You know you have problems when...
Tonight I realized something about myself that the rest of the world probably noticed years ago. I am not normal. I came to this conclusion when I was studying for blood banking class. I was looking at some of the notes and noticed that in African Americans, B is A LOT more common than it is in whites. However, I would think that this would make AB also more common, but it isn't. This didn't make sense to me. So, I did something very nerdy. I calculated Hardy-Weinberg.
A little bit of an explanation. ABs come about when an A and a B have a baby and the baby gets both an A and a B. Since only 11% of whites are B, and most As and Bs are actually AO and BOs, ABs are really, really, rare. Only 4% of the population. Even when an A and a B have a baby together, only 25% of their children even come out AB. But, in African Americans, the B frequency jumps to 20%. You would think that this would result in more ABs, but it doesn't. I was baffled by this during class, and I was baffled again while studying. The population genetics just didn't work out.
Luckily, there is an equation for this. Theoretically, the frequency of a gene should be predictable by the frequency of the phenotype. This is assuming Hardy-Weinberg equilibrium, which assumes large population, no genetic drift, no natural selection, random mating, no mutation, and no migration. Since we're talking all of an entire race, and a large race at that, number one is met. We haven't been able to find any kind of natural selection in blood groups, meaning that no blood group is any more advantageous than another. ABO doesn't tend to be a mutation-heavy gene. There doesn't tend to be a lot of migration (meaning, they tend to marry within their race). Theoretically, we should have Hardy-Weinberg equilibrium.
But why am I calculating this? I don't know. That's why I'm wondering about myself. By the way, for African Americans, the Hardy-Weinberg equation equals 0.975. It should equal 1. Maybe I wasn't as crazy as I thought. It is out of equilibrium.
A little bit of an explanation. ABs come about when an A and a B have a baby and the baby gets both an A and a B. Since only 11% of whites are B, and most As and Bs are actually AO and BOs, ABs are really, really, rare. Only 4% of the population. Even when an A and a B have a baby together, only 25% of their children even come out AB. But, in African Americans, the B frequency jumps to 20%. You would think that this would result in more ABs, but it doesn't. I was baffled by this during class, and I was baffled again while studying. The population genetics just didn't work out.
Luckily, there is an equation for this. Theoretically, the frequency of a gene should be predictable by the frequency of the phenotype. This is assuming Hardy-Weinberg equilibrium, which assumes large population, no genetic drift, no natural selection, random mating, no mutation, and no migration. Since we're talking all of an entire race, and a large race at that, number one is met. We haven't been able to find any kind of natural selection in blood groups, meaning that no blood group is any more advantageous than another. ABO doesn't tend to be a mutation-heavy gene. There doesn't tend to be a lot of migration (meaning, they tend to marry within their race). Theoretically, we should have Hardy-Weinberg equilibrium.
But why am I calculating this? I don't know. That's why I'm wondering about myself. By the way, for African Americans, the Hardy-Weinberg equation equals 0.975. It should equal 1. Maybe I wasn't as crazy as I thought. It is out of equilibrium.
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