/ 2 Y, 1 B
TTT, (.5)(.5)(.5)=.125 /all blacks
TTH, (.5)(.5)(.5)=.125 / 2 B, 1 Y
THT, (.5)(.5)(.5)=.125 / 2 B, 1 Y
THH, (.5)(.5)(.5)=.125 / 1 B, 2 Y

So 12.5% chance litter of 3 pups is all yellow; 12.5% chance all black; 37.5% chance 2 Yellow one black; 37.5% chance 2 blacks one yellow.
 
#20 ·
Another way to ask the question would be, how many puppies would need to be born to be 90% sure you would observe one homozygous bb from a Bb x Bb.

I believe the number is 6. It depends on the distribution you are using to make your assumptions. You can google up Fischer's exact test to find more about this stuff.
 
#22 ·
Another way to ask the question would be, how many puppies would need to be born to be 90% sure you would observe one homozygous bb from a Bb x Bb.

I believe the number is 6. It depends on the distribution you are using to make your assumptions. You can google up Fischer's exact test for a 2 x 2 contingency table to find more about this stuff (and actually calculate the probability of observing any individual litter combination).
 
#23 ·
Another way to ask the question would be, how many puppies would need to be born to be 90% sure you would observe one homozygous bb from a Bb x Bb.

I believe the number is 6. It depends on the distribution you are using to make your assumptions. You can google up Fischer's exact test for a 2 x 2 contingency table to find more about this stuff (and actually calculate the probability of observing any individual litter combination).

http://www.danielsoper.com/statcalc3/calc.aspx?id=29
 
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