Lab 3 Population Genetics
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ANTH 110L
LAB # 3: GENETICS PART 2
28 February 2024
Kenya DE Loza
Anthropology 102 – LABORATORY EXPERIENCE 3
1
GENETICS PART 2: POPULATION GENETICS THE THEORY AND PRACTICE OF THE
HARDY-WEINBERG EQUILIBRIUM MODEL
Lab report begins on page 13.
In 1908, Godfrey Hardy, an English mathematician, and Wilhelm Weinberg, a German physician, simultaneously published papers dealing with a misconception was that dominant
genes would eventually replace recessive genes. Both Hardy and Weinberg demonstrated mathematically that gene frequencies
, or the percent of genes in a breeding population
, would not change from one generation to another if
the following conditions
were met
:
(1)
random mating across the population (panmixia)
(2)
equal number of males and females
(3)
the population is infinitely large
(4)
there is no “in” or “out” migration
(5)
natural selection, mutation, and genetic drift are not acting upon the population
This is the Hardy-Weinberg Equilibrium Model. To conceptualize this model, it is easiest to think of it as the Mendelian genetics of breeding. Using a Punnett square it is possible to calculate genotype frequencies or the percent of a certain genotype (dominant homozygotes, heterozygotes, recessive heterozygotes) in a breeding population if the genotypes of both parents are known. The study of genetics of breeding populations is called population genetics and the Hardy-Weinberg Equilibrium Model is central to understanding and doing research in population genetics. DETERMINING GENE FREQUENCIES
Determining gene frequencies is not very difficult. One method simply involves counting identifiable genes. For example, in the human MN
blood-group system (this is similar to the ABO system) there are three identifiable genotypes and three identifiable phenotypes because the MN
system is a co-dominant system. If two alleles are co-dominant both will be expressed when both are present. One allele is not dominant and the other is not recessive. In a co-dominant system the heterozygote is distinguishable from the dominant homozygote. The three identifiable phenotypes are M
, MN
, and N
. These are identifiable phenotypes so it is less difficult to determine genotypes. Phenotype M has the genotype MM 2
or two copies of the M
allele; phenotype MN
has the genotype MN
because both alleles are expressed; and phenotype N
has the genotype NN
, or two copies of the N
allele. Example 1:
PHENOTYPE
NUMBER OF
INDIVIDUALS
GENOTYPE
NUMBER OF M
ALLELES
NUMBER OF N
ALLELES
M
600
MM
1200
0
MN
300
MN
300
300
N
100
NN
0
200
1000 individuals
1500 genes
500 genes
By doing a census in this hypothetical population of 1000
, it was determined that 600
individuals were blood type M
, 300
were blood type MN
, and 100
were blood type N
. Using this information, gene frequencies of the percent of genes at this locus that are M
and the percent that are N
can be determined. Remember that each individual has two genes at each locus, so for every 1000 individuals
there will be 2000 genes
(
1000 x 2 = 2000
). If 600
individuals are blood type M
, then each has two copies of M
allele or 1200
total M
alleles. If 300
individuals are blood type MN
, then each has one copy, or 300
total copies of the M
allele and one copy or 300
total copies of the N
allele. If 100
individuals are blood type N
, then each has two copies of the N
allele or 200
total N
alleles. Using this information, it is then a simple matter to determine the frequency of the M
allele and the frequency of the N
allele in this population. The frequency of the M
allele in this population equals 1500/2000,
which equals 0.75
or 75
percent
. The total number of M
alleles is 1500
and the total number of M
and N
alleles is 2000
. The frequency of the N
allele in this population equals 500/2000,
which equals 0.25
or (
25 percent)
. Notice that the frequency of the M
plus the frequency of N
equals 1.0
or (
100 percent)
. It is impossible
to have more than 100
percent of the alleles at a given gene locus.
Example 2:
3
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PHENOTYPE
NUMBER OF
INDIVIDUALS
GENOTYPE
NUMBER OF M
ALLELES
NUMBER OF N
ALLELES
M
75
MM
150
0
MN
75
MN
75
75
N
50
NN
0
100
200 individuals
225 genes
175 genes
The frequency of M
in this population equals 225/400,
which equals 0.56
(or 56
percent). The frequency of the N
allele equals 175/400,
which equals 0.44 (or 44
percent). RANDOM AND NON-RANDOM MATING
There is much interest in defining the genetic population because this is a convenient way to
describe a population that shares a gene pool. In order to share a gene pool equally, mating must be random. Random mating in a sexually reproducing species, such as humans, means
that the statistical chance of mating with any individual of a particular genotype is proportional to the frequency of that genotype among members of the opposite sex. For example, blood groups are not visible characteristics so they should not play any role in mate selection. To use Example 1 above, the likelihood that a woman will mate with a man who is blood type M
is 60 percent
because 60 percent
of the male breeding population is blood type M
. If mating is non-random, then women of blood type M will mate with men of blood type M
either more or less than 60 percent
of the time. If this occurs then it is necessary to look for the reasons for the departure from random mating. Non-random mating is frequently the result of social factors, especially for traits that are outwardly visible such as skin color; mating among humans is not random with respect to skin color. Another term for non-random mating is assortative mating. Assortative mating with respect to one trait usually leaves mating with respect to other traits essentially random. For
example, tall men may mate with tall women but since stature is not related to blood type, mating with respect to the ABO and MN
blood systems should be random. There are two types of assortative mating: positive and negative
. Positive assortative mating refers to preferential matings between like genotypes as reflected through their phenotypes 4
while negative assortative mating refers to preferential mating between different or opposite genotypes as reflected through their phenotypes. A consistent pattern of assortative mating
can lead to changes in genotype and ultimately gene frequencies.
THE HARDY-WEINBERG EQUILIBRIUM FORMULA
If random mating is assumed, then there are certain conclusions that can be drawn about the distribution of genotypes. The most basic is that the proportion of genotypes in a population is basically stable. Once equilibrium has been reached, this means that gene and genotype frequencies do not change, they will not change unless some force is applied to the system. This is the entire idea behind the Hardy-Weinberg Equilibrium Model.
(1)
The Hardy-Weinberg Equilibrium Formula
is used to determine whether or not evolution is taking place in a given population with respect to a particular gene locus.
It tests whether or not the gene and genotype frequencies are changing from one generation to the next; so it tests whether or not the genetic make-up of a population
is stable.
(2) The Hardy-Weinberg Equilibrium Formula
is only a simple, basic model demonstrating idealized gene behavior; it does not tell us why change is occurring – only that change is or is not occurring. (3)
Any population, until proven otherwise, is assumed to be in Hardy-Weinberg Equilibrium
. For a population to be in Hardy-Weinberg equilibrium we assume the following
: (a) mating is random; (b) equal numbers of males and females, (c)
the population is infinitely large, and (d) no migration “in” or “out” of the population; and (e) no forces of evolution (natural selection, mutation, genetic drift, gene flow) are acting on the population.
(4)
Hardy-Weinberg demonstrates two
important concepts:
(a)
genes do not change in and of themselves; there must be some pressure or force to produce the change (four forces of evolution and non-random mating)
(b)
there is a consistency within the system; without some type of pressure being applied the system will remain stable and static – it will remain in a state of equilibrium
(5)
Four practical uses
of the Hardy-Weinberg Equilibrium Formula are to: 5
(a)
estimate the relative frequency of heterozygotes in cases of dominance and recessiveness at a single gene locus;
(b)
count specific alleles in a population;
(c)
count genotypes; and
(d)
count phenotypes
(6)
It is important to remember that Hardy-Weinberg
is a hypothetical case
of no evolution
. For evolution to be an impossibility, populations would have to meet all the criteria (
see No. 3
) so evolution is not only real but must take place.
The Hardy-Weinberg Equilibrium Formula: p + q = 1 (gene frequencies)
If we square both sides: (p + q)
2
=1
2
If we expand the formula: p
2 +
2pq + q
2 = 1
(genotype frequencies)
p always equals the frequency of the dominant gene
q always equals the frequency of the recessive gene
To determine gene
frequencies of the two alleles
:
p = p
2
+ ½(2pq) q = q
2
+ ½(2pq)
To determine genotype
frequencies:
dominant homozygote = p
2
heterozygote = 2pq
recessive homozygote
= q
2
To determine phenotype
frequencies:
dominant phenotype =
p
2 + 2pq
recessive phenotype =
q
2
Remember:
(1)
Whether determining gene, genotype, or phenotype frequencies, they must always
add up to 1.0
. If they do not, you have made a mistake.
6
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(2) The heterozygote
is phenotypically indistinguishable
from the dominant homozygote
in cases of simple dominance and recessiveness
. Example 3:
In a hypothetical population, a genetic anthropologist has gone in and determined the following genotypes and gene frequencies for earlobe attachment. From these data, this scientist wants to determine gene frequencies at this locus for the dominant and recessive alleles. Unattached earlobes (
E
) are the dominant form of the trait while attached earlobes (
e
) are the recessive form of the trait. At this particular gene locus, the following genotype distribution was determined:
(1)
Dominant homozygote (EE)
= 0.16
(or 16 percent)
of the individuals are dominant homozygotes for this trait and have unattached earlobes.
(2)
Heterozygote (Ee) = 0.48
(or 48 per
cent) of the individuals in this population are heterozygous at this locus and have unattached earlobes.
(3)
Recessive homozygote (ee)
= 0.36
(or 36 percent)
of the individuals in this population are recessive homozygotes and have attached earlobes.
The first step in determining the frequency of the E
and e
alleles is to place the genotype frequencies in the expanded Hardy-Weinberg Formula:
p
2
+ 2pq +
q
2
= 1.0
0.16
+ 0.48 + 0.36 = 1.0
The second step is to use the formula above for determining p
and q
. Remember, since “
E”
is the dominant allele it equals the frequency of “
p”
while “
e”
equals the frequency of “
q”
because it is the recessive allele.
p =
p
2 + ½ (2pq)
= 0.16 + ½ (0.48)
= 0.16 + 0.24
= 0.40 (or 40%)
q
=
q
2
+ ½ (2pq) = 0.36 + ½ (0.48)
=
.36
+
.24
7
=
0.60 (or 60%)
The frequency of the dominant allele equals 0.40
meaning that 40 percent
of the alleles at this locus are dominant. The frequency of the recessive allele equals 0.60 meaning
that 60
percent
of the alleles at this locus are recessive. Together they add up to 100 percent of the alleles at this locus: 0.4 + 0.6 = 1.0 or 40 + 60 = 100
. It is important to remember that these are alleles, not individuals.
Example 4:
In another population, another genetic anthropologist examines the same trait as in Example 3. This scientist is also interested in determining the gene frequencies for both alleles at this locus. Initial research has determined the following genotype distribution:
EE
Ee
ee
p
2
+
2pq
+
q
2
=
1.0
0.25
+
0.50
+
0.25
=
1.0
p
=
p
2
+
½ (2pq)
=
0.25
+
½ (0.50)
=
0.25
+
0.25
=
0.50 (or 50%)
q
=
q
2
+
½ (2pq)
=
0.25
+
½ (0.50)
=
0.25
+
0.25
=
0.50 (or 50%)
In this population, 50 percent
of the alleles are dominant (
E
), and 50 percent
of the alleles
are recessive (
e
). It is also possible to determine both genotype and phenotype frequencies from gene frequencies.
Example 5:
In another population, other researchers interested in the attachment of earlobes have determined that the frequency of E
equals 0.80
(
80 percent
of the alleles are dominant
) 8
and that the frequency of e
equals 0.20
(
20 percent
of the alleles are recessive
). Their goal is to determine genotype and phenotype distributions in this population. The first step is to substitute the above data into the expanded Hardy-Weinberg formula. Remember
, p = E = 0.80
and q = e = 0.20
p
2
+
2pq
+
q
2
=
1.0
(0.80)
2
+
2(0.80 x 0.20)
+
(0.20)
2
=
1.0
0.64
+
0.32
+
0.04
=
1.0
In this population, 64 percent
of the individuals are dominant homozygotes
, 32 percent
are heterozygotes
, and 4 percent
are recessive homozygotes
. In order to determine phenotype frequencies in this system of simple dominant and recessive genes the dominant phenotype equals 0.64 + 0.32 = 0.96
(or 96 percent)
of this population will exhibit the dominant phenotype, or unattached earlobes. The recessive phenotype equals the frequency of the recessive homozygote. In this case 0.04
(or 4 percent)
of this population will exhibit the recessive trait, or attached earlobes. Example 6:
In yet another population it was determined that the frequency of the E allele
is 90 percent
and the frequency of the e allele
equals 10 percent
. The researcher here was also interested in determining genotype and phenotype frequencies with respect to this locus in this population. p
2
+
2pq
+
q
2
=
1.0
(0.90)
2
+
2(0.90 x 0.10)
+
(0.10)
2
=
1.0
0.81
+
0.18
+
0.01
=
1.0
Genotype distribution:
EE = p
2
= 0.81
Ee = 2pq = 0.18
ee = q
2
= 0.01
Under the Hardy-Weinberg Equilibrium Model
these genotype frequencies are what we would expect in the next generation. If the gene and genotype frequencies change from one
generation to the next then evolution has taken place.
The dominant phenotype
= 0.81 + 0.18 = 0.99
(or 99 percent) of this population has unattached earlobes.
9
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The recessive phenotype
= 0.01
(or 1 percent)
of this population has attached earlobes.
In order to determine if evolution with respect to this trait has occurred in this population, the scientists then calculated gene frequencies
from the genotype frequencies of the next generation:
EE
Ee
ee
p
2
+
2pq
+
q
2
=
1.0
0.81
+
0.18
+
0.01
=
1.0
p
=
p
2
+
½(2pq)
=
0.81
+
½(0.18)
=
0.81 +
0.09
=
0.90 = p
q
=
q
2
+
½(2pq)
=
0.01
+
½(0.18)
=
0.01
+
0.09
=
0.10
=
q
The gene frequencies for these alleles did not change so evolution has not taken place with
respect to this gene locus.
THE PRACTICAL USES OF HARDY-WEINBERG
Anthroposcopy
Humans show a lot of variation in their morphological, physiological, and biochemical traits. Some of these traits, such as the ABO blood group are under the relatively simple genetic control of single genes with two or more alleles. Other traits such as skin color, stature, weight, and intelligence are inherited in a complex manner controlled by a large number of loci and genes with multiple alleles. When many alleles and loci are involved in the inheritance of a trait, this is called a polygenic trait. Exactly how many genes and loci are involved in inheritance of genetically complex traits is not known. As a result of this genetic complexity, the inheritance of polygenic traits is analyzed differently than those traits under simple genetic control. Since the variation of simple genetic traits is discontinuous (i.e., either one trait or the other is expressed), the inheritance of simple genetic traits can be studied through Mendelian ratios and pedigree analysis. The variation of polygenic traits, however, is continuous. For example, skin color varies continuously from light to dark with all shades in between. Due to this continuous variation, the inheritance of polygenic traits must be studied using statistical techniques. 10
Another approach to polygenic studies uses twins. Polygenic traits are often investigated among identical twins that are raised in the same or different environments. In this manner the effects of the environment can be determined because the genotypes of the twins are identical. Differences in the traits of identical twins are due to genotype-environmental interactions in different environments. One example of a genotype-environmental interaction in a polygenic trait is that of a poor diet during childhood resulting in individuals who are shorter than their genetic potential.
Since a primary focus of human biology research is the description and explanation of group similarities and differences, physical anthropologists have devised many methods for collecting data on the phenotypic expression of single genes and polygenic traits. For example, anthropometry is the technique of measuring and recording quantitative differences in body form or morphology. Anthroposcopy
, on the other hand, is a non-
metrical technique of visual observation and description of traits such as tongue rolling and earlobe attachment. Traits which are inherited in a simple fashion as well as those that are inherited in a complex manner may be subjected to both quantitative and non-quantitative techniques of analyses. Anthropometry and anthroposcopy play important roles in the studies of growth and development patterns and in research on the effects of environmental
factors such as nutrition, disease, and climate. Anthropometric and anthroposcopic research by industry and the military are conducted for solutions to problems of the design of clothing
and equipment.
HARDY WEINBERG IN A NUTSHELL:
Conditions for a Hardy Weinberg Equilibrium:
1) random mating across the population (panmixia; 2)equal number of males and females; 3) the population is infinitely large; 4) there is no “in” or “out” migration; 5) natural selection, mutation, and genetic drift are not acting upon the population.
Variables and concepts for mathematical computations:
p
=
the frequency of the dominant allele
q
=
the frequency of the recessive allele
p
+
q
=
1
p
2
+
2pq
+
q
2
=
genotype frequencies
p
2
+
2pq
=
the frequency of the dominant phenotype
q
2
=
the frequency of the recessive phenotype
11
For example
, if 25%
of your lab section cannot roll their tongue, then the frequency of the recessive phenotype and genotype is 0.25
. The frequency of the recessive allele (
q
) is equal
to the square root of q
2
. In this case, q is equal to the square root of 0.25,
which equals 0.5
.
Since p + q = 1
, the frequency of the dominant allele (
p
) is also equal to 0.5
because 1 – 0.5 = 0.5
. Genotype frequencies are p
2 + 2pq + q
2
= 1
or 0.25 + 0.50 + 0.25 = 1. According to Hardy-Weinberg, in a class of 200 students we would have the following genotype distribution:
0.25
*
200
=
50 students
=
p
2
=
dominant homozygote
0.50
*
200
=
100 students
=
2pq
=
heterozygote
0.25
*
200
=
50 students
=
q
2
=
recessive homozygote
In order to determine gene and genotype frequencies you must begin
with the frequency of the recessive phenotype,
which equals
the frequency of the recessive homozygote
. Once you have taken the square root of the frequency of the recessive phenotype
, the rest is simple. ___________________________________________________
For example, suppose in our hypothetical class of 200 students, we have determined that 160 have a widow’s peak while 40
do not
. The 40 who do not have this trait are all recessive homozygotes
.
In order to determine the frequency of the recessive homozygotes
we divide by 40 by 200, which equals 0.20 (or 20 percent). In other words, .20 are homozygous recessive for the trait, which means
.20 = q
2
. In order to determine the gene frequency of q we take the square root of q
2
, which, in this case, equals .20. The square root of 0.20 is 0.45. So q equals 0.45. If q equals 0.45 then p equals 0.55 since 1 – 0.45 = 0.55. These are our gene frequencies. Once we have gene frequencies we can determine genotype frequencies. Remember, the formula for genotype frequencies is p
2 + 2pq + q
2
= 1, so in this case, p
2
= (0.55)
2
, 2pq = 2(0.55)(0.45), and q
2 = (0.45)
2
. The dominant homozygote frequency = 0.3025; the heterozygote frequency = 0.496, and the frequency of recessive homozygotes equals 0.2025 (which is close to 0.20).
___________________________________________________
+++++++++++++++++++++++++++++++++++++++++++++++++++++++
++
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POPULATION GENETICS LAB REPORT
Your Lab Report
should include the following:
1)
A cover sheet that includes the LAB number, your instructor’s name, your name, and the date
2)
Answers to the theoretical exercises
3)
Answers to the practical exercises, including:
a)
A copy of your lab data sheets
b)
Calculations of gene and genotype frequencies for your lab section
c)
Calculations of gene and genotype frequencies for the entire class
d)
A concise, well-written answer to the essay question
_________________________________________
QUESTIONS: Answer the following Questions. (Worth 5 points each):
1.
What are the five
assumptions of the Hardy-Weinberg Equilibrium Model?
1.) The population is infinitely large.2.) Equal number of males and females.3.) Random mating across the population.4.) There is no “in” or “out” migration.5.) Natural selection, mutation, and genetic drift are not acting upon the population.
2a.
Gene counting: In a hypothetical population of 500
individuals, a genetic anthropologist observed the genotype distribution for the MN
blood system that follows. Remember that M and
N
are co-dominant where both M and N are expressed. Complete this table determining the frequency of both the M and N alleles
(show all your calculations).
PHENOTYPE
NUMBER OF
INDIVIDUALS
GENOTYPE
NUMBER OF
M ALLELES
NUMBER OF
N ALLELES
M
300
MM
600
0
MN
150
MN
150
150
N
50
NN
0
100
13
Total
500 individuals
750 genes
250 genes
2b. Briefly explain the results frequencies of blood in this population:
To find phenotypic frequencies in population we take that specific pheotypic population and divide it by total number phenotype possible in that population.
Frequency of M allele: 750/1000=75% Frequency of N allele: 250/1000=25%
3. Briefly explain the difference between the following terms:
Recessive-
Only expressed in the phenotype when an individual has two copies of the allele.
Dominant-
Expressed in the phenotype even with just one copy of the allele.
Co-Dominant- Both alleles are fully expressed in the phenotype when present together.
+++++++++++++++++++++++++++++++++++
Hardy-Weinberg Equilibrium Problems
.
4. Determining Genotype and Phenotype Distributions from Gene Frequencies.
In a
hypothetical population, a scientist has determined the following frequencies for the alleles that govern the presence or absence of a widow’s peak
; p = W = 0.60 and q =
w = 0.40
. The dominant allele (
W
) codes for the presence of this trait while the recessive allele (
w
) codes for the absence of this trait. A)
Determine the genotype
and phenotype
distributions for this trait in this population (show all your calculations
):
Genotype frequencies:
Dominant homozygote (WW) = p^2 = (0.60)^2 = 0.36
Heterozygote (Ww) = 2pq = 2(0.60)(0.40) = 0.48
Recessive homozygote (ww) = q^2 = (0.40)^2 = 0.16
14
Phenotype frequencies:
Dominant phenotype (presence of widow's peak) = p^2 + 2pq = 0.36 + 0.48 = 0.84
Recessive phenotype (absence of widow's peak) = q^2 = 0.16
B) In a few sentences, explain your results.
In this hypothetical population, the genotype distribution shows that 36% of individuals are homozygous dominant (WW), 48% are heterozygous (Ww), and 16% are homozygous recessive (ww) for the widow's peak trait. Regarding phenotype distribution, 84% of the population exhibits the dominant phenotype (presence of widow's peak), while 16% exhibit the recessive phenotype (absence
of widow's peak). These results indicate that the trait is predominantly expressed in the population due to the high frequency of the dominant allele.
5. Determining Gene Frequencies from Genotype Distribution.
In a hypothetical population, a genetic anthropologist has determined the following genotypes and genotype frequencies for the presence or absence of a widow’s peak. From these data, this scientist wants to determine gene frequencies at this locus for the dominant and recessive alleles. Presence of a widow’s peak (
W
) is the dominant form of the trait while absence of a widow’s peak (
w
) is the recessive form of the trait. At this particular gene locus, the following genotype distribution was determined:
(1)
dominant homozygote (
WW
) = 0.16
(or 16 percent
of the individuals are dominant homozygotes for this trait and have widow’s peaks)
(2)
heterozygote (
Ww
) = 0.48
(or 48 percent
of the individuals in this population are heterozygous at this locus and have widow’s peaks)
(3)
recessive homozygote (
ww
) = 0.36
(or 36 percent
of the individuals in this population are recessive homozygotes and do not have widow’s peaks)
5a) Determine the frequency of the W and w alleles in this population (show all your calculations): W = 0.16 + 0.5 * 0.48 = 0.16 + 0.24 = 0.40
w= 0.36 + 0.5 * 0.48 = 0.36 + 0.24 = 0.60
15
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5b) Explain your results in a few sentences. In this population, the frequency of the dominant allele (W) is 0.40, and the frequency of the recessive allele (w) is 0.60. This means that 40% of the alleles at this gene locus are for the presence of a widow's peak (W), while 60% are for the absence of a widow's peak (w). These
allele frequencies provide insights into the genetic makeup of the population regarding this specific trait.
+++++++++++++++++++++++++++++++++++++++++++
IN THE NEXT SECTION YOU WILL USE THE FOLLOWING LIST OF MENDALIAN TRAITS
Practically Using Hardy Weinberg:
You will examine twelve easily observed traits in order to look at human variation. Your will determine whether or not you possess each of these traits and then determine your probable genotype (dominant homozygote, heterozygote, recessive homozygote). Nest you will collect counts of these traits from your classmates. You will then determine gene,
genotype, and phenotype frequencies for your lab section, and then you will do this for the all 102 student.
TRAIT 1: Widow’s Peak
In some people the hairline drops downward and forms a distinct point in the center of the forehead. It results from the action of a certain dominant gene (
W
). With the help of your lab
section determine your phenotype for this trait.
TRAIT 2: Attached Earlobes
Most people have earlobes that hang free but when a person is homozygous
for a certain gene (
e
) the earlobes are attached directly to the side of the head so that there is no lobe hanging free. With the help of your lab section determine your phenotype for this trait. You will find that there is considerable variation in the size and appearance of earlobes; this is due to other genes, so concentrate only on whether the lobes hang free or not.
TRAIT 3: Darwin’s Point
The Darwin’s point is a small projection of cartilage on the helix of the ear. The presence of the point is dominant (
D
) and the absence is recessive (
d
). Some individuals may have the point in only one ear and the size may vary. Check both ears for this trait. TRAIT 4: Hair Whorl Pattern
16
Hair whorl pattern refers to the direction that hair naturally falls from the “part” on the top of your head. A pattern that is clockwise is dominant (
H
) and a pattern that is counterclockwise (
h
) is recessive.
TRAIT 5
: Pigmented Iris
When a person is homozygous for a certain recessive gene (
p
) there is no pigment in the front part of the eye and the iris, which reflects blue light, appears blue. A dominant allele (
P
) of this gene causes pigment to be deposited in the layer, which masks this blue reflection to various degrees. Other genes determine the exact nature and density of this pigment and thus we have brown, hazel, green, grey, or blue eyes. We will concern ourselves only with the presence or absence of pigment. Grey eyes should be considered un-
pigmented. With the help of your lab section determine your phenotype.
TRAIT 6: Tongue Rolling
A dominant gene (R) gives some people the ability to roll their tongue into a U-shape when the tongue is extended from the mouth. Others who do not possess this gene can do no more than produce a slight downward curve of the tongue when it is extended. Determine your phenotype for this trait.
TRAIT 7: Tongue Folding
The ability to turn the tip of the tongue back upon the body of the tongue is inherited as a recessive trait (
f
). Those people who can fold their tongues have homozygous recessive genotypes. Those people who cannot fold their tongue on the dominant phenotype. Tongue rolling and tongue folding are inherited independently.
TRAIT 8
: Hitchhiker’s Thumb
The presence or absence of this trait can be determined by bending the distal joint of the thumb as far back as possible. It is found that some persons can bend it back until there is almost a 45 degree angle. A recessive gene (
t
) seems to determine this trait.
TRAIT 9: Bent Little Finger
A dominant gene (
B
) causes the last joint of the little finger to bend inward toward the fourth finger. Lay both hands flat on the table, relax the muscles and note whether you have
a bent or strait little finger. 17
TRAIT 10: Long Palmar Muscle
Persons homozygous for a certain recessive gene (
l
) have a long palmar muscle, which can be detected by examination of the tendons that run over the insides of your wrist. Clench your fist tightly and flex your hand. Now feel the tendons. If there are three, you have the long palmar muscle. If there are only two, you do not have this muscle. Examine both wrists as this muscle is sometimes present in only one because of variations in other genes.
TRAIT 11: Mid-Digital hair
Observe the middle segment of all your fingers and note the presence of absence of hair on this segment. The presence of hair is dominant (
M
) and the absence is the recessive (
m
) phenotype.
TRAIT 12: Interlocking Fingers and Thumbs
Fold your hands interlocking the fingers. If the left thumb is over the right thumb this is the dominant (
I
) position. If the right thumb is over the left thumb this is the recessive (
i
) position.
6. Record observation about your own traits
on Data Sheet I (worth 10 points):
DATA SHEET I – Your Observations
Trait
All Possible
Genotypes Dominant
Phenotype
(check if yes)
Recessive
Phenotype
(check if yes)
Your
Possible
Genotype
Widow’s Peak
(Dominant)
WW, Ww, ww
✔ ww
Attached
Earlobes
(recessive)
EE.Ee, ee
✔ EE, Ee
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Darwin’s Point
domin
(Dominant)
DD,Dd,dd
✔ dd
Hair Whorl
(Dominant)
HH, Hh, hh
✔ HH, Hh
Pigmented Iris
(Dominant)
PP, Pp, pp
✔ pp
Tongue Rolling
(Dominant)
RR, Rr, rr
✔ RR,Rr
Tongue Folding
(recessive)
FF,FF,ff
✔ FF,Ff
Hitchhiker’s
Thumb
(recessive) TT,Tt,tt
✔ TT,tt
Bent Little Finger
(Dominant)
BB, Bb, bb
✔ bb
Palmar Muscle
(recessive)
LL, Li, ii
✔ Ll,ll
Mid-Digital Hair
(Dominant)
MM, Mm, mm
✔ mm
Interlocking Fingers
& Thumb (left over
right: Dominant)
II, Ii, ii
✔ ii
7. Record observations about your lab section
: Note: you will receive data for other lab sections provided on additional Data Sheet IIs from your instructor (worth 10 points).
DATA SHEET II – Observations for Your Lab Section
(Section ____):
Trait
Total
Number
# of
Dominant
Phenotypes
% of
Total
# of
Recessive
Phenotypes
% of
Total
19
Widow’s Peak
(Dominant)
5
1
20% 4
80%
Attached Earlobes
(recessive)
5
3
60%
2
40%
Darwin’s Point
(Dominant)
5
4
80%
1
20%
Hair Whorl
(Dominant)
5
2
40%
3
60%
Pigmented Iris
(Dominant)
5
5
100%
0
0%
Tongue Rolling
(Dominant)
5
3
60%
2
40%
Tongue Folding
(recessive)
5
0
0%
5
100%
Hitchhiker’s
Thumb
(recessive) 5
3
60%
2
40%
Bent Little Finger
(Dominant)
5
1
20%
4
80%
Palmar Muscle
(recessive)
5
4
80%
1
20%
Mid-Digital Hair
(Dominant)
5
3
60%
2
40%
Interlocking Fingers &
Thumb (left over
right: Dominant)
5
2
40%
3
60%
20
8. In Data Sheet III (below) calculate the gene frequencies and the genotype frequencies for your lab section
from the phenotype frequencies you recorded on you section’s Data Sheet II (worth 10 points).
DATA SHEET IIIA – Gene and Genotype Frequencies for Your Lab Section
(Section__)
REMEMBER: p +
q = 1
AND
p
2
+
2pq
+ q
2
=1
Trait
Gene
Frequenc
y (p)
Gene
Frequenc
y (q)
Genotype
Frequenc
y (p
2
)
Genotype
Frequenc
y (2pq)
Genotype
Frequenc
y (q
2
)
Widow’s Peak
.20
.80
0.04
0.32
0.64
Attached
Earlobes
.60
.40
0.36
0.48
0.16
Darwin’s Point
.80
.20
0.64
0.32
0.04
Hair Whorl
.40
.60
0.16
0.48
0.36
Pigmented Iris
1.00
0.00
1.00
0.00
0.00
Tongue Rolling
.60
0.40
0.36
0.48
0.16
Tongue Folding
0.00
1.00
0.00
0.00
1.00
Hitchhiker’s
Thumb
0.60
0.40
0.36
0.48
0.16
Bent Little Finger
0.20
0.80
0.04
0.32
0.64
Palmer Muscle
0.80
0.20
0.64
0.32
0.04
Mid-Digital Hair
0.60
0.40
0.36
0.48
0.16
Interlocking
Fingers & Thumb
0.40
0.60
0.16
0.48
0.36
9. Calculate the gene and genotype frequencies for the entire class by first combining phenotype frequencies from ALL lab sections
(All Data Sheet IIs) (worth 10 points).
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DATA SHEET IV – Class Observations (Combine Data for ALL Lab Sections
)
Trait
Total
Number
# of
Dominant
Phenotypes
% of
Total
# of
Recessive
Phenotypes
% of
Total
Widow’s Peak
37
22
59.5%
15
40.5%
Attached Earlobes
37
20
54.1%
17
45.9%
Darwin’s Point
37
10
27.0%
27
73.0%
Hair Whorl
37
19
51.4%
18
28.6%
Pigmented Iris
37
25
67.6%
12
32.4%
Tongue Rolling
37
31
83.8%
6
16.2%
Tongue Folding
37
8
21.6%
29
78.4%
Hitchhiker’s
Thumb
37
23
62.2%
14
37.8%
Bent Little Finger
37
9
24.3%
28
75.7%
Palmer Muscle
37
14
37.8%
23
62.2%
Mid-Digital Hair
37
5
13.5%
32
86.5%
Interlocking
Fingers & Thumb
37
24
64.9%
13
35.1%
10. Calculate the gene frequencies and the genotype frequencies for the entire class (
all lab sections
) from the phenotype frequencies on Data Sheet IV (worth
10 points). DATA SHEET V – Gene and Genotype Frequencies for ALL lab sections
REMEMBER: p +
q = 1
AND
p
2
+
2pq
+ q
2
=1
22
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Trait
Gene
Frequenc
y (p)
Gene
Frequenc
y (q)
Genotype
Frequenc
y (p
2
)
Genotype
Frequenc
y (2pq)
Genotype
Frequenc
y (q
2
)
Widow’s Peak
0.595
0.405
0.354025
0.48015
0.1656
Attached
Earlobes
0.541
0.459
0.292681
0.495579
0.211739
Darwin’s Point
0.270
0.730
0.0729
0.1971
0.7300
Hair Whorl
0.514
0.486
0.264196
0.499884
0.235924
Pigmented Iris
0.676
0.324
0.457376
0.438048
0.104576
Tongue Rolling
0.838
0.162
0.702244
0.271332
0.026424
Tongue Folding
0.216
0.784
0.046656
0.338688
0.614656
Hitchhiker’s
Thumb
0.622
0.378
0.386884
0.470196
0.142936
Bent Little Finger
0.243
0.757
0.059049
0.366651
0.574801
Palmer Muscle
0.378
0.622
0.142884
0.469716
0.387396
Mid-Digital Hair
0.135
0.865
0.018225
0.233775
0.747600
Interlocking
Fingers & Thumb
0.649
0.351
0.421201
0.454299
0.124800
11. ESSAY:
Please remember: An essay is a group of paragraphs composed of sentences with internal consistency and flow, all written in the same verb tense. An essay includes 1)
an introductory paragraph providing a thesis statement or proposal, 2)
a central portion typically, three paragraphs, providing three lines of argument or examples to support the thesis statement, and finally 3)
a summary/concluding paragraph. Laundry lists of numbered points are NOT acceptable
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ESSAY QUESTION (worth 25 points): In a well thought out essay, explain your results from this lab. How do your traits compare to your lab section and to the class as a whole? How does your lab section, as an example of a sub-population, compare with the class as a whole (which is an example of a population)? Are there significant differences between your sub-population and the population? What do you think this variation or lack of variation means?
In this lab, I looked at different traits that people have, like the shape of their hairline or whether they can roll their tongue. I got to check my own traits and then compared them with others in my lab group and the whole class. While analyzing the results of this lab. I made a couple key observations about the distribution of traits within my own genotype, my lab section, and the entire class.
I will first examine my own traits. Based on my results, I now know that I possess certain
recessive phenotypes such as a widow's peak, Darwin's point, pigmented iris, bent little finger, mid-digital hair, and right over left and I also have dominant phenotypes like attached earlobes, hair whorl, tongue rolling, tongue folding, hitchhikers thumb, and palmar muscle. This mix of dominant and recessive traits that I have reflects the genetic diversity inherited. Now moving on to the observations I made in my lab section, it is evident that trait expression varies. For example, some characteristics, such as a pigmented iris and Darwin’s point are highly prevalent dominant phenotypes, whereas other characteristics, such as widows’ peak and tongue folding, are primarily recessive. This variation in the lab section demonstrates the range of genetic diversity among people in a kind of smaller group.
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When I Compare my lab section to the entire class, it becomes apparent that there are some similarities but also notable differences. For instance, certain traits like pigmented iris and tongue rolling show similar patterns of dominance across both the lab section and the class as a whole. However, there are significant differences in the prevalence of other traits such as Darwin's point and mid-digital hair. These differences show that although certain characteristics might be more common in nature, others show more diversity.
The differences that I saw between my lab group and the class as a whole serve to emphasize the idea that genetic variety exists within populations. Subpopulations, like the people
in my lab, can have distinct characteristic distributions because of things like natural selection, gene flow, and genetic drift operating within smaller groups. A wider range of genetic diversity, however, becomes clear when the entire class is viewed as a population, illustrating the complex relationship between environmental and genetic influences on phenotypic variation.
In conclusion, this lab shows how complex genetic variation is both inside and between groups. Certain features are very diverse between populations and subpopulations, whereas others can demonstrate consistency. This variance highlights the significance of genetic diversity in preserving healthy and solid populations and is an essential trait of evolution.
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