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Lecture 4

BIOL 112 Lecture Notes - Lecture 4: Thomas Hunt Morgan, Homologous Chromosome, Y Chromosome


Department
Biology
Course Code
BIOL 112
Professor
Joseph Dent
Lecture
4

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Mendelian Genetics: Linkage and Continuous Variation
To review, Mendel proposed a simple theory of exquisite predictive power that explains
the rules governing heredity in terms of chromosomes and genes. The problem is, Mendel
still has some explaining to do. First of all, our theory has explained the exception,
discrete traits, not the rule, continuous traits. Second, in order for all genes to segregate,
they must all be on different chromosomes. But there are only 23 pairs of human
homologs. Does that mean there are only 23 genes?
Turns out there are several traits that run in families but predominantly affect males.
These are called sex-linked traits. Let's look closely at one example: color blindness.
Consider; a normal woman and a normal man who produce some color blind children.
What will the ratios be? Colorblindness must be recessive since it turns up in children of
unaffected parents (think F1 x F1 cross). Both parents must be heterozygous or you
wouldn't see it at all (think test cross). If sex and colorblindness followed the law of
independent assortment, you should get the following ratios: 1/8c, 3/8C; 1/8c,
3/8C. (where C is full color vision and c is colorblind)
But what you actually see is 1/2C 1/4C 1/4c.
Among the progeny of non-affected parents, only the males are ever color blind. How can
we explain this? Suppose if in the male, the recessive color blindness allele were always
on the same chromosome as, and therefore attached to, the dominant maleness allele (i.e.
on the Y chromosome). Then it makes sense. Indeed, if you cross a colorblind man with a
normal homozygous woman, all children are normal but females will be heterozygous
(also called carriers in the context of human genetics). And if a rare, colorblind woman
is crossed with a normal man, all male children are colorblind and all female are
heterozygous. So here is a case where independent assortment does not apply. The genes
are linked because they are on the same chromosome.

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Now, you might ask, how come the colorblindness allele on the Y chromosome is always
recessive? The reason is that the Y is small and doesn't have a gene for colorblindness.
How can the Y have a recessive allele for a gene that doesn't exist? The absence of a gene
is a type of allele. Indeed, the vast majority of recessive alleles of genes that geneticists
study are in fact genes that have been mutated to become non-functional, sometimes
completely deleted from the DNA. This is the case for colorblindness. The colorblindness
gene makes the protein (opsin) that detects green light. The colorblindness trait results
from a gene on the X chromosome that is nonfunctional and therefore cannot make this
protein properly; it has been mutated. Y has no gene, consequently it is always recessive
for the colorblindness trait. There is a special name for the situation where a gene is
missing from one of the chromosomes, hemizygous. The functional allele is called wild
type the defective allele is called a mutant allele. Obviously, "wild type" is a somewhat
relative term. If multiple alleles are common in a population a gene is called
polymorphic.
Are there other examples of linkage? Yes and not just associated with sex chromosomes.
Consider an organism that is the darling of geneticists, the fruit fly, popularized by
Thomas Hunt Morgan (1866-1945). Morgan discovered linkage and correctly surmised
that it was the result of two genes being on the same chromosome. The book gives the
example of vestigial and black genes.
Morgan did the following cross which we will call a dihybrid testcross:
Po BbVv x bbvv
Independent assortment predicts progeny will have the genotypes:1/4BbVv 1/4 Bbvv
1/4bbVv 1/4bbvv
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