# Easy genetics corn



## pants125 (Jan 30, 2009)

Ok I'm still getting my head round genetics 
I got the basic idea for het to het,albino to het ect 
And I'm sure this will be easy too but could someone show me how to work it out please using the punnet square 
I got a anery het amel and I breed it to normal het amel 
I know they both must be het amel as the babies have come out normal and amel 
It's the anery het amel that's confusing me 
Ta


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## paulh (Sep 19, 2007)

Normal looking het amelanistic x anerythristic het amelanistic

Assign symbols:
An = normal gene
an = anerythristic mutant gene


A = normal gene
a = amelanistic mutant gene


This problem involves two gene pairs in the anerythristic het amelanistic snake:
gene pair #1 = two anerythristic mutant genes = an/an
gene pair #2 = a normal gene and an amelanistic mutant gene = A/a

Every gamete (sperm or egg) gets one gene from each gene pair. As gene pair #1 has only anerythristic mutant genes, every gamete has one anerythristic mutant gene. Gene pair #2 has a normal gene and an amelanistic gene. Half the gametes gets the normal gene, and half get the amelanistic gene. Therefore, half the gametes have the genes an A, and the other half of the gametes have the genes an a.

We need the contents of the two matching gene pairs in the het amelanistic snake:
gene pair #1 = two normal genes = An/An
gene pair #2 = a normal gene and an amelanistic mutant gene A/a

As gene pair #1 has only normal genes, every gamete has one normal gene. Gene pair #2 has a normal gene and an amelanistic gene. Half the gametes gets the normal gene, and half get the amelanistic gene. Therefore, half the gametes have the genes An A, and the other half of the gametes have the genes An a.

Put one set of gametes across the top, and the other set of gametes down the left side:

_____|_an A__|__an a
-----|-------|--------
An A_|_______|
-----|-------|--------
An a_|_______|
-----|-------|--------

Fill in the four blanks in the Punnett square, add identical genotypes together, assign phenotypes, and you are done.

Clear as mud?


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## pants125 (Jan 30, 2009)

Cheers I will study it when I get back from work


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## pants125 (Jan 30, 2009)

Ok I think I've done it 
Will punnet look something like this 
AnanAA AnanAa
AnanaA. Ananaa 

Meaning 
Normal het anery Normal het anery amel
Normal het anery amel. Amel het anery 

Is that correct


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## pants125 (Jan 30, 2009)

I have just noticed there are also some anery in the clutch so the normal must be het amel and het anery 
How would this look please

It's ok I have figured it all out only took me 5 hrs of non stop reading and practising lol


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## REDDEV1L (Nov 27, 2008)

It was all foreign to me, until I started messing with the corn calculator.


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## pants125 (Jan 30, 2009)

I like the working out side of it tho I think I got it now aswell


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## paulh (Sep 19, 2007)

pants125 said:


> Ok I think I've done it
> Will punnet look something like this
> AnanAA AnanAa
> AnanaA. Ananaa
> ...


The Punnett square result is correct. But you still need to do a few things to get the final result.

Readability -- A genotype like Ananaa is hard to read. And that is a relatively simple one. It is easier to read if the gene pairs are separated by a space and the two genes in a gene pair are separated by a "/" character or by "//" characters. Instead of Ananaa, I would use An/an a/a.

Add identical genotypes together and insert probabilities -- you got four genotypes in the Punnett square
AnanAA = 1/4 An/an A/A.
AnanAa = 1/4 An/an A/a.
AnanaA = 1/4 An/an a/A.
Ananaa = 1/4 An/an a/a.

An/an A/a = An/an a/A. These two gene orders are the same. It does not matter which gene is on the left and which is on the right in a gene pair. The convention is the more dominant gene on the left and the more recessive gene on the right (A/a rather than a/A). But that is only a human convention which means nothing to the cell containing the genes.

Adding genotypes together gives
1/4 An/an A/A.
2/4 An/an A/a.
1/4 An/an a/a.

We assign phenotypes to each genotype (--> means produces)
1/4 An/an A/A. --> 1/4 normal looking
2/4 An/an A/a. --> 2/4 normal looking 
1/4 An/an a/a. --> 1/4 amelanistic

Add like phenotypes together gives
3/4 normal looking
1/4 amelanistic

We know all the babies are het anerythristic, and the Punnett square tells us that the normal looking babies have a 66% (fraction rounded down) probability of being het amelanistic. So the final result is 
3/4 normal looking, het anerythristic, 66% probability het amelanistic
1/4 amelanistic, het anerythristic. 

Congratulations on the anerythristic babies. You are right; the normal looking parent is both het amelanistic and het anerythristic An/an A/a. The gametes are An A, An a, an A, and an a. The An/an A/a x an/an A/a mating would produce a 2 x 4 box Punnett rectangle.

The Punnett square is a good learning tool, but it takes too much time for more than a two gene pair mating. You ought to learn to use the branching system, which takes less than half the time. See posts 63, 66, and 67 at http://www.reptileforums.co.uk/forums/genetics/2-learning-genetics-7.html


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## pants125 (Jan 30, 2009)

Cheers for that I'm gonna master this first then I will look at other methods


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## pants125 (Jan 30, 2009)

20 eggs hatched I got amel anery and normal but no snow corn 
Thought I would get at least 1


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## paulh (Sep 19, 2007)

Congratulations on the hatching!

Probability of an egg hatching out a non-snow is 7/8. Probability of 20 eggs in a row of hatching out non-snows is (7/8)^20 = 0.069. That's about 1/15. So it is somewhat disappointing but is by no means unheard of. If you keep trying, one will come along.


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## pants125 (Jan 30, 2009)

Another strange hatch result Anery x Anery 
Got 11 Anery and 5 snow 
No amel 
And I know for a fact I haven't mixed clutches as they was all stuck together


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## bothrops (Jan 7, 2007)

pants125 said:


> Another strange hatch result Anery x Anery
> Got 11 Anery and 5 snow
> No amel
> And I know for a fact I haven't mixed clutches as they was all stuck together



This is actually a very explainable result!


Both animals are het amel.

The expected result here is 

75% anery (poss het amel)
25% snow


Note that you can't get amel from this pairing as ALL offspring will be at least anery as both parents have two copies of the anery gene to give and no copies of the normal gene at that locus. All offspring can only therefore receive anery.

The snow's are the anerys that also received two copies of the amel gene (one from each parent).

With a litter of sixteen, your expected results are 12 anery 4 snow. Therefore a result of 11 anery and 5 snow is absolutely expected!


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## pants125 (Jan 30, 2009)

Ahh ya just did a punnet square to make a bit more sense of what you was saying ( I understand things better when I write it down ) 
And I get it now I would have 1/4 Anery 2/4 Anery het amel and 1/4 snow 
I just assumed there would be amels with there being snows but I see where I went wrong cheers ( for now)


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