Lab 8:
Population Genetics and Evolution
This may leave a bad taste in your mouth…
Pre-Lab Orientation
• Recall that the Hardy-Weinberg Equation
helps us identify allele frequencies throughout
a population.
– Given certain assumptions like “large population
size,” “random mating,” et cetera…
• A great example for a classroom is ability to
taste PTC (phenylthiocarbamide).
– This stuff is, oddly, used to grow transparent fish
by inhibiting melanin production.
PTC
• Not everyone can taste PTC.
• In North America, 55% of people can taste it (it
tastes quite bitter), while 45% cannot.
• Which one are you?
– Let’s take a taste, shall we? (you don’t have to)
• I’m going to wash my hands and give you a small piece.
• Place it on the tip of your tongue and wait a few seconds.
• If you are a PTC taster, you’ll get a bitter taste. If not, you’ll
just taste something papery.
– Don’t swallow it. Just awkwardly peel it off your
tongue and toss it in the trash.
Starting the lab…
• Take some time to read the Introduction and
Exercise 8A while I’m coming around with
PTC.
• Once we all taste the PTC strips, we’ll take
some classroom data and record it in Table
8.1, then answer the Topics for Discussion on
Page 2.
– Save Exercise 8B – Case Studies for another time.
Exercise 8B
• Now we start the Case Studies…
• …so I’m sorry if this seems a little forward, but
you need to go find a mate.
– A random mate – remember this is simulating HardyWeinberg equilibrium conditions. Find someone with
whom you wouldn’t normally pair.
– Gender has no role here.
– Need a team of three? One of you is going to have to
two-time it and act as partner to two others. Only
write one result down, just like everyone else.
Case I
• Start by turning to the back page of your lab
packet – the Data Page.
• Under Case I, note that we are simulating Initial
Class Frequencies of:
– AA: 25%
– Aa: 50%
– aa: 25%
• Your Initial Genotype: Aa
• On your lab desks is a bag full of letters on card
stock.
• These letters represent those same alleles that
can be inherited – either A or a.
Case I
• Start by taking four total for each of you – two
A and two a.
– These are the products of meiosis.
– Meiosis, if you forgot, is the production of
gametes – haploid cells destined for reproduction.
– From one cell, four are produced, but they only
have one set of chromosomes so there’s only one
allele each.
– These are the only alleles you can pass on.
Case I
• Now put the cards into a pile, face down, and
shuffle them.
• Take one card off the top. That’s one of the
two alleles for your offspring. Have your
partner do the same.
• These two alleles comprise the first offspring.
One of you (and only one) should write the
genotype in the data section on the last page
(F1 Genotype).
Case I
• Now put the letters back, repeat the process,
develop a second offspring, and have the
other partner write that down (F1 Genotype).
• Let’s record our data. (you don’t need to do
this – just me)
Case I
• Okay…now the tricky part…
• Each of you needs to assume the role of the F1
generation.
– Leave the cards at the table but remember your genotype.
– Find a new mate and settle down at another lab table.
• Now that you have found a new mate, take a new set
of cards.
– Remember that the letters represent the products of
meiosis, so if you’re AA, take four A cards.
– If you’re Aa, take two A cards and two a cards.
– aa? Take four a cards.
Case I
• Class data time! What’s your genotype?
• Now you’re going to (randomly) find a new
mate with the alleles you’ve determined.
• Find someone, then repeat the process.
• After each generation, pause so we can collect
genotype data.
• Once we’re done with five generations, record
the data in #4 after Case 1 on Page 3.
Complete the questions and the Data Page
section.
Case I Discussion
• Did our allele frequencies change?
Importantly, would we expect them to
change?
• Is our population size large enough?
• Are we at Hardy-Weinberg equilibrium?
FYI
• Wondering why our initial frequencies are
0.25 AA, 0.50 Aa, and 0.25 aa if everyone
starts Aa?
• The answer is because you need to look
forward a little.
– The Hardy-Weinberg equation tells us that, since
p=0.5 and q=0.5 (we’re all heterozygous), p2
should be 0.25, 2pq is 0.50, and q2 is 0.25.
– Further, take a look at a Punnett Square for your
first cross.
Punnett Square
A
a
A
AA
Aa
AA: 0.25
Aa: 0.50
aa: 0.25
a
Aa
aa
Case II
• Now we’re going to repeat Case I but change a
condition – as in reality, not every genotype will have
an equal chance of survival.
– An example is sickle-cell anemia, which can kill humans
prior to reproduction if they are homozygous recessive.
• Each time you draw aa in this process, don’t record it;
it doesn’t reproduce.
• You and the other parent must keep trying until you get
a non-homozygous recessive offspring.
• Again, after five generations, we’re going to count
frequencies, and then you’ll complete data and
questions.
Case III
• By now you know that sickle-cell anemia, while a
disease, helps prevent a far worse disease in
malaria.
• Individuals that are heterozygous for sickle-cell
anemia have some sickle cells, but not enough to
make them ill.
• At the same time, having those sickle-cells
increases resistance to malaria because the
parasites can’t infect the erythrocytes (red blood
cells).
Case III
• We’re going to simulate this heterozygote
advantage with Case III.
• The procedure’s the same except:
– If your offspring is AA, flip a coin (or card).
• Heads = Does not survive (try again).
• Tails = Does survive.
– aa still doesn’t survive, by the way, and allele
frequencies start the same (two A, two a, et cetera).
– We’ll do this for five generations, then record data,
then do it for five more generations and record.
• And answer the questions as usual.
Case IV
• Case IV further explores the concept of
genetic drift.
• Use the rules of Case I, but we’ll do so with a
smaller population size.
– We will divide the class into three populations –
no gene flow between groups!
Hardy-Weinberg Problems
• To finish the lab, complete the HardyWeinberg Problems starting after Case IV.