Download Lesson 13: Polygenic Inheritance student notes

POLYGENIC INHERITANCE
The very first personality trait that was ascribed to a known gene is “novelty seeking”. People can be
high novelty seekers – they like bungee jumping and risky behaviors, or they can be low novelty
seekers – they like reading and solving Sodoku puzzles. While as many as 10 genes are estimated to
play a role in this trait, the first one identified was the gene that encodes a dopamine receptor. DRD2
(sometimes referred to as D2DR) was the second dopamine receptor identified and is found in the
nucleus accumbens among other places. The idea is that the rewarding feeling of reading is enough
for some people, but others need more stimulation.
This lesson is as much about learning how to deal with complex probability as it is to learn about traits
that are controlled by many genes. Almost all of the interesting traits are controlled by many genes.
Novelty seeking, hair color, height, and response to alcohol are all controlled by more than one gene.
MOST of these genes also have alleles that work by incomplete dominance. Making predictions about
the probability of inheriting one of these complex traits requires thinking about several rules of
computing probability. Reminding yourself about HOW to add and subtract fractions will be a very good
idea if you have not worked with this type of math recently.
Objectives
After successfully completing this lesson you will be able to:
 To appreciate how most behavioral traits, including response to drugs, are attributable to
many genes
 To learn how to make predictions about probabilities for given polygenic phenotypes
 To understand the principles of independent assortment that underlie these predictions
 To appreciate that polygenic traits are “quantitative” and bell-curve distributions of
phenotypes are generally observed in a population and even in a mating
Before you begin!
o Your ideas
 What are some personality traits besides the ones we’ve used as examples, that is controlled
by many genes?
 By now you know that searching for something online, such as the history of Apple Cup
games, is most efficient if you use all the terms, Apple – cup – history, rather than just using
one term. Which logical term “AND” or “OR” allows for the restriction in the number of
matching results?
 What is 3/8 x 1/4? What is 3/8 + 1/4?
 What is the chance of throwing two dice and obtaining a combination of a “1” and a “6”?
o Previously learned material
 When dealing with gene “B” that shows incomplete dominance and controls basketball
dribbling ability, a BB genotype makes someone expert at dribbling. What is the phenotype
of someone who has the genotype Bb?
 What would be the phenotype of Bb if dribbling showed Mendelian dominance?
 People sometimes say they have “slow metabolism”. They are usually talking about calorie
use but how might this phrase pertain to drug response?
Lesson 13: Polygenic Inheritance
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POLYGENIC INHERITANCE
Guiding Questions
1.
2.
3.
4.
What types of traits are controlled by many genes?
How do these genes work together to manage these traits?
How can one make predictions about polygenic traits from known matings?
How do the rules of probability and logic help with making these predictions?
Key Terms
 Polygenic
 Independent Assortment
 Boolean logic
 Dihybrid cross
Activity One: Addition and Multiplication with Fractions
Probabilities are often expressed as fractions. Especially when dealing with Punnett squares, it is much
more convenient and common to say that 1/4 of the possible children will have a genotype, instead of
converting to percentages. Probabilities are expressed as fractions ranging between 0 and 1. 1 = 100%
probability (no other option).
Thus dealing with fractions ad probability requires you to know how to multiply and add them.
To multiply fractions – you just multiply the numerators as well as the denominators. 1/5 x 1/5 = 1 x 1
over 5 x 5, or 1/25.
To add fractions – you first need to ensure that you have common denominators. Adding 1/5 + 1/5 is
easy – no need to compute the common denominator (5). So 1 + 1 / 5 = 2/5. But if you had to add 1/2 +
1/5, the common denominator would be 10. You would convert your 1/2 into 5/10 ad 1/5 into 2/10.
5/10 + 2/10 = 7/10.
Activity Two: Boolean Logic: And vs. Or
As hinted at with the internet search analogy question, when expressing probability of multiple events,
said probability gets smaller as one makes more restrictions (and uses the word “and”). If 1/4 of my
students speak Spanish, and 1/3 of the same class wears glasses, number much smaller than 1/4 speaks
Spanish AND wears glasses. When dealing with probabilities of INDEPENDENT events (such as glasses
wearing and speaking Spanish), saying “and” requires you to MULIPLY your fractions. 1/4 x 1/3 of my
students both speak Spanish and wear glasses. 1/12 of my class has both of these traits.
Saying “or” is like saying that there are many ways to get the right answer. If I want to reward my
students who either speak Spanish OR wear glasses, then I can reward those with one trait or the other
(and students with both traits). When we say “or” for independent events, we ADD the fractions. 1/4 +
1/3 of my students has at least one of these traits, which means that 7/12 of them can be rewarded.
TEST OF CONTENT
If 1/2 of my kids can have the AA genotype, and 1/4 of them can have the BB genotype, what
fraction of them can be AABB?
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POLYGENIC INHERITANCE
If I have two dice (numbered 1 – 6) what is the chance that one (or both) of my dice will show a 1
when I roll them?
Activity Three: Web Site Review
At: http://anthro.palomar.edu/mendel/mendel_3.htm read sections on polygenic traits, multiple-allele
series, modifying and regulator genes, and environmental influences.
At: http://www.buzzle.com/articles/polygenic-traits.html read the page with the following guiding
questions:
 Why is blood pressure an example of a polygenic trait? What genes are involved?
 What kind of phenotypic distribution is expected from polygenic traits (two answers)?
 Why are polygenic traits often referred to as quantitative traits?
At: http://www.learnnc.org/lp/pages/3070 feel free to try the activities described with non-genetic
objects to get better familiar with polygenic probability problem solving strategies.
At: http://en.wikipedia.org/wiki/Mendelian_inheritance focus on the portion pertaining to independent
assortment (other material may serve as good review).
Finally, at: http://www.sumanasinc.com/webcontent/animations/content/independentassortment.html
work through the tutorial on independent assortment, all four tabs. But don’t be discouraged if you find
the quiz to be a bit too challenging, we still have to work through this!
Activity Four: What did Mendel Find when Dealing with Two Genes?
Mendel studied 7 traits with contrasting phenotypes, such as wrinkled seed vs. round seed. He was
thorough in studying so many. He was also thorough in studying more than one generation and in
being highly quantitative. But for this activity, we’ll focus on another reason Mendel won the title
“Father of Genetics”. He was not content to just watch a single trait separate from the others. He did
numerous tests of two traits in a cross. For instance, he mated round seed, purple flowered plants by
wrinkled seed, white flowered plants. If both parents were homozygous for their version of the trait
(and they were – Mendel used “pure breeding lines” to start all his crosses), then the offspring of this
mating would be heterozygous at both the seed texture and flower color genes, RrPp. He referred to
these offspring as “dihybrid”.
Mendel’s second law came from these two-gene studies. He noticed that just because round and
purple came from one parent, they were transmitted to the “grandbabies” independently of each
other. While round were more common than wrinkled (3 : 1), round and white also outnumbered
wrinkled white 3 to 1. You might have expected to either just find white with wrinkled or MORE
white/wrinkled than white/round. But this was not the case. A mathematical analysis of the offspring
frequencies suggested that the texture alleles segregated independently of the flower color alleles, like
two dice when thrown. His second law is called “Independent Assortment”.
Recall that before doing a Punnett square, one must understand how chromosomes and the alleles on
those chromosomes move during meiosis. If Round = RR, wrinkled = rr, and Purple = PP, white = pp,
then the RrPp offspring can make gametes whose alleles come from two different genes we are
following. RP is ONE kind of gamete that this dihybrid individual can make. What are the other three?
Rp, rP, and rp. There is a 1/2 chance of the dihybrid making a gamete with the “R” genotype, and 1/2
chance of the same individual making a gamete whose genotype is “p”. Using Boolean logic, the
probability that she will make a gamete with the genotype “Rp” is 1/2 x 1/2, or 1/4. This makes sense
since there are four gamete types she can make.
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POLYGENIC INHERITANCE
Activity Five: Independent Assortment of Multiple Chromosomes.
The chromosomal basis of independent assortment is that while maternal and paternal chromosomes
separate during meiosis I, the chromosomes “1” do not care which way the chromosomes “2” go.
Recall the symbols used in lesson 14. But add to them another chromosome with the “B” gene.
DNA sequence for “A” allele
DNA sequence for “a” allele
Paternal 3
Maternal 3
DNA sequence for “B” allele
DNA sequence for “b” allele
Paternal 13
Maternal 13
Because chromosome 3 and 13 are independent of each other (because each is its own DNA double
helix), during meiosis I, they can line up one of two ways, each of which is equally likely.
The line up on the left would yield AB and ab gametes while the line up on the right would yield Ab and
aB gametes.
With three independent genes, there are 8 different gametes possible if an individual is heterozygous for
all three genes. Do you see a pattern? Recall that novelty seeking was probably the result of 10 or more
genes. If it is 10 genes, what mathematical representation shows you the number of different gametes
possible for a person heterozygous for all 10?
Activity Six: Probability of Outcomes for Multiple Genes Using Punnett Squares
Once you figure out your gamete types and probabilities, you are now able to apply the Punnett square
approach to solving for offspring genotypes and proportions. For a dihybrid cross, the Punnett Square
would look like this:
AB
Ab
aB
ab
AB
AABB
AABb
AaBB
AaBb
Ab
and so forth….
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POLYGENIC INHERITANCE
aB
Ab
Keep in mind that if meiosis is working properly, each gamete should have ONE allele of each gene.
TEST OF CONTENT
What would be one gamete made by an individual with genotypes AaBbCc?
From the Punnett square given previously (dihybrid), what is the probability of having a child with
genotype AaBB?
From the same Punnett square, what is the probability of having the dominant phenotype for
gene “A” and the recessive phenotype for gene “B”?
Activity Seven: Alternate Method for Solving Multiple Gene Problems
You might be able to see that the complexity of solving genetics problems grows as the number of genes
grows. In fact, enough transcribing errors are made with 16-box Punnett squares that making a Punnett
square for even a three gene trait is not recommended. So going back to the mathematical rules of
dealing with independent probabilities, we have a much more simple strategy to recommend.
That strategy requires you to make only 1-gene Punnett squares, and make them for each gene in your
problem. Once these separate probabilities are solved, combine them using the “and” / “or” strategies
described in activities 1 and 2.
If 1/4 of my progeny are AA, and 1/4 of my progeny are bb based on each separate Punnett square, then
the probability of AAbb is 1/4 x 1/4 or 1/16. Convince yourself but solving the same “test of content”
questions without the large Punnett square. But recall, when you say “or” in the last problem (about AA
or Aa), you must ADD the probability of AAbb to the probability of Aabb).
TEST OF CONTENT
From the strategy of using two separate Punnett squares, one for gene “A” and another for
gene “B”, in the dihybrid cross, what is the probability of having a child with genotype AaBB?
From the same 2-Punnett square technique, what is the probability of having the dominant
phenotype for gene “A” and the recessive phenotype for gene “B”?
Activity Eight: Multiple Genes, Each with Contributing (and non-contributing) Alleles
While it is over-simplistic to say that each gene responsible for the genetic control of a polygenic trait
works equally to all other genes controlling that trait, polygenic inheritance is already complicated
enough if we start there. And generally it is true that the genes controlling polygenic traits can act
additively, with each gene having alleles that add to the trait (contributing alleles) and others that do not
(non-contributing alleles). Recall that in some plants, red colored flowers result when there are two red
pigment making alleles, and pink colored flowers result when only one allele is able to make red
pigment. This same idea of contributing alleles follows for many of the traits that are polygenic.
By this logic, if genes X, Y, and Z are all equally responsible for, for instance weight, and each of these
genes has contributing alleles that will add ounces to the organisms “base line” weight and alleles that
do not, then an organism with genotype Xxyyzz will have the same weight as an organism with xxYyzz.
While gene X may manufacture growth hormone, and gene Y regulates appetite, and gene Z determines
the digestive efficiency for the organism, contributing alleles for each gene have the ability to increase
the organisms weight-gaining power.
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POLYGENIC INHERITANCE
We generally will call upper case letters the contributing allele and lower case letters the noncontributing version of the allele. Phenotypes are also often named numerically based on the number of
contributing alleles an organism has. On a scale of 0 to 6, the xxYyzz organism is a light weight, but not
the lightest, therefore is a “1”. But be careful. Because an xxyyZz organism is also phenotypically a “1”,
when solving for probabilities with regard to phenotypes, you must note all the genotypes that will
make the phenotype in question and add their probabilities together to find the total probability of
obtaining a phenotype of “1”.
Here is one practice problem we can work through together. A farmer wishing to obtain a pumpkin
more enormous than his neighbor has taken flowers from two of his fairly large pumpkins to generate a
cross. One of his “fairly large” pumpkins has the genotype XxYYZz and the other has the genotype
XxYyZZ. Predict the probability that he will create this enormous (a phenotype of “6” on a 0 – 6 scale)
pumpkin.
What is the chance he will be fooled into thinking that a really large (“5”) pumpkin is as big as he can
get? In other words, what is the chance of getting a “5” pumpkin?
Activity Nine: “Laboratory” Practice Session
Perform the Genetics Workshop laboratory activity to solidify your understanding of lessons 14 and 15.
Activity Ten: Reading
Required Reading
Introductory Biology Textbook
 Read about independent assortment and Mendel’s dihybrid crosses.
Internet
 http://en.wikipedia.org/wiki/Mendelian_inheritance focus on the portion pertaining to
independent assortment (other material may serve as good review).
 http://anthro.palomar.edu/mendel/mendel_3.htm read sections on polygenic traits, multipleallele series, modifying and regulator genes, and environmental influences.
Supplemental Reading
Internet
 http://anthro.palomar.edu/mendel/mendel_3.htm read sections on sex linked inheritance. We
did not cover this, but it also poses some strategic challenges for solving genetics problems.
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