Download Exploration 13 - Warner Pacific College

Genes, Pedigrees, and Populations
Investigations Into the World of Genetics
BIO 101
Name_________________
Objectives
The purpose of this exploration is to . . .
1. review the terms gene, allele, phenotype, genotype, heterozygous, and homozygous, and practice using
them correctly;
2. analyze several human pedigrees to determine the type of inheritance exhibited by the trait;
and
3. predict the likelihood that particular individuals in the families will have the trait that is the
focus of the pedigrees.
4. understand the significance of the Hardy-Weinberg law in predicting that allele frequencies
remain the same from generation to generation;
5. determine gene frequencies under the conditions of codominance and recessive inheritance;
and
6. explore the impact of effects such as selection and migration on allele frequencies.
Part 1 Genes and You.
The example we will explore in class today, PTC taste testing, is an example of alleles that
exhibit dominance/recessiveness. Pedigree studies of human families have indicated that the
ability to taste phenylthiocarbamide (PTC) is due to the presence of a dominant allele (T) and
that the inability to taste the chemical occurs in homozygous recessive individuals (tt). There are
two phenotypes (tasters and non-tasters), but three genotypes (TT, Tt, and tt).
Procedure Part 1
You will be given a small piece of filter paper that has been lightly saturated with PTC. Before
you place the PTC paper on your tongue, take a piece of untreated paper and taste it. This wills
serve as a control so that you can readily determine the difference between the taste of the paper
and the taste of PTC.
Fill out the questions for “Part 1” on your worksheet now.
Most hereditary traits in humans and other species are controlled by multiple genes with multiple
alleles that utilize complex gene interactions to result in the organisms phenotype. The
following human characteristics are controlled by a single gene with two alleles. Determine
your phenotype for each of the following traits. Also determine your genotype if possible. If
you have the dominant phenotype, you will not know whether you are heterozygous or
homozygous. What procedures could you use to determine whether or not you are
homozygous? Other traits that are easily observed and controlled by a single gene are listed
below. Try to identify both your genotype and phenotype for each of these.
Trait
1. Bent Pinky (B,b) Dominant allele causes distal segment of fifth finger to bend distinctly
inward toward fourth (ring) finger
2. Tongue Rolling (R,r) Dominant allele in
heterozygous or homozygous condition
allows people to roll their tongues into a
tube-like shape.
3. Blue eyes (E,e) Homozygous recessive
individuals lack pigment in the iris, others have iris pigment color determined by other genes.
4.Widows Peak (W,w) Dominant allele causes
individuals to have V-shaped front hairline.
5. Thumb crossing (T,t) In relaxed interlocking of fingers, the left thumb over the right
indicates the dominant allele is present; homozygous recessives naturally place right thumb over
left.
6. Attached Earlobe (A,a) Dominant allele causes
individuals to have detached ear lobes.
7. Hitchhiker Thumb (H,h) Homozygous recessives can bend the distal joint of the thumb
backward to nearly a 90°° angle.
8. Red-green colorbind (C,c)Sex linked trait (on the X chromosome). Recessive allele on the X
chromosome causes the color blindness.
9. Hairy Ears (M,m) Sex linked trait (on the Y chromosome). Dominant allele causes individual
to have hair growing on the outside edge of the ears.
10. Pattern Baldness (P,p) A sex influenced trait. Dominant allele causes premature loss of hair
on top and front of head in heterozygous or homozygous dominant males.
11. Finger length (F,f) A sex influenced trait that involves the length of the index finger relative
to the ring finger. In males, the allele for short index finger is dominant; in females it is
recessive
Part 2 Human Pedigree Analysis
Introduction
The inheritance of human traits is typically determined using a technique called pedigree
analysis. Pedigrees are “family trees” that show which individuals in a family exhibit a
particular trait and how they are related to other affected and non-affected family members.
This information, plus a basic understanding of Mendelian genetics, is used to make hypotheses
about the inheritance of the trait and to make predictions about the probability that a child will
have the trait. Genetic counselors use pedigree analysis, among other skills, in their work.
In a pedigree, circles represent females, while squares represent males. A diamond indicates that
the sex of the individual is unknown. Shaded symbols indicated that the individual exhibits the
phenotype under consideration. For example, in a pedigree that examines the inheritance of
sickle cell disease in a family, a shaded symbol indicates an individual with this disease. A
horizontal line connects parents; the vertical line from them leads to their offspring. All of their
offspring are “sibs” and are connected by a horizontal “sibship line.” They are placed from left
to right in order of birth. Connected diagonal lines indicate twins.
Analyzing inheritance patterns revealed in pedigrees
A pedigree is typically used to determine the type of inheritance involved for a trait and the
probability that a specific family member will exhibit the trait in question. Inherited traits due to
a single gene may be inherited in one of four general ways:
•
Autosomal recessive inheritance. The trait is due to a gene on an autosomal
chromosome (chromosomes 1 to 22) and is expressed only in the homozygous recessive
condition. Pedigree “tip-offs” for this kind of inheritance: both males and females are
affected; unaffected parents have an affected child.
•
Autosomal dominant inheritance. The trait is due to a gene on an autosomal
chromosome and is expressed whenever the gene is present (either homozygous
dominant or heterozygous condition). As a rule of thumb, most individuals with a
dominant trait are heterozygous (unless information in the pedigree reveals otherwise).
Pedigree “tip-offs:” both males and females are affected; affect individuals have at least
one parent who also exhibits the trait.
•
X-linked inheritance (also known as “sex-linked inheritance”). The trait is due to a
gene on the X chromosome. Most traits of this type are recessive and are expressed in
males who receive an X chromosome carrying the gene from their mothers. Traits of
this type are expressed in females only if they are homozygous, that is, if they received X
chromosomes carrying the gene from both parents. Thus, males exhibit the trait far more
often than females—in fact, most pedigrees of this type will show only males with the
trait.
•
Y-linked inheritance. This type of inheritance is due to a gene carried on the Y
chromosome and, except for genes that cause an individual to be male, is very rare.
Pedigrees for traits with this type of inheritance will show that only males exhibit the
trait and that all affected males have fathers who also have the trait.
In this lab, you will work in pairs to analyze several pedigrees. This is an opportunity for you to
review several genetics terms and concepts. Keep in mind that “real life” situations are often
more ambiguous that those described above. For example, penetrance is a phenomenon that
makes pedigree interpretation difficult. It refers to the likelihood that an individual who has the
genotype for a particular trait will actually have the phenotype for the trait. If a dominant trait
has 100 percent penetrance, then every individual who has a dominant allele will exhibit the trait
associated with it. If a dominant trait has 90 percent penetrance, then 90 percent of the
individuals who have the allele will have the corresponding phenotype, but 10 percent of them
will not show the trait (it does not “penetrate” in them).
Procedure Part 2
1. Obtain a set of pedigrees from your instructor.
2. In pairs, discuss the pedigrees and answer the following questions for each one:
a. Is the gene associated with the trait on an autosomal chromosome or the X
chromosome? Explain.
b. Is the trait conferred by a dominant or recessive gene? Explain.
c. What is the probability that individual A will have the trait? Explain your reasoning.
d. What is the probability that individual B will have the trait? Explain your reasoning.
Fill out the table for “Part 2” on your worksheet now.
Part 3. Genes in Populations, the Hardy-Weinberg Equilibrium
Introduction
The English mathematician G.H. Hardy, and the German physician, W. Weinberg formulated
the basic law of population genetics independently in 1908. It is called the Hardy-Weinberg
equilibrium in their honor. This principle allows population geneticists to calculate the
frequency of alleles of a gene in a randomly interbreeding group of plants or animals. Hardy and
Weinberg demonstrated that in such natural populations allele frequencies reach equilibrium
(that is, the frequencies of the different alleles of a gene remain constant generation after
generation), provided that no disturbing effects occur such as those caused by mutation,
migration, selection pressure, non-random mating, or random genetic drift (which occurs
in small populations).
For example, consider a gene that has two alleles, A and a. Let p equal the frequency, or
percent, of the A alleles in a population and q equal the frequency, or percent, of the a allele.
Then, p + q = 1. If random mating occurs, the population should consist of p2 AA individuals,
2pq Aa individuals and q2 aa individuals (see the Punnett square below).
Alleles carried by sperm
in the population
Alleles carried by
ova in population
p=A
q=a
p=A
p2 = AA
pq = Aa
q=a
pq = Aa
q2 = aa
If the table above is confusing to you, try substituting values for p and q. For example, instead
of p sperm/ova with the A allele and q with the a allele, substitute 0.7 sperm/ova with A and 0.3
with a. Then, what percent of offspring with the genotype aa would you expect?
The generation offspring represented in the Punnett square will consist of p2 (AA) + 2pq (Aa) +
q2 (aa) individuals. According to the Hardy-Weinberg law, the next generation should consist of
exactly the same frequencies of each genotype.
If a population is in Hardy-Weinberg equilibrium, it will stay in equilibrium, generation after
generation. Neither the allele frequencies nor the genotype frequencies will change as long as
the following apply: random mating occurs; there is no selection for particular genotypes; there
is no migration into or out of the population; and the population is relatively large.
Determining gene frequencies when alleles exhibit dominance/recessiveness
When dominance and recessiveness affect a pair of alleles, it is impossible to detect all three
genotypes by their phenotypes and to calculate gene frequencies directly. However, if you
assume the population to be in Hardy-Weinberg equilibrium, you can estimate gene frequencies
using the Hardy-Weinberg equations.
You know that q2 represents that frequency of homozygous recessive individuals. The square
root of this frequency is q; knowing q you can determine p using the p + q = 1 equation. Once
you have determined the values for p and q, you can calculate the frequencies of homozygous
dominant individuals (p2) and heterozygous individuals (2pq).
For example, if you know that 16 percent of a particular population is made up of Rh-negative
people (dd):
q2
q
Then,
=
=
p = 1–q
0.16;
0.40
=
= 1 – 0.4
frequency of the recessive allele d.
= 0.60
=
frequency of the dominant allele D.
The 84 percent of the population that is Rh-positive can be divided as follows:
2pq = 2 x 0.6 x 0.4 = 0.48 = 48% of the population expected to be heterozygous Dd;
and
p2 = 0.6 x 0.6
= 0.36 = 36% of the population expected to be homozygous DD.
Procedure Part 3
Using the information gained about PTC tasting in Part 1 complete the table for your lab
section.
Fill out the questions for “Part 3” on your worksheet now.
Optional Extension:
Substitute your data from the table on the previous page into the table below and calculate
the chi-square value, where the “expected” number reflects the expected 70 percent
tasters/30 percent non-tasters of the North American population.
Phenotype
O (number
observed)
E (number
expected)
O–E
(O – E)2
(O – E)2
E
Taster
Non-taster
TOTAL
---------------- ------------------------------- ----------------
=χ2
a. How many degrees of freedom do you have in interpreting this chi-square?
b. Does the class appear to be a representative sample of the North American White
population? Explain your answer, based on the chi-square test results.
c. If the class is not a representative sample, give some reasons why it may not be:
Determining gene frequencies when alleles exhibit codominance
The Hardy-Weinberg equations are not needed to calculate gene frequencies in the case of
codominant inheritance. In these cases, gene frequencies can be determined directly from the
phenotypes of the individuals involved. Suppose you wish to determine the frequencies of the
alleles A (for normal hemoglobin) and S (for sickle cell hemoglobin) in an African American
population consisting of 343 AA (normal), 294 AS (sickle cell trait), and 63 SS (sickle cell
disease) individuals. (Heterozygous individuals—those with sickle cell trait—can be determined
by microscopic examination of their red blood cells; under conditions that reduce the oxygen
tension in the environment of the cells, some proportion of their cells will exhibit the
characteristic sickle shape. These conditions do not ordinarily occur within the body of a
heterozygous individual, although heterozygotes may experience some sickle cell disease
symptoms at high altitudes.)
Suppose you want to know whether the population is in Hardy-Weinberg equilibrium. There are
1400 alleles for the hemoglobin gene in the total population of 700 people (each individual
carries two alleles of the gene). The number of A alleles is 980: 343 people have two A alleles
and 294 have one A allele. The frequency of A is 0.7:
(343 x 2) + 294
686 + 294
980
1400
=
1400
=
1400
=
0.7
Similarly, the frequency of S is 0.3:
(63 x 2) + 294
126 + 294
420
1400
=
1400
=
1400
=
0.3
If this population is in Hardy-Weinberg equilibrium, then the frequency of homozygous
dominant (AA) individuals is 0.7 (the probability of having one A allele) x 0.7 (the probability of
having a second A allele) = 0.49. Thus, you would expect that 49 percent of the population
would have the AA genotype. Similarly, 2pq, or 2 x 0.7 x 0.3 = 0.42, or 42 percent of the
population should be heterozygous. You would expect only 9 percent (0.3 x 0.3 = 0.09) of the
population to have sickle cell disease (SS). Applying these percentages to the population of 700
individuals reveals that it is in perfect Hardy-Weinberg equilibrium:
49% x 700 = 343
42% x 700 = 294
9% x 700 = 63
Part 4 Revisiting the Micro-evolutionary Process and the
Hardy Weinberg Equilibrium
In 1950 Atlantic Puffins were introduced to two islands that are 763
miles apart, San Carlos and Nola Rei. Every 6 years more puffins were
released on the San Carlos, but not on Nola Rei due to lack of funding.
No puffins were observed moving between islands. In 2000 researchers
noticed that nearly all the puffins on Nola Rei had completely orange
beaks instead of the striped ones puffins usually have, both populations
appear to be stable with 75 breeding pairs
Procedure Part 4
Using the information above and the H-W equations, complete the information on the
worksheet.
Fill out the questions for “Part 4” on your worksheet now.
Part 5 Multiple Genes in Action
The following gel electrophoresis is for Labrador retrievers. Coat color arises through
interactions among alleles of two gene pairs. One gene pair codes for melanin production (B)
the dominant form allows production of black pigment and the recessive form(b)allows for
brown or chocolate color. The other gene codes for melanin deposition (E) in order to place
pigment in the hair the dominant form must be present, if only the recessive allele (e) is present
the dog will have yellow coat color.
B= Black labs, Y=Yellow Labs, and C= Chocolate Labs:
B1
B2
B3
Y1
Y2
C1
C2
U1
U2
1
2
3
4
5
6
7
8
9
10
Procedure Part 5
Using the information above, complete the information on the worksheet.
Fill out the questions for “Part 5” on your worksheet now.
BIO 101 GENETICS
Name _____________________
Part 1 Genes and You
1. Are you a taster of PTC?
If you are a taster, what did it taste like? Was it sweet, sour, salty, bitter, etc.?
2. You now know your phenotype. Do you know your genotype?
a. If no, why not?
b. If no, how might you be able to determine what your genotype is?
3. Fill out the chart below for yourself and one or more people.
Trait
Bent
Pinky
Tongue
rolling
Blue Eyes
Widows
Peak
Thumb
Crossing
Attached
Earlobe
Hitchhiker
Thumb
Redgreen
Colorblind
Hairy
Ears
Pattern
Baldness
Finger
length
Your
Traits
Other
Phenotype
Possible Genotype Phenotype
Person
Possible Genotype
Part 2 Human Pedigree Analysis
Complete the table below for the traits exhibited in the six pedigrees you analyzed in class.
Autosomal or X- Dominant or
Probability A will Probability B will
Pedigree linked?
Recessive?
have trait
have trait
1
2
3
4
5
6
Part 3. Genes in Populations, the Hardy-Weinberg Equilibrium
1. Record your phenotype on the table on the board or overhead projector. After all the data
for the class is recorded, copy the total call data below:
Phenotype
Number
Frequency (Percent)
Taster
Non-taster
TOTAL
2.Calculate the frequency of the recessive allele, t, in your class by applying the
Hardy-Weinberg equations to the class data:
3. Use the equations to determine the frequency of the T allele:
4. According to the Hardy-Weinberg equations, the frequency of TT individuals is:
5. . . . and the frequency of Tt individuals is:
6. Use the chi-square test to determine whether your lab section is a representative sample of
the North American White population (the majority of our class population). Studies of the
frequency of tasters and nontasters have shown that among this population, about 70
percent can taste PTC, while about 30 percent cannot. Other populations show different
frequencies of the two phenotypes. For example, some studies shown that about 90 percent
of African Americans are tasters and only 10 percent are nontasters.
Part 4. Microevloutionary Process and the Hardy Weinberg Equilibrium
1. As you may expect, the conditions for Hardy-Weinberg equilibrium do not always exist five
factors disturb genetic equilibrium in natural populations they are:
1. ________________________________________________________
2. ________________________________________________________
3. ________________________________________________________
4. ________________________________________________________
5.________________________________________________________
2. If we assume a single dominant gene controls that beak-stripping allele (S) and that the
recessive allele is (s) causes a non-striped completely orange beak. Given the following data
calculate the Hardy-Weinberg Equilibrium:
Phenotype
Stripped
Beaks
Orange
Beaks
p freq
Total
Initial
Releases
80
0
80
San
Carlos
149
1
150
Nola
Rei
13
137
150
q freq
Part 5. Multiple Genes in Action
1. What Bands give important information about coat color? _______________________
2. What Band(s) determine Black Coats? _______________________________________
3. What Band(s) determine Yellow Coats? ______________________________________
4. What Band(s) determines Brown Coats? _____________________________________
5. What is the phenotype of U1? _____________________________________________
6. What is the phenotype of U2? _____________________________________________
7. What is the genotype of B1? _______________________________________________
8. What is the genotype of Y2? _______________________________________________
9. What is the genotype of C2? _______________________________________________
10.What is the genotype of U2? _______________________________________________
NAMES
DATE
Analysis Question
Examine the pedigree below for a rare trait found in a particular family:
Analyze this pedigree below:
Which of the following statements is the best interpretation of the pedigree? Accept or reject
each choice, with explanation.
A.
B.
C.
D.
E.
Both males and females have the trait in this family.
Individual I-1 is homozygous dominant.
Individual II-2 is heterozygous.
The gene associated with this trait is on the X chromosome.
The gene associated with this trait is recessive.
The Hardy-Weinberg law applies not only to human populations, but also to plant populations
and other animal populations. T and N represent two alleles of the gene for venom found in a
population of gaboon vipers. A study of 2000 vipers from the population revealed the
following: 100 were genotype NN (non-venomous), 800 were genotype TN (mildly venomous),
and 1100 were TT (deadly venomous).
Analyze this group of gaboon vipers, from a population genetics perspective.
Which of the following is the best interpretation of the data? Accept or reject each statement,
with explanation.
A.
B.
C.
D.
E.
The frequency of the T allele is 0.95.
The T allele is recessive to the N allele.
The population has not reached Hardy-Weinberg equilibrium.
The population is in Hardy-Weinberg equilibrium.
Vipers with the NN genotype have greater fertility than vipers with the TT or TN genotypes.