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Extending Mendelian
Genetics
23 October, 2002 Text Chapter 14
Incomplete Dominance
Alleles of some genes show
incomplete dominance. In
these cases, the heterozygote
has an intermediate
phenotype - a mixture of the
phenotypes conferred by its
alleles.
Is this an example of blending?
Why or why not?
No. Two pink individuals
can mate to produce red,
white offspring.
Codominance
In a set of codominant alleles, the heterozygote does not show an
intermediate phenotype. The phenotypes of both alleles are individually
expressed.
What is a Dominant Allele?
•Usually, dominant alleles are recipes for functional proteins.
•Recessive alleles are altered recipes for non-functional proteins.
Think about flower color in pea plants.
Substrate (colorless)
Enzyme P
Product (purple)
The P allele is a recipe for a functional enzyme. The p allele is a
recipe for a non-functional enzyme. Purple is dominant because one
copy of a functional recipe is enough.
In the analogous situation in snapdragons, one copy is not enough,
And an intermediate phenotype is seen.
At the molecular level, both functional and non-functional proteins
are present. This is more like codominance.
Epistasis
In epistasis, the alleles of
one gene alter the
expression of alleles of
another gene.
Polyfactorial
Characters
Most interesting characters
are influenced not by one or
two genes, but by dozens or
hundreds. In this example,
alleles at three loci control
skin color.
The distribution shown here is
characteristic of quantitative
characters. These characters
vary continuously, rather than
in a few discrete states.
Environmental factors can
also influence these
polyfactorial characters
Pleiotropy
Pleiotropy is the converse of
the concept of polyfactorial
characters. In this case, one
allele causes many different
phenotypes.
Pedigree Analysis
Pedigree analysis is a useful
tool for studying inheritance
in families.
Mendelian Inheritance in Humans
In humans, many disorders follow Mendelian patterns of inheritance.
Cystic fibrosis, sickle cell disease, Tay-Sachs syndrome, and many
others are inherited as recessive alleles. The recessive allele codes
for a non-functional protein. Usually, one functional allele is
enough, so heterozygotes are asymptomatic.
Why are cystic fibrosis and sickle cell alleles so common? In both
cases, heterozygotes are protected from other diseases - recessive
CFTR alleles protect against typhoid infection, while HbS alleles
protect against malaria.
Lethal dominant alleles are (for obvious reasons) less common.
Huntingtons disease is caused by a dominant allele whose effect on
phenotype is not obvious until after age 40.
Problem 1
What is the mode of inheritance for this disorder?
Fill in as many genotypes as possible.
Use “A-” when you are not sure of the genotype of a dominantphenotype individual.
A man with type O blood has a sister with type AB blood.
What are the genotypes and phenotypes of their parents?
Both parents must have i allele.
Sister must have gotten IA from one parent. IB from the other.
So parents must be IAi and IBi, type A and type B blood.
Four babies, of blood types A, B, AB, and O, were born in a
hospital one night. Two of these babies were twins (non-identical
twins, produced from two different fertilized eggs). The three sets of
parents had the following blood types:
P1) A and B
P2) O and B <-- parents of the twins
P3) O and AB
Assuming that the twins were born to the parents with blood types
O and B, which baby belonged to which parent? What were the
genotypes of the parents and babies in each case?
AB must get A from one parent, B from the other, so AB baby goes
home with P1. O baby has to get I from both, so she is one of the
twins. A baby has to get A from somewhere, so congratulations P3.
This leaves B baby as the twin brother. Wasn’t that fun?!
A form of blindness in humans, retinitis pigmentosa, is caused by
either a dominant allele R, or a recessive allele a. Hence, only rrApersons are normal. A man with the abnormality whose parents were
normal marries a woman of the genotype RrAa. Assume independent
assortment.
a. What proportion of their children would be expected to have the
abnormality?
Since the man’s parents were normal, his genotype must be rraa.
All his gametes will be ra.
The woman’s gamtetes will be 1/4 RA, 1/4 Ra, 1/4 rA. 1/4 ra
Three ways to be blind: rraa, RrAa, Rraa, each 1/4 probability.
Add to give 3/4 children blind.
b. What are the genotypes of the man's parents?
Since the man’s parents are normal, they must be rrA-.
But, since the man is blind, he must be rraa.
This means that both of his parents must be rrAa.
c. If this couple has 3 children, what is the probability
that 2 are blind and 1 has normal vision?
For any child, 3/4 chance of being blind.
There are three ways to get 2 blind, 1 normal: BNB, BBN, NBB.
The probability of each way is (3/4)(3/4)(1/4) = 9/64. Add to 27/64.
d. If the parents have 3 children, what is the probability that the
first 2 are blind and the youngest child has normal vision?
This is just the middle way from part c. The answer is 9/64.
Summer squash can be found in three shapes: disk, spherical,
and elongate. In one experiment, two squash plants with diskshaped fruits were crossed. The first 160 seeds planted from
this cross produced plants with fruit shapes as follows: 89 disk,
61 sphere, and 10 elongate. What is the mode of inheritance of
fruit shape in summer squash?
This is very close to a 9:6:1 ratio. This ratio could result from
two genes in an epistatic relationship. In this case, the
genotype of the parents would be AaBb. Then A-B- would be
disks, any plant with either aa or bb would be spherical, and
aabb plants would be elongate.