Advanced Genetics

[Macage]

I have learned my punnett square for basic genetics like pastel x spider ect, but I was wondering about more advance punnett squares like pastave x queenbee and the same withe recessives x doms ect. Could anyone show me or give me a video to watch on it?

[John1982]

It's the same, just add an extra row or column of squares for each gene you add.

[Ba11er]

I just use world of ball pythons website. They have pictures of most all the morphs, and a genetic calculator to do punnett squars for you. It's a great tool I use all the time and highly recommend checking it out. If your looking for more on how to do your own squares I imagine youtube has some videos or someone here can turn you in the right direction.

http://www.worldofballpythons.com/wizard/

[OhhWatALoser]

I wouldn't bother with "advanced" punnett squares as they take forever to do. Forked Line method only requires you to know your basic punnett square (aka monohybrid) and is far faster to do. It is the method the genetic calc uses. recessive co-dom and dom are just labels we put on how genes express themselves. genes only come in heterozygous and homozygous pairs, they don't care what you label them and all work the same.

For the sake of my typing WT = Wild Trait aka Normal, Het = Heterozygous, Homo = Homozygous
To use the forked line you need to know your basics:
WT x WT = 1/1 WT
Het x WT = 1/2 Het 1/2 WT
Homo x WT = 1/1 Het
Het x Het = 1/4 Homo 1/2 Het 1/4 WT
Homo x Het = 1/2 Homo 1/2 Het
Homo x Homo = 1/1 Homo

from there you basically just multiply each gene probability. without me spending too much time explaining it, check out this video and ask any question you might have.

[paulh]

Question: Why do all the example genetics problems use recessive mutant genes, like albino? Why not dominant or codominant mutant genes? For example, what is the result of mating a pinstripe ball python to another pinstripe ball python?

Answer: Most example genetics problems use recessive mutant genes because most mutant genes are recessive to the corresponding normal (AKA wild type) gene. And the method for solving a problem with a codominant or dominant mutant gene is the same as for a problem with a recessive mutant gene. If someone can do a problem with a recessive mutant gene, that person can also do a problem with a dominant or codominant mutant gene.

The standard steps in solving a genetics problem with a Punnett square are as follows:

1. Determine the genes in the father's gene pair(s) and in the mother's gene pair(s).

2. Assign a symbol to each gene.

3. Determine the gene(s) in all possible sperm and in all possible eggs.

4. Draw the Punnett square and fill the boxes.

5. If any two or more boxes in the square hold identical genotypes, add them together.

6. Label each genotype with its phenotype.

7. Add identical phenotypes together.

Let's do an example using a recessive mutant gene (albino) and a dominant normal gene:

1. Determine the genes in the father's gene pair(s) and in the mother's gene pair(s).
Father is het albino. (He has a gene pair made up of a normal gene and an albino mutant gene). Mother is also het albino.

2. Assign a symbol to each gene.
A = normal gene
a = albino mutant gene

Father has the A/a genotype, and mother has the A/a genotype

3. Determine the gene(s) in all possible sperm and in all possible eggs.
Half of the sperm have the A gene, and half of the sperm have the a gene.
Half of the eggs have the A gene, and half of the eggs have the a gene.

4. Draw the Punnett square.

___|__A___|__a____
_A_|_A/A__|_A/a___
_a_|_a/A__|_a/a___


5. If any two or more boxes in the square hold identical genotypes, add them together.
The order of the genes in a gene pair does not matter. The A/a genotype and the a/A genotype are identical. Adding those two together produces
1/4 A/A
2/4 A/a
1/4 a/a

6. Label each genotype with its phenotype.
1/4 A/A -- normal phenotype
2/4 A/a -- normal phenotype (same as the A/A snake's phenotype)
1/4 a/a -- albino phenotype.

7. Add identical phenotypes together.
Both the A/A and A/a genotypes produce the same phenotypes. This produces
3/4 normal phenotype
1/4 albino phenotype


Next, let's do an example using a dominant mutant gene (pinstripe) and a recessive normal gene. Note that the phenotype results are the mirror image of the results of the same mating with a recessive mutant gene:

1. Determine the genes in the father's gene pair(s) and in the mother's gene pair(s).
Father has a gene pair made up of a normal gene and a pinstripe mutant gene. Mother has a gene pair made up of a normal gene and a pinstripe mutant gene.

2. Assign a symbol to each gene.
A = pinstripe mutant gene
a = normal gene

Father has the A/a genotype, and mother has the A/a genotype

3. Determine the gene(s) in all possible sperm and in all possible eggs.
Half of the sperm have the A gene, and half of the sperm have the a gene.
Half of the eggs have the A gene, and half of the eggs have the a gene.

4. Draw the Punnett square.

___|__A___|__a____
_A_|_A/A__|_A/a___
_a_|_a/A__|_a/a___


5. If any two or more boxes in the square hold identical genotypes, add them together.
The order of the genes in a gene pair does not matter. The A/a genotype and the a/A genotype are identical. Adding those two together produces
1/4 A/A
2/4 A/a
1/4 a/a

6. Label each genotype with its phenotype.
1/4 A/A -- pinstripe phenotype
2/4 A/a -- pinstripe phenotype (same as the A/A snake's phenotype)
1/4 a/a -- normal phenotype.

7. Add identical phenotypes together.
Both the A/A and A/a genotypes produce the same phenotypes. This produces
3/4 pinstripe phenotype
1/4 normal phenotype


Finally, let's do the same sort of mating using a codominant mutant gene (mojave) and a codominant normal gene:

1. Determine the genes in the father's gene pair(s) and in the mother's gene pair(s).
Father has a gene pair made up of a normal gene and a mojave mutant gene. Mother has a gene pair made up of a normal gene and a mojave mutant gene.

2. Assign a symbol to each gene.
A = mojave mutant gene
a = normal gene

Father has the A/a genotype, and mother has the A/a genotype

3. Determine the gene(s) in all possible sperm and in all possible eggs.
Half of the sperm have the A gene, and half of the sperm have the a gene.
Half of the eggs have the A gene, and half of the eggs have the a gene.

4. Draw the Punnett square.

___|__A___|__a____
_A_|_A/A__|_A/a___
_a_|_a/A__|_a/a___


5. If any two or more boxes in the square hold identical genotypes, add them together.
The order of the genes in a gene pair does not matter. The A/a genotype and the a/A genotype are identical. Adding those two together produces
1/4 A/A
2/4 A/a
1/4 a/a

6. Label each genotype with its phenotype.
1/4 A/A -- super mojave phenotype (blue-eyed white)
2/4 A/a -- mojave phenotype (not the normal phenotype and not the same as the A/A snake's phenotype)
1/4 a/a -- normal phenotype.

7. Add identical phenotypes together.
Each genotype has its own phenotype. There is no change from the list in step 6.




StepsSteps to determine genotypes and phenotypes using a tree (AKA forkline, branching system, arithmetic method, etc.):

1. Determine the genes in the father's gene pair(s) and in the mother's gene pair(s).

2. Assign a symbol to each gene in the first gene pair. Repeat for each additional gene pair.

3. Determine the gene in each of the possible sperm and in each of the possible eggs for one gene pair.

4. Draw the Punnett square for that gene pair.

5. If any two or more boxes in the square are identical, add them together.

6. Label each genotype with its phenotype.

7. Add identical phenotypes together.

8. Repeat steps 3 through 7 for each additional gene pair.

9. List the genotypes from all the gene pairs, using the genotype tree diagram. Multiply the fractions while going out each branch to get the final fraction.

10. List the phenotypes from all the gene pairs, using the phenotype tree diagram. Multiply the fractions while going out each branch to get the final fraction.

[paulh]

Pastave x queenbee is a three gene pair genetics problem.

1. Determine the genes in the father's gene pair(s) and in the mother's gene pair(s).

Pastave gene pairs:
gene pair 1 = mojave mutant gene and normal gene
gene pair 2 = pastel mutant gene and normal gene
gene pair 3 = two normal genes

Queenbee gene pairs:
gene pair 1 = two normal genes
gene pair 2 = two pastel mutant genes
gene pair 3 = spider mutant gene and normal gene


2. Assign a symbol to each gene in the first gene pair. Repeat for each additional gene pair.

M = Mojave mutant gene
m = normal gene

P = pastel mutant gene
p = normal gene

S = spider mutant gene
s = normal gene

Pastave genotype = M/m P/p s/s
Queenbee genotype = m/m P/P S/s

3. Determine the gene in each of the possible sperm and in each of the possible eggs for one gene pair.

4. Draw the Punnett square for that gene pair.

M/m x m/m --> 1/2 M/m, 1/2 m/m
P/p x P/P --> 1/2 P/P, 1/2 P/p
s/s x S/s --> 1/2 s/s, 1/2 S/s

5. If any two or more boxes in the square are identical, add them together.

No extra additions needed.

6. Label each genotype with its phenotype.

M/m produces mojave phenotype.
m/m produces normal phenotype.

P/P produces super pastel phenotype.
P/p produces pastel phenotype.

S/s produces spider phenotype.
s/s produces normal phenotype.

7. Add identical phenotypes together.

No extra additions needed

8. Repeat steps 3 through 7 for each additional gene pair.

Done

9. List the genotypes from all the gene pairs, using the genotype tree diagram. Multiply the fractions while going out each branch to get the final fraction.

The eight branches in the tree:
1/2 M/m -- 1/2 P/P -- 1/2 S/s = 1/8 M/m P/P S/s
1/2 M/m -- 1/2 P/P -- 1/2 s/s = 1/8 M/m P/P s/s
1/2 M/m -- 1/2 P/p -- 1/2 S/s = 1/8 M/m P/p S/s
1/2 M/m -- 1/2 P/p -- 1/2 s/s = 1/8 M/m P/p s/s
1/2 m/m -- 1/2 P/P -- 1/2 S/s = 1/8 m/m P/P S/s
1/2 m/m -- 1/2 P/P -- 1/2 s/s = 1/8 m/m P/P s/s
1/2 m/m -- 1/2 P/p -- 1/2 S/s = 1/8 m/m P/p S/s
1/2 m/m -- 1/2 P/p -- 1/2 s/s = 1/8 m/m P/p s/s

10. List the phenotypes from all the gene pairs, using the phenotype tree diagram. Multiply the fractions while going out each branch to get the final fraction.

All mutant genes in this problem are codominant to their corresponding normal genes.

The eight branches in the tree:
1/2 M/m -- 1/2 P/P -- 1/2 S/s = 1/8 M/m P/P S/s (1/8 mojave super pastel spider phenotype)
1/2 M/m -- 1/2 P/P -- 1/2 s/s = 1/8 M/m P/P s/s (1/8 mojave super pastel phenotype)
1/2 M/m -- 1/2 P/p -- 1/2 S/s = 1/8 M/m P/p S/s (1/8 mojave pastel spider phenotype)
1/2 M/m -- 1/2 P/p -- 1/2 s/s = 1/8 M/m P/p s/s (1/8 mojave pastel phenotype)
1/2 m/m -- 1/2 P/P -- 1/2 S/s = 1/8 m/m P/P S/s (1/8 super pastel spider phenotype)
1/2 m/m -- 1/2 P/P -- 1/2 s/s = 1/8 m/m P/P s/s (1/8 super pastel phenotype)
1/2 m/m -- 1/2 P/p -- 1/2 S/s = 1/8 m/m P/p S/s (1/8 pastel spider phenotype)
1/2 m/m -- 1/2 P/p -- 1/2 s/s = 1/8 m/m P/p s/s (1/8 pastel phenotype)

[paulh]

I got a PM that I had mixed up queen bee and killer bee in my example (post 6 in this thread). (Thanks, Dennis.) He's right. A queen bee is a lesser pastel spider, while a killer bee is a super pastel spider. So I will try again, with the real queen bee genotype this time. The method is the same though.

Pastave x queenbee is a three gene pair genetics problem.

1. Determine the genes in the father's gene pair(s) and in the mother's gene pair(s).
Pastave gene pairs:
gene pair 1 = mojave mutant gene and normal gene
gene pair 2 = pastel mutant gene and normal gene
gene pair 3 = two normal genes

Queenbee gene pairs:
gene pair 1 = lesser mutant gene and normal gene
gene pair 2 = pastel mutant gene and normal gene
gene pair 3 = spider mutant gene and normal gene

2. Assign a symbol to each gene in the first gene pair. Repeat for each additional gene pair.

M = Mojave mutant gene
M^l = M with lower case L as a super script = Lesser mutant gene
m = normal gene

P = pastel mutant gene
p = normal gene

S = spider mutant gene
s = normal gene

Pastave genotype = M/m P/p s/s
Queenbee genotype = M^l/m P/p S/s

3. Determine the gene in each of the possible sperm and in each of the possible eggs for one gene pair.

4. Draw the Punnett square for that gene pair.

M/m x M^l/m --> 1/4 M/M^l, 1/4 M/m, 1/4 M^l/m, 1/4 m/m
P/p x P/p --> 1/4 P/P, 1/4 P/p, 1/4 p/P, 1/4 p/p
s/s x S/s --> 1/2 s/s, 1/2 S/s

5. If any two or more boxes in the square are identical, add them together.

P/p x P/p --> 1/4 P/P, 1/4 P/p, 1/4 p/P, 1/4 p/p
P/p and p/P genotypes are identical. Therefore,
P/p x P/p --> 1/4 P/P, 2/4 P/p, 1/4 p/p


6. Label each genotype with its phenotype.

M/M^l produces blue-eyed white phenotype
M^l/m produces lesser phenotype
M/m produces mojave phenotype.
m/m produces normal phenotype.

P/P produces super pastel phenotype.
P/p produces pastel phenotype.
p/p produces normal phenotype

S/s produces spider phenotype.
s/s produces normal phenotype.

7. Add identical phenotypes together.

See step 10.

8. Repeat steps 3 through 7 for each additional gene pair.

Done

9. List the genotypes from all the gene pairs, using the genotype tree diagram. Multiply the fractions while going out each branch to get the final fraction.

1/4 M/M^l--1/4 P/P-- 1/2 S/s = 1/32 M/M^l P/P S/s (blue-eyed white, possible spider wobble)
1/4 M/M^l--1/4 P/P-- 1/2 s/s = 1/32 M/M^l P/P s/s (blue-eyed white)
1/4 M/M^l--2/4 P/p-- 1/2 S/s = 2/32 M/M^l P/p S/s (blue-eyed white, possible spider wobble)
1/4 M/M^l--2/4 P/p-- 1/2 s/s = 2/32 M/M^l P/p s/s (blue-eyed white)
1/4 M/M^l--1/4 p/p-- 1/2 S/s = 1/32 M/M^l p/p S/s (blue-eyed white, possible spider wobble)
1/4 M/M^l--1/4 p/p-- 1/2 s/s = 1/32 M/M^l p/p s/s (blue-eyed white)

1/4 M^l/m--1/4 P/P-- 1/2 S/s = 1/32 M^l/m P/P S/s (lesser, super pastel, spider)
1/4 M^l/m--1/4 P/P-- 1/2 s/s = 1/32 M^l/m P/P s/s (lesser, super pastel)
1/4 M^l/m--2/4 P/p-- 1/2 S/s = 2/32 M^l/m P/p S/s (lesser, pastel, spider)
1/4 M^l/m--2/4 P/p-- 1/2 s/s = 2/32 M^l/m P/p s/s (lesser, pastel)
1/4 M^l/m--1/4 p/p-- 1/2 S/s = 1/32 M^l/m p/p S/s (lesser, spider)
1/4 M^l/m--1/4 p/p-- 1/2 s/s = 1/32 M^l/m p/p s/s (lesser)

1/4 M/m--1/4 P/P-- 1/2 S/s = 1/32 M/m P/P S/s (mojave, super pastel, spider)
1/4 M/m--1/4 P/P-- 1/2 s/s = 1/32 M/m P/P s/s (mojave, super pastel)
1/4 M/m--2/4 P/p-- 1/2 S/s = 2/32 M/m P/p S/s (mojave, pastel, spider)
1/4 M/m--2/4 P/p-- 1/2 s/s = 2/32 M/m P/p s/s (mojave, pastel)
1/4 M/m--1/4 p/p-- 1/2 S/s = 1/32 M/m p/p S/s (mojave, spider)
1/4 M/m--1/4 p/p-- 1/2 s/s = 1/32 M/m p/p s/s (mojave)

1/4 m/m--1/4 P/P-- 1/2 S/s = 1/32 m/m P/P S/s (super pastel, spider)
1/4 m/m--1/4 P/P-- 1/2 s/s = 1/32 m/m P/P s/s (super pastel)
1/4 m/m--2/4 P/p-- 1/2 S/s = 2/32 m/m P/p S/s (pastel, spider)
1/4 m/m--2/4 P/p-- 1/2 s/s = 2/32 m/m P/p s/s (pastel)
1/4 m/m--1/4 p/p-- 1/2 S/s = 1/32 m/m p/p S/s (spider)
1/4 m/m--1/4 p/p-- 1/2 s/s = 1/32 m/m p/p s/s (normal phenotype)

10. List the phenotypes from all the gene pairs, using the phenotype tree diagram. Multiply the fractions while going out each branch to get the final fraction.

Done above. The blue-eyed white phenotype masks the effect of the pastel gene (as far as I know). The blue-eyed white phenotype also masks the pattern effect of the spider gene, but not the wobble. Of course, some spiders may not show the wobble immediately. The 8/32 blue-eyed whites can be added together as 8/32 (= 1/4) blue-eyed white, with a 50% probability of also being spider.