Transcript Review of Hardy Weinberg

Contemplate the wonders of the age in which we live, this
greatest of all ages in the history of humanity. More
inventions and scientific discoveries have been made during
my lifetime than in all the previous centuries of human
history combined. This is the remarkable function of the
efforts of thinking men and women who have applied their
inquisitive and dedicated thought processes in the fields of
medicine, industrial safety, hygiene and sanitary measures,
chemistry, and research in genetics, microbiology, the
environment, and other disciplines, all involving the
processes of the human mind. How can we help but be
grateful for such miracles?
Gordon B. Hinckley Standing for Something pg.92
Review of Hardy Weinberg
Here is what you need to know
already.
When working with one polymorphic gene....
• There are only two alleles, A and a, possible for a gene
locus.
• Let the original A allele frequency be represented by p
and the original a allele frequency be represented by q
• Possible genotypes are: AA; Aa; aa
• AA = p2 Aa = 2pq
aa= q2
• In any population
– the frequencies of p + q =1.0
– Genotype frequencies can be represented as
p2+2pq+q2=1.0
– Allele frequencies can be calculated from genotype frequencies
as follow:
– A = p = p2 + ½ (2pq) (see page 178-79)
– a = q = q2 + ½ (2pq)
• Only in Hardy Weinberg equilibrium
we can also calculate genotype frequencies from allele frequencies
– Genotypic frequencies =
AA or p2 = (p x p); Aa or 2pq= (2 x p x q); and aa or q2 = (q x q)
A gene pool
• The sum total of all alleles for all genes which exist in
a population.
• This means that genotype numbers can be used to
calculate allele frequencies.
• Example:in a population of 100 people (with 200 alleles):
36 are AA; 48 are Aa; 16 are aa.
• How many A alleles are in this population’s gene
pool? _____
• How many a alleles?_____
(36*2)+48)= 120 A alleles so
A frequency =120 A /200 total alleles= .6
(16*2) +48 = 80 a alleles so
a frequency =80 a /200 total alleles= .4
Hardy Weinberg Conclusions
1. The allele frequencies in a population will not
change from generation to generation.
2. The genotype frequencies can be calculated
from the allele frequencies as follows:
AA= p x p; Aa= 2pq ; aa=q x q.
Alternatively if
AA cannot be calculated by p2 (pxp),
Aa cannot be calculated by 2pq and
aa cannot be calculated by q2 (qxq)
then the population is not in equilibrium
What is necessary for a population to
be in Hardy-Weinberg Equilibrium?
• There can be no situations in which the
current distribution of alleles in the
population will change over time.
• This would be violated under any of 5
assumptions …..
There are 5 assumptions which must be met
in order to have a population in equilibrium
1. There is no selection. In other words there is
no advantage for the survival of one genotype
over another.
2. There is no mutation. None of the alleles in a
population will change over time. No alleles get
converted into other forms already existing and
no new alleles are formed
3. There is no migration (gene flow) New
individuals do not bring new alleles into nor take
existing alleles out of a population.
Exceptions to Hardy Weinberg cont.
4. There are no chance events (genetic drift)
Populations are sufficiently large to ensure
that chance alone does not dictate which
alleles are passed on and which are lost.
5. There is no sexual selection or mate
choice The determination of who mates with
whom is totally random.
Checking for Hardy Weinberg
Equilibrium
• Working with the person sitting beside
you, read and work out problem #6 on
page 219 on your text.
How to attack this problem
• First figure actual genotype frequencies
from the actual numbers given in the
problem.
• Then count the actual A alleles in the
population and the actual a alleles in the
population to get the allele frequencies.
• Then calculate the genotype frequencies
predicted by the Hardy Weinberg
equilibrium equations and compare to the
actual numbers.
Directly from the numbers given The genotype frequencies are:
SS =1194/1778 = .67 Ss= 526/1778 = .30 ss= 58/1778 = .03
There are two ways that you can calculate the allele frequencies
from this data:
(1) Directly from the numbers given the allele frequencies of
p (S) and q (s) are: 2914/ 3556 S alleles = 0.82 S
frequency
(2) We can use genotype frequencies to get allele frequencies
such that p = p2 + 2pq :
so S= p = .67+1/2(.30) = 0.82
then also for q…………..
(1) 642 / 3556 s alleles = 0.18 s frequency OR
(2) .03 + ½(.30) =0.18 OR
(3) q= 1-p = 1-0.82 = 0.18
If the population were in Hardy-Weinberg
equilibrium we can apply conclusion #2
• SS genotype frequency would be
predicted by p2 = (.82)2= 0.67 and
• Ss frequency would be 2pq or
2 (.82)(.18) = 0.30; and
• ss frequency would be q2 = (.18)2 = 0.03.
• These numbers almost exactly match the
measured genotype frequencies - so this
population appear to be in HardyWeinberg equilibrium.
What else must we also know to
be sure?
• We would need to check in future
generations to make sure that the allele
frequencies are not changing.
• So in our calculations we confirmed
conclusion #2 (that we can predict
genotype frequencies from allele
frequencies)
• but have not yet verified conclusion#1
(that allele frequencies will not change
from generation to generation.
Using Chi Square to determine H-W
• Use numerical values NOT frequencies or percentages
• df = For Mendelian Dominant recessive =
# phenotypes - 1
• df for incomplete dominance, codominnace etc.
# phenotypes - #alleles
• X2 =  (O-E)2
E
• Null hypothesis stated as
The difference between observed and expected values is
simply due to chance.
• We want a .05 level for acceptance of Null hypothesis. This
means
There is a 95% chance that a LARGER Chi Square value
indicates that the difference is REAL and we REJECT our
null hypothesis
One more problem adding Chi square test
The ability of certain people to taste the chemical
phenylthiocarbamide (PTC) is governed by the dominant
allele T, and the inability to taste PTC by its recessive
allele t.
The frequency of the T allele in a human population = 0.8,
and a sample of 200 yields 90% tasters (T_) of a
chemical called PTC and 10% nontasters (tt).
(a) Does the sample conform to the equilibrium
expectations?
(b) What is the chi-square value?
(c) How many degrees of freedom exist?
(d) What is the probability that the observed deviation is
due chance?
End of H-W review