Transcript A allele - cloudfront.net

NATURAL SELECTION
IN:
Teddy Grahams
DARWIN PROPOSED A MECHANISM FOR
EVOLUTIONARY CHANGE:
NATURAL SELECTION IS THE
DIFFERENTIAL SURVIVAL AND
REPRODUCTION OF INDIVIDUALS IN A
POPULATION.
ALLELE FREQUENCY IS A MEASURE OF THE
RELATIVE FREQUENCY OF AN ALLELE ON A
GENETIC LOCUS IN A POPULATION. IT IS
EXPRESSED AS A PROPORTION OR A
PERCENTAGE.
LOCUS (PLURAL LOCI) IS THE SPECIFIC
LOCATION OF A GENE OR DNA SEQUENCE
OR POSITION ON A CHROMOSOME.
IN POPULATION GENETICS, ALLELE
FREQUENCIES SHOW THE GENETIC
DIVERSITY OF A SPECIES POPULATION OR
THE RICHNESS OF ITS GENE POOL.
POPULATION GENETICS STUDIES THE
DIFFERENT "FORCES" THAT MIGHT LEAD TO
CHANGES IN THE DISTRIBUTION AND
FREQUENCIES OF ALLELES - OR EVOLUTION.
A gene pool is the total
number of genes of every
individual in an
interbreeding population.
A large gene pool indicates
high genetic diversity,
increased chances of
biological fitness, and
survival in a
changing
environment.
A small gene pool indicates
low genetic diversity,
reduced chances of
acquiring biological fitness,
and increased possibility of
extinction.
THE POPULATION IS THE BASIC
UNIT OF EVOLUTION.
POPULATIONS EVOLVE,
BUT INDIVIDUALS DO NOT
EVOLVE.
THE HARDY–WEINBERG PRINCIPLE
STATES THAT BOTH ALLELE AND
GENOTYPE FREQUENCIES IN A
POPULATION REMAIN CONSTANT—
THAT IS, THEY ARE IN EQUILIBRIUM—
FROM
GENERATION TO GENERATION
UNLESS SPECIFIC DISTURBING
INFLUENCES ARE INTRODUCED.
THOSE DISTURBING INFLUENCES INCLUDE:
 non-random mating
 Mutations
 Selection
 Limited population size
 Random genetic drift
 Gene flow
THE HARDY-WEINBERG THEOREM
STATES THAT THE FREQUENCY OF
ALLELES IN A POPULATION WILL
REMAIN THE SAME REGARDLESS OF
THE STARTING FREQUENCIES.
THIS THEOREM IS VALID ONLY IF CERTAIN CONDITIONS ARE MET:
1) THE POPULATION IS VERY LARGE.
2) MATINGS ARE RANDOM.
3) THERE ARE NO NET CHANGES IN THE GENE
POOL DUE TO MUTATIONS IN THE DNA.
4) THERE IS NO MIGRATION OF INDIVIDUALS INTO
AND OUT OF THE POPULATION.
5) ALL GENOTYPES ARE EQUAL IN
REPRODUCTIVE SUCCESS (NO SELECTION).
The frequency of the possible
combinations of alleles (AA, Aa,
and aa) in this population is
expressed as:
THE HARDY WEINBERG
EQUILIBRIUM IS IMPOSSIBLE IN
NATURE.
GENETIC EQUILIBRIUM IS AN
IDEAL STATE THAT PROVIDES A
BASELINE TO
MEASURE GENETIC CHANGE
AGAINST.
REAL LIFE APPLICATION:
A patient's child is a carrier of a recessive mutation that causes cystic
fibrosis in homozygous recessive children. The parent wants to know
the probability of her grandchildren inheriting the disease.
In order to answer this question, the genetic counselor must know
the chance that the child will reproduce with a carrier of the
recessive mutation. This fact may not be known, but disease
frequency is known. We know that the disease is caused by the
homozygous recessive genotype; we can use the Hardy–Weinberg
principle to work backward from disease occurrence to the frequency
of heterozygous recessive individuals.
INTRODUCTION:
You are a bear-eating monster. There are two
kinds of bears that you like to eat: happy bears
and sad bears. You can tell the difference
between them by the way they hold their
hands. Happy bears hold their hands high in the
air, and sad bears hold their hands down low.
Happy bears taste sweet and are easy to
catch. Sad bears taste bitter, are devious and
hard to catch. Because of this you only eat
happy bears.
The happy trait in bears is caused by the
expression of a recessive allele. The
homozygous recessive condition is being
happy. The sad trait is caused by a dominant
allele.
NEW BEARS ARE BORN
EVERY YEAR (WHEN THEY
ARE HIBERNATING IN
THEIR DEN: THE
CARDBOARD BOX).
THE BIRTH RATE IS
ONE NEW BEAR FOR
EVERY OLD BEAR LEFT
FROM LAST YEAR.
FORM A HYPOTHESIS ABOUT
WHAT YOU EXPECT TO HAPPEN
TO THE NUMBER OF HAPPY
AND SAD BEARS OVER TIME.
PROCEDURE:
1) Obtain a population of 10 bears. Record the
number of happy and sad bears, and the total
population number. Using the equation for
Hardy-Weinberg equilibrium, calculate the
frequencies of both the dominant and
recessive alleles and the genotypes that
are represented in the population.
PROCEDURE:
p2 + 2pq + q2 = 1
Example:
If 5 of the 10 bears are happy.
Each bear has 2 alleles, then…
* 10 out of 20 alleles would be happy alleles,
* or q2 (10/20) is 0.5.
* To determine the q number, find the square of 0.5.
PROCEDURE:
2) Now, go hunting! Eat 3 happy bears. (If you
do not have 3 happy bears then eat the
difference in sad bears.)
PROCEDURE:
3) Once you have consumed the bears, obtain
a new generation by removing seven additional
bears from the den . Add these new bears to
your old ones and chart happy, sad and total
populations in your data table. (This is
generation 2). You should only have a
total of 14 bears – remember birth rate.
PROCEDURE:
4) Repeat steps 2 and 3 until you have
four generations recorded in your data
table. Be sure to record the number of
each type of bear and the total
population.
THE PERCENTAGE OF
HAPPY AND SAD BEARS
Determine the percentage of Happy and Sad bears for each generation and record
the percentages in Table 2.
To determine the percentage take the number of happy or Sad bears and divide by
the total number of bears for that generation and multiply the answer by 100.
For example, if there were 12 Sad bears and 4 Happy bears in a generation, then
there were 16 bears total.
To obtain the percentage of Sad bears you would divide 12 by 16 and then multiply
the answer by 100.
To obtain the percentage of Happy bears you would divide 4 by 16 and then multiply
the answer by 100.
TABLE 1. THE NUMBER OF BEARS FOR EACH GENERATION
TABLE 2. THE PERCENTAGE OF BEARS FOR EACH GENERATION
Generation
Number of Happy
Bears
Number of Sad
Bears
Total Population of
Bears
1
2
3
4
Generation
1
2
3
4
Percentage of Happy bears
Percentage of Sad bears
THE PERCENTAGE OF
HAPPY AND SAD BEARS
Graph what happens to the bear population over time.
Graph the percentage data for both the Happy and Sad bears on the same graph.
USING YOUR ACTUAL POPULATION DATA FROM THE ABOVE DATA TABLE AND THE HARDY WEINBERG
FORMULAS, COMPLETE THE FOLLOWING DATA TABLE TO DETERMINE THE PERCENTAGE FOR EACH
GENOTYPE IN THE POPULATION:
ALLELE FREQUENCIES:
Make a second line graph with two lines. One line should compare the “p” vs.
generations and the second line should graph the “q” vs. generations.
QUESTIONS:
1. Describe what is happening to the genotype and allele frequencies
in the population of Teddy Grahams?
2. What would you expect to happen if you continued the selection
process for additional generations?
3. How would the frequencies change if you were to now select for the
sad bears?
4. Why doesn’t the recessive allele disappear from the population?
How is it protected?
QUESTIONS:
5) Is the Hardy-Weinberg theorem valid for the population in this
experiment? Which, if any, of the conditions of the theorem did this
population violate?
6) Do you think this population was undergoing evolution? Why or
why not?
7) In what way(s) is the Teddy Graham ecosystem not like a real world
ecosystem? (Hint: read the 5 conditions that must be met by a
population)
SAMPLE CALCULATIONS USING THE HARDY-WEINBERG EQUATION
Albinism is a rare genetically inherited trait that is only expressed in
the phenotype of homozygous recessive individuals (aa). The most
characteristic symptom is a marked deficiency in the skin and hair
pigment melanin. This condition can occur among any human group
as well as among other animal species. The average human
frequency of albinism in North America is only about 1 in 20,000.
The frequency of homozygous recessive individuals (aa) in a
population is q². Therefore, in North America the following
must be true for albinism:
 q² = 1/20,000 = .00005
 By taking the square root of both sides of this equation, we
get: (Note: the numbers are rounded off for simplification.)
 q = .007
The frequency of the recessive albinism allele (a) is .00707 or
about 1 in 140 (or 20000 x 0.007 = 140) Knowing one of the
two variables (q) in the Hardy-Weinberg equation, it is easy to
solve for the other (p):
p=1–q
p = 1 - 0.007
p = 0.993
The frequency of the dominant, normal allele
(A) is, therefore, .993 or about 99 in 100.
The next step is to plug the frequencies of p and q
into the Hardy-Weinberg equation:
p² + 2pq + q² = 1
(.993)² + 2 (.993)(.007) + (.007)² = 1
.986 + .014 + .00005 = 1
This gives us the frequencies for each of the three genotypes for this trait in the population:
p² = predicted frequency of homozygous dominant individuals = .986 = 98.6%
2pq = predicted frequency of heterozygous individuals = .014 = 1.4%
q² = predicted frequency of homozygous recessive individuals (the albinos) = .00005 = .005%
Interpret Results:
With a frequency of .005% (about 1 in 20,000), albinos are extremely
rare. However, heterozygous carriers for this trait, with a predicted frequency
of 1.4% (about 1 in 72), are far more common than most people
imagine. There are roughly 280 times more carriers than albinos
(1.4/.005). Clearly, though, the vast majority of humans (98.6%) probably are
homozygous dominant and do not have the albinism allele.
EXAMPLE 2:
Consider a population of 1000 individuals with a locus and alleles
described below. Assume that you have no information on the
presence or absence of evolutionary mechanisms in this population.
You find that the population consists of:
90 individuals homozygous for the A allele(AA genotype)
490 individuals homozygous for the a allele (aa genotype)
420 heterozygotes (Aa genotype)
90 individuals homozygous for the A allele(AA genotype)
490 individuals homozygous for the a allele (aa genotype)
420 heterozygotes (Aa genotype)
1000
The frequency of the A allele will equal:
Total number of A alleles in the population / (Total number of alleles in population)
(90 x 2) + 420 / (1000 x 2 ) = 0.30
The frequency of the a allele will equal :
Total number of a alleles in the population/ (Total number of a alleles in the
population)
(490 x 2 ) + 420 / (1000 x 2) = 0.7
Or
1 – 0.30 = 0.07
90 individuals homozygous for the A allele(AA genotype)
490 individuals homozygous for the a allele (aa genotype)
420 heterozygotes (Aa genotype)
1000
90 individuals homozygous for the A allele(AA genotype)
490 individuals homozygous for the a allele (aa genotype)
420 heterozygotes (Aa genotype)
1000
q² = 490/1000 = 0.49
By taking the square root of both sides of this equation, we get q = .7
p=1–q
p = 1 - 0.7
p = 0.3
p² + 2pq + q² = 1
(0.3)2 + 2(0.3)(0.7) + (0.7)2 = 1
0.09 + 0.42 + 0.49 = 1
Since the observed genotype frequencies equal those predicted by the
Hardy – Weinberg Equilibrium Theory, we may (tentatively) conclude that
no evolutionary mechanisms operate on this locus in this population. The
population meets the assumptions of the Hardy-Weinberg theory:
1)The population is very large.
2)Matings are random.
3)There are no net changes in the gene pool due to mutations in the
DNA.
4)There is no migration of individuals into and out of the population.
5)All genotypes are equal in reproductive success (no selection).