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Genetics Info & Tutorial: Is It Het?
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In the previous example, we ended up with “possible het” animals. But what if you really need to know which ones are the hets and which ones are not? How can you tell which of these really is het?

The answer lies in a few rules that work with simple recessive mutations. You now know that each parent gives one of its genes to each baby, and each baby gets half of its genes from each parent. You also know that in order for an animal to display a recessive trait (such as amelanism) it must have both copies of the mutated gene (it must be “aa”). If either one of the parents is “AA” then the baby will always get that one “A” from that parent, and no babies from that parent will ever display the mutation.

If you add these things together, it means that in order for any animal to display a recessive trait (meaning it is “aa”), both parents must be carrying at least one copy of that mutation. In other words, each parent must have one of those mutant “a” genes to give to the baby.

Knowing this, you can determine whether an animal is carrying a mutant gene by mating it to another animal that you know is carrying that same (or a compatible) mutation.

The easiest way to do this is to mate the animal in question to a mutant. The mutant will be “aa” as you recall. Knowing what that one parent is, what gene will the babies get from it? The “a” gene, of course.


a
A
?
In this simple punnett square, the mutant parent will always contribute a mutant gene. The unknown parent can contribute one “wild type” gene, which we know because it looks “normal.” The other gene from that parent is unknown.
---------->
a
A
Aa
?
a?
The results in the offspring will tell us whether that other “?” gene is “A” or “a.” Half of the babies will always get “Aa.” The other half, depending on the unknown, will be either “Aa” or “aa.”

There are two possibilities of what the other “unknown” parent might be. We already know that it’s not “aa” or else it would display the trait. The other two remaining combinations are (het for amel) “Aa” and (homozygous normal) “AA.”

If the animal in question is “AA” then what will the babies get? They will get one “a” from the mutant parent and one “A” wild-type gene from the other parent. So all of the babies will be “het.”

a
A
Aa
If the other parent is not carrying the mutation, all offspring will be normal-looking.


If, however, the animal in question is actually het, it will be “Aa” and will give the “A” gene to half the babies and the “a” mutation to the other half. Remember, these babies are all going to get the “a” mutation from the one parent, so the half that gets the “a” gene from our “unknown” parent will end up with “aa” and will display the mutation. When this happens, the parent is no longer unknown, because we will then know that this parent is a carrier!

a
A
Aa
a
aa
<--- These will be normal-looking.

<--- These will show the mutation.


So by mating an animal to another that is known to carry the mutation in question, you can determine if this “unknown” is carrying that mutation or not by the offspring they create. If any mutants of that kind appear, it is then known to be a carrier. If no mutants appear, it is probably not a carrier, depending on how many offspring were produced (and it would no longer be right to call it a “possible het”)

Be aware that these outcomes are probabilities, so if you only have one offspring and it’s not a mutant, then you really haven’t determined anything, because there was only a 50/50 chance of producing a mutant anyway. (which would be like proving that a coin doesn’t have “tails” because I flipped it and it came up heads.) However, if you have 7 offpsring, all non-mutants, the odds of the parent being a carrier are less than 1%. If you have 16 offspring and none of them are mutants, the odds that this parent is carrying the mutation are one in several billion, so you can safely say you’re sure. :)



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