A karyotype

Question 1 (5 points)

 

Question 1 Saved

 

A karyotype ______.

 

Question 1 options:

 

compares one set of chromosomes to another.
of a normal human cell shows 48 chromosomes.
is a photograph of cells undergoing mitosis during anaphase.
cannot be used to identify individual chromosomes beyond the fact that two chromosomes are homologues.
is a visual display of chromosomes arranged according to size.

 


 

Question 2 (5 points)

 

Question 2 Saved

 

At which stage of mitosis are distinct chromatids visible and moving towards the opposite poles of the cell?

 

Question 2 options:

 

Metaphase
Telophase
Interphase
Anaphase
Prophase

 

Question 3 (5 points)

 

Question 3 Saved

 

Mitosis in humans usually results in the formation of _____.

 

 

 

Question 3 options:

 

2 diploid cells

 

4 diploid cells

 

2 haploid cells

 

4 haploid cells

 

Sperm or egg cells

 

Page 1 of 7

 

All of the following are characteristics of cancer cells EXCEPT _____.

 

Question 4 options:

 

angiogenesis
unregulated mitosis
unmutated DNA
ability to metastasize
no apoptosis

 

Question 5 (5 points)

 

Question 5 Unsaved

 

What stage does cytokinesis generally overlap with in the typical cell cycle?

 

Question 5 options:

 

S phase
prophase
telophase
anaphase
metaphase

 


 

Question 6 (5 points)

 

Question 6 Unsaved

 

An important result of meiosis is that _____.

 

Question 6 options:

 

each gamete receives one member of each pair of homologous chromosomes and gametes are formed that are haploid.
each gamete receives one member of each pair of homologous chromosomes.
gametes are formed that are diploid.
gametes receive one copy of each member of each pair of homologous chromosomes.
gametes are formed that are haploid.

 

Question 5 (5 points)

 

Question 5 Unsaved

 

What stage does cytokinesis generally overlap with in the typical cell cycle?

 

Question 5 options:

 

S phase
prophase
telophase
anaphase
metaphase

 


 

Question 6 (5 points)

 

Question 6 Unsaved

 

An important result of meiosis is that _____.

 

Question 6 options:

 

each gamete receives one member of each pair of homologous chromosomes and gametes are formed that are haploid.
each gamete receives one member of each pair of homologous chromosomes.
gametes are formed that are diploid.
gametes receive one copy of each member of each pair of homologous chromosomes.
gametes are formed that are haploid.

 

Question 10 (5 points)

 

Question 10 Unsaved

 

For anaphase to begin, which of the following must occur?

 

Question 10 options:

 

Chromatids must lose their kinetochores.
Cohesin must attach the sister chromatids to each other.
Cohesin must be cleaved enzymatically.
Kinetochores must attach to the metaphase plate.
Spindle microtubules must begin to depolymerize.

 

Question 11 (5 points)

 

Question 11 Unsaved

 

Spindle microtubules attach to kinetochores during this phase of mitosis.

 

Question 11 options:

 

metaphase
prometaphase
anaphase
prohase
telophase

 

Question 12 (5 points)

 

Question 12 Unsaved

 

Four of the five answers listed below are related by a common phase of mitosis. Select the exception.

 

Question 12 options:

 

chromosomes separate
chromosomes decondense
spindle microtubules disappear
nuclear envelope re-forms
nucleolus reappears

 

Question 13 (5 points)

 

Question 13 Unsaved

 

A woman is found to have 47 chromosomes, including three X chromosomes. Which of the following describes her expected phenotype?

 

Question 13 options:

 

A) masculine characteristics such as facial hair
B) enlarged genital structures
C) excessive emotional instability
D) normal female
E) sterile female

 

Question 14 (5 points)

 

Question 14 Unsaved

 

Which of the following is an example of polygenic inheritance?

 

Question 14 options:

 

A) pink flowers in snapdragons
B) the ABO blood group in humans
C) Huntington’s disease in humans
D) white and purple flower color in peas
E) skin pigmentation in humans

 

Question 15 (5 points)

 

Question 15 Unsaved

 

Why did the F1 offspring of Mendel’s classic pea cross always look like one of the two parental varieties?

 

Question 15 options:

 

A) No genes interacted to produce the parental phenotype.
B) Each allele affected phenotypic expression.
C) The traits blended together during fertilization.
D) One phenotype was completely dominant over another.
E) Different genes interacted to produce the parental phenotype.

 

Page 5 of 7

 

Page 6 of 7

 

Note: It is recommended that you save your response as you complete each question.

 


 

Question 16 (5 points)

 

Question 16 Unsaved

 

Use the following information to answer the questions below.

 

Tallness (T) in snapdragons is dominant to dwarfness (t), while red (R) flower color is dominant to white (r). The heterozygous condition results in pink (Rr) flower color.

 

A dwarf, red snapdragon is crossed with a plant homozygous for tallness and white flowers. What are the genotype and phenotype of the F1 individuals?

 

Question 16 options:

 

A) ttRr—dwarf and pink
B) ttrr—dwarf and white
C) TtRr—tall and red
D) TtRr—tall and pink
E) TTRR—tall and red

 

Question 17 (5 points)

 

Question 17 Unsaved

 

National Basketball Association (NBA) scientists have identified the gene for superior athletic ability.  The allele “A” represents the dominant form of the gene, while the allele “a” represents the recessive form of the gene.  The scientists find that the superior athletic ability gene is optimally expressed in individuals who possess the homozygous dominant form.

 

If Michael Jordan, who has the Aa genotype, has a child with a woman who has the aa genotype, what are the chances that his child will inherit the optimal superior athletic ability genotype?

 

Question 17 options:

 

0%
25%
50%
75%
100%

 

Question 18 (5 points)

 

Question 18 Unsaved

 

There is evidence that a certain color in cats is sex-linked. Yellow is recessive to black.
A heterozygous condition results in tortoise shell or calico color. A calico cat has a litter
of 8 kittens: 1 yellow male, 2 black males, 2 yellow females, and 3 calico females. What
was the male parents probable color?

 

Question 18 options:

 

yellow
black
calico
yellow and black
albino

 

Page 6 of 7

 

Page 7 of 7

 

Note: It is recommended that you save your response as you complete each question.

 


 

Question 19 (5 points)

 

Question 19 Unsaved

 

Imagine you are performing a cross involving seed color in garden plants.  What F1 offspring would you expect if you cross true-breeding parents with green seeds with true-breeding parents with yellow seeds?

 

Question 19 options:

 

25% white seeds, 25% yellow seeds, and 50% green seeds
25% green seeds, and 75% yellow seeds
50% yellow seeds, and 50% green seeds
100% yellow seeds
100% yellow-green seeds

 

Question 20 (5 points)

 

Question 20 Unsaved

 

All of the genes located on a given chromosome constitute a

 

Question 20 options:

 

linkage group.
karyotype.
none of these
wild-type allele.
bridging cross.

 

Page 7 of 7

 
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BIO 102 Lab 04: ELISA and Immunology

BIO 102 Lab 04: ELISA and Immunology

 

Instructions: Submit, pages 6 and 8 of this document. Print, complete both lab activities and answer the questions. Scan your lab pages using the free phone app AdobeScan, and upload your PDF to Canvas. Please be sure to write your name on your first page of work.

 

Background

The human immune system contains several layers of defense, but before explaining them in depth, we must review some terminology. Starting with the answer sheet in this lab, and using your book, fill in the definitions from question #1 before continuing to read the remainder of this lab. You may also find it helpful to review diagrams in the book relating to antigen binding.

The broadest part of the immune system responds in the same manner to every antigen (ie., invader) it encounters. It is called the nonspecific immune system. It includes things like our skin, mucus membranes, ear wax, stomach acid, sweat, tears, vaginal secretions, antimicrobial proteins, and internal cellular defenses. The nonspecific immune system will mount the same response regardless of the nature of the antigen. It has no memory and doesn’t have the ability to recognize that a specific defense may have been ineffective against a pathogen in the past. The nonspecific immune system’s primary function is to prevent us from getting sick by attacking anything foreign and if that fails, to contain the pathogen until our adaptive immune system is activated.

The adaptive immune system “studies” each pathogen and learns how to effectively kill it. It also has memory of the pathogens it has faced in the past and will improve its effectiveness every time it encounters the same pathogen, meaning that the person doesn’t get sick from later encounters with the same pathogen, provided that pathogen is recognized. It includes two types of white blood cells (all types of white blood cells are called leukocytes), the B & T cells (B & T cells are also called B lymphocytes and T lymphocytes respectively). The B cells will begin releasing antibodies into the interstitial fluid and blood after they encounter a specific antigen. Each antibody is specific not for just that antigen, but a single epitope on the surface of the antigen. Once a B cell begins producing antibodies, they can remain present in the body for years. Vaccines stimulate our B cells into making antibodies so we don’t get sick if we encounter certain viruses. Some vaccines only require a few administrations and the immunity lasts for life, while others need to be reminded about that antigen, hence the need for vaccine “boosters.”

In this lab, we will investigate an important test called ELISA. ELISA stands for enzyme linked immunosorbent assay. It is used in many different ways, from diagnostic lab tests used by doctors to measure a patient’s exposure to a virus, to research lab investigations that and separate a specific protein among many. Though very specific and powerful, it can be easily performed and is a commonly found experiment in college biology courses. ELISA functions detect the presence of antibodies or a pathogen. ELISA has several varieties, but two of the most common are: direct ELISA, which uses artificially made antibodies to bind to the antigen, while indirect ELISA uses the person’s antibodies against a specific pathogen to determine if a person has encountered the disease before. In indirect ELISA the artificial antibodies bind to a person’s antibodies against a specific disease. In both direct and indirect ELISA, the artificial antibodies are engineered to change color when added to a special chemical. Direct ELISA is explained in the image on the next page.

 

 

 

 

Direct ELISA

Image from Wikipedia.org, Reteived 6/12/19

 

 

 

 

BIO 102 Lab 04: ELISA and Immunology 7

 

 

 

In indirect ELISA parts of the suspected pathogen are anchored to the sides of a well plate (a well plate is a small circular clear plastic dish). The person’s blood serum is allowed to sit in the plate long enough for any antibodies present (if the person has encountered that pathogen recently) to stick to the antigen that is part of the well plate. In the diagram to the left, the middle representation shows a green antibody attached to a purple enzyme cluster; the antibody is bound to the red viral antigen. The serum is then drained away, but the attached antibodies will remain, stuck. Then, artificial antibodies are added, which will only stick if the person’s antibodies are attached to the antigen in the well plate. A color change will occur if artificial antibodies remain stuck (the bottom diagram to the left shows the blue colored molecules which contain the dye and the artificial antibody). Thus, a clear solution means the person is not sick and a color change means the person has encountered the disease before.

 

 

 

 

 

 

The Live Lab ELISA Procedure (Performed When Meeting in Lab) – READ ONLY

Read through the procedure below. This document includes sample ELISA data. You will use these results to answer the questions.

 

Every person gets 1 “serum” sample tube (we will avoid using real human bodily fluids in lab). One of these samples contains the antibodies against a sexually transmitted or blood borne disease.

Everyone will then share “fluids” three times. DO NOT start a round of sharing until instructed to do so!!! Each fluid sharing will be done by transferring ½ of the contents from one person’s tube into the other person’s tube. The cap off the receiving tube & swirl, then ½ of the receiving tube’s fluid is transferred back to the original tube, thus each tube is roughly ½ original and ½ new sample. Every time you change serum solutions you need to change pipette tips or you will contaminate the samples.

 

Record your tube # here____________________

 

Round of fluid transfer Partner’s name Partner’s tube #
1    
2    
3    

 

Again, no one should have a 2nd (or 3rd) partner until their instructor tells them to find one. At each round you only share with 1 other person!

Everyone should record their information on the spreadsheet on the projector/board while doing the ELISA test.

Now that everyone has done the fluid transfer it is time to find out who has the disease & see if you can figure out where the disease started.

Using a pen, mark one row of the well plate with +, the second with -, and then each person at your table gets a row with their initials. Each row should have 3 wells.

1. Transfer 1/3 of your serum sample into each of the well plates with your initials. The positive control goes into the 3 wells with the “+”, and the negative control goes into the plate with the “-”.

2. Allow the samples to remain in the well plate depressions for 3 minutes

3. Empty the plate in the sink and wash the sample depressions 3 times with ELISA wash solution, tap plate against a paper towel on the counter each time. Be careful not to allow fluid to spill from one well to another while washing & rinsing!

4. Add antibody (AB) to the washed out sample depressions and allow to sit for 3 minutes

5. Repeat #2

6. Add color substrate (CS) and allow to sit for 3-5 minutes

7. Positive reaction is blue, negative reaction is clear

8. Record who was “sick” on the projector/board.

9. Determine from the sharing & who was sick, what couple started off the disease.

10. Answer questions 2-5 on the answer sheet.

 

Sample ELISA results:

Each of the circular discs is a well, this would be called a 24-well plate because it contains 24 wells and a different sample can be run in each well. The blue color changes indicates a positive result, the clear (ie., see through) indicates a negative result.

Use the data in this table to answer the ELISA questions on the worksheet.

 

Sample Class ELISA Results Data Table
Patient’s Sample # Patient’s Name + or – First Partner’s # Second Partner’s # Third Partner’s #
1 Cary 5 14 24
2 Chris 4 12 15
3 Ryan + 6 10 14
4 Bo 2 9 23
5 Tim 1 11 13
6 Lei + 3 7 19
7 Vashti + 10 6 22
8 Geeta 11 13 21
9 Vijaya 12 4 11
10 Xin + 7 3 17
11 Jacob 8 5 9
12 Fred 9 2 16
13 Diane 14 8 5
14 Tiffany + 13 1 3
15 Thy 16 24 2
16 Yukti 15 18 12
17 Mary + 18 23 10
18 Michel 17 16 20
19 Vincent + 24 22 6
20 Yan yan 23 21 18
21 Beatrice 22 20 8
22 Swati + 21 19 7
23 Kirsten 20 17 4
24 Alex 19 15 1

 

*Disclaimer: the names are randomly chosen from instructor names at NVCC, spring 2020 semester. No matches or +/- results are actually true, all data was randomly selected and assigned.

ELISA Worksheet

1) Define the following terms dealing with the immune system

 

Antigen: _________________________________________

 

Pathogen: ________________________________________

 

Epitope: _________________________________________

 

2) Why did you run both positive & negative controls?

 

 

 

 

3) What is a false positive? How do you think one could come about using a test like the ELISA?

 

 

 

 

4) At the end of 3 rounds of fluid transfers, what percentage of the class had the disease?

 

 

 

 

5) You can narrow the initial outbreak down to 2 people, who are they?

 

 

 

 

6) Describe how an organization like the CDC could use results like those you obtained to track down “patient zero” for a disease outbreak.

 

 

 

 

 

7) Do you think this method shows a direct or indirect ELISA procedure? Why?

 

 

8) With an indirect ELISA, it tests not for the antigen, but antibodies against the antigen, does a positive test then mean the person currently has the disease in question? If not, what does a positive

Immunity & Vaccinations

Background

Vaccinations are given to stimulate the production of antibodies without the animal actually getting the sick with a disease. Vaccinations come in many different forms, but the most common types are inactivated viruses or bacteria (also called attenuated bacteria or viruses). Attenuated means the pathogen has been disabled so it can’t cause the full-blown illness, but is still figuratively “alive” so the animal may get a very mild form of the illness. You can think of it as a recognizable, but harmless form of the pathogen. Inactivated vaccines, which are “dead,” often contain fragments of the original pathogen which are attached to another particle to attract B cells. As the body removes the attenuated (= inactivated) pathogen, the adaptive immune system learns, with an eye toward remembering this encounter with the “pathogen” (note that your immune system does not “know” that it only encountered a harmless form of the pathogen). It remembers by making antibodies against that disease which should prevent future infections.

In the last decade, a movement has started in the U.S., supported by Hollywood celebrities among others, who claim that vaccines are unnecessary. Some even claim that vaccines even cause autism. The claim that vaccines cause autism has been conclusively proven false through many studies conducted all over the world. One initial study, which caused the initial concern, was filled with many mistakes and after close examination, was determined to be invalid by many different scientists. See https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2831678/ and https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3136032/ for more information on this.

As to the first claim that vaccines are unnecessary against certain diseases, we will closely examine exactly how vaccines protect many people not just those who receive the vaccine.

In every population there will be people whose immune system is weakened. It could be weakened for many reasons: genetic (missing or damaged genes); the person is elderly, very young, allergic; or undergoing chemo or radiation therapy for cancer (which can cripple the immune system). These people often can’t receive vaccinations or if given, they won’t produce antibodies or sufficient numbers of antibodies to prevent them from getting sick.

 

Procedure

Visit https://fred.publichealth.pitt.edu/measles

This is a program that simulates the spread of the measles. On the left side of the screen will assume that only 80% of the people are vaccinated. The right side will assume that 95% of the people are vaccinated.

1. Select District of Columbia & for the city, select Washington D.C. Answer questions 1 & 2 about this simulation.

 

2. Now change the state to North Carolina. For the city select Hickory. Run the simulation again. Answer question 4 on the answer sheet.

 

3. Now change the state to New York & the city to New York. Run the simulation again. Answer question 5 on the answer sheet.

 

 

Disease Spread Simulation Answer Sheet

1) How long did it take for the disease to disappear on the vaccinated side for D.C.?

 

 

 

2) Make a hypothesis on why the 80% vaccinated side of D.C. Eventually saw the disease begin to slow down? About how many days did it take to slow down?

 

 

3) How does each side compare to D.C.?

 

 

4) Hickory has a population of about 40,000 people & D.C. Has about 630,000 people. What does this tell you about how population density affects disease spread?

 

 

5) How does New York city compare to D.C & Hickory? Explain why you think that is the case.

 

 

6) Based on these simulations and given that both measles and COVID-19 are both airborne proximity diseases, why are all the governors capping the size of gatherings?

 

 

7) For a blood borne (or STD/STI) based disease (like that simulated with the ELISA part of this lab), would limiting the size of gatherings be as effective at stopping the spread of those diseases? Explain.

 
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BIO – INET Lab Genetics Worksheet Report Tempate

INET Lab
Genetics Worksheet Report Template
Student:
Email:
Date:

For each correctly conducted genetics problem, you receive 5 points for a maximum 20 points.  All Punnett squares must
be shown.
1. There are 3 alleles controlling the ABO blood types. IA and IB are codominant genes so that the combination IAIB
produces the AB blood type. The third allele, (i) is recessive to the other two alleles.  Show your work for each cross.
Indicate which of these parents could produce the given child.
The first one is provided with answers as an example.
(HINT: Do not forget that a heterozygous versus a homozygous blood type may yield different results. )
a)

Parent
A x AB

Child
IB
B

IA

IA

IA IA

IA IB

IA

IA IA

IA IB

IA

IB

IA

IA IA

IA IB

i

IA i

IB i

Produce Child of Type?

No

Produce Child of Type?

© Access Learning Systems

2

b)

Parent

Child

A x O

A

Produce Child of Type?

Produce Child of Type?

© Access Learning Systems

3
c)

Parent

Child

A x B

O

Produce Child of Type?

Produce Child of Type?

Produce Child of Type?

Produce Child of Type?

d)

Parent

Child

A x AB

O

© Access Learning Systems

4

Produce Child of Type?

Produce Child of Type?

© Access Learning Systems

5
e)

Parent

Child

B x B

O

Produce Child of Type?

Produce Child of Type?

Produce Child of Type?

f)

Parent

Child

AB x AB

A

Produce Child of Type?

© Access Learning Systems

6
2. The allele for albinism (a) is recessive to the allele for normal pigmentation (A). A normally pigmented  woman whose
father is an albino marries an albino man whose parents are normal. They have three children, two normal and one
albino. Give the genotypes for each person listed. Prove your answer.

8

Maternal Grandmother
Father

6

9

Maternal Grandfather

Mother
Child 1

8

7

Paternal Grandmother

5

6

Paternal Grandfather

7

5

4

1

4

3

Relationship

3

2

#
2

1

Child 2

9

Genotype(s)

Child 3

a) Show the cross of the paternal grandparents.

(HINT: What must their genotypes be to produce and albino child? )

b) What are the possible genotypes of the maternal grandmother?

© Access Learning Systems

7

© Access Learning Systems

8
Show the cross of the mother and father.

3. In horses, black coat color is influenced by the dominant allele (B), and chestnut coat color by the recessive allele (b).
a) What color horse would you use to find out the genotype of a black trotter?

b) Give the genotype and phenotype. Show your work and discuss the reason you would know the genotype of the
black horse.

Possible Allele Pairings with Homozygous Dominant Trotter

F1 offspring probabilities:

Possible Allele Pairings with Heterozygous Trotter

F1 offspring probabilities:

Reasoning:

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9
4. In Drosophila, the fruit fly, white eyes are determined by a recessive X­linked gene, and the wild­type or normal brick­
red eyes are due to its dominant allele. Use symbols of the following types: X rY = a white­eyed male; XRXR = a
homozygous normal red female.
a) What offspring can be expected from a cross of a white­eyed male and a homozygous normal female?

b) Show the genotypes and list the phenotypes of the F1 offspring.
F1 Cross

F1 offspring probabilities:
Male

Female

Red­Eyed
White­Eyed

c) Now, cross the F1 offspring.  Show the genotypes and list the phenotypes of the F 2 offspring.
F2 Cross

F2 offspring probabilities:
Male
Red­Eyed
White­Eyed

Female

INET Lab

Genetics Worksheet Report Template

1

 

© Access Learning Systems

 

Student:

Email:

Date:

 

 

 

For each correctly conducted genetics problem, you receive 5 points for a maximum 20 points. All Punnett squares must be shown.

 

1. There are 3 alleles controlling the ABO blood types. IA and IB are codominant genes so that the combination IAIB produces the AB blood type. The third allele, (i) is recessive to the other two alleles. Show your work for each cross.

 

Indicate which of these parents could produce the given child.

 

The first one is provided with answers as an example.

(HINT: Do not forget that a heterozygous versus a homozygous blood type may yield different results.)

 

a) Parent Child
  A x AB B

 

 

  IA IB
IA IA IA IA IB
IA IA IA IA IB
  IA IB
IA IA IA IA IB
i IA i IB i

 

 

Produce Child of Type? No
Produce Child of Type? Yes

 

b) Parent Child
  A x O A

 

 

 

     
     
     
     
     
     

 

 

Produce Child of Type?  
Produce Child of Type?  

 

 

 

c) Parent Child
  A x B O

 

 

 

     
     
     
     
     
     

 

 

Produce Child of Type?  
Produce Child of Type?  

 

     
     
     
     
     
     

 

Produce Child of Type?  
Produce Child of Type?  

 

 

d) Parent Child
  A x AB O

 

 

 

 

     
     
     
     
     
     

 

 

Produce Child of Type?  
Produce Child of Type?  

 

 

 

e) Parent Child
  B x B O

 

 

 

     
     
     
     
     
     

 

 

Produce Child of Type?  
Produce Child of Type?  

 

     
     
     

 

Produce Child of Type?  

 

 

 

f) Parent Child
  AB x AB A

 

 

 

 

     
     
     

 

 

Produce Child of Type?  

 

 

2. The allele for albinism (a) is recessive to the allele for normal pigmentation (A). A normally pigmented woman whose father is an albino marries an albino man whose parents are normal. They have three children, two normal and one albino. Give the genotypes for each person listed. Prove your answer.

 

 

 

# Relationship Genotype(s)
1 Paternal Grandfather  
2 Paternal Grandmother  
3 Maternal Grandfather  
4 Maternal Grandmother  
5 Father  
6 Mother  
7 Child 1  
8 Child 2  
9 Child 3  

 

3 4 1 2 5 6 7 9 8

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

a) Show the cross of the paternal grandparents.

(HINT: What must their genotypes be to produce and albino child?)

 

 

 

     
     
     

 

 

 

b) What are the possible genotypes of the maternal grandmother?

 

 

 

     
     
     
     
     
     

 

 

 

c) Show the cross of the mother and father.

 

 

 

     
     
     

 

 

 

3. In horses, black coat color is influenced by the dominant allele (B), and chestnut coat color by the recessive allele (b).

 

a) What color horse would you use to find out the genotype of a black trotter?

6

 

 

 

 

 

5

 

 

 

b) Give the genotype and phenotype. Show your work and discuss the reason you would know the genotype of the black horse.

 

 

Possible Allele Pairings with Homozygous Dominant Trotter

 

     
     
     

 

 

F1 offspring probabilities:
 

 

 

Possible Allele Pairings with Heterozygous Trotter

 

     
     
     

 

 

F1 offspring probabilities:
 

 

 

 

Reasoning:

4

 

 

 

 

 

5

 

 

 

4. In Drosophila, the fruit fly, white eyes are determined by a recessive X-linked gene, and the wild-type or normal brick-red eyes are due to its dominant allele. Use symbols of the following types: XrY = a white-eyed male; XRXR = a homozygous normal red female.

 

a) What offspring can be expected from a cross of a white-eyed male and a homozygous normal female?

6

 

 

 

 

 

7

 

 

 

b) Show the genotypes and list the phenotypes of the F1 offspring.

 

 

F1 Cross

 

     
     
     

 

 

F1 offspring probabilities:
  Male Female
Red-Eyed    
White-Eyed    

 

 

 

c) Now, cross the F1 offspring. Show the genotypes and list the phenotypes of the F2 offspring.

 

 

F2 Cross

 

     
     
     

 

 

F2 offspring probabilities:
  Male Female
Red-Eyed    
White-Eyed    
 
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Foundations of Biological Sciences I Evolutionary Agents

Foundations of Biological Sciences I Evolutionary Agents – 1

A quick recap…. There are several terms that need to be clarified so that you can more easily follow the exercise. A gene is a

piece of DNA that directs the expression of a particular characteristic (trait). Genes are located on

chromosomes, and the location where a particular gene is found is referred to as the locus (plural: loci) of that

gene. An allele is a gene for which there is an alternative expression, which can lead to the alterative form of a

trait. For example, a diploid organism carries the allele “A” on one homologous chromosome, and the allele “A”

on the other. The genotype of this organism is then AA and it is said to be homozygous. An organism may also

carry two different alleles. For example on one chromosome it could carry the allele “A” and on the other it

could carry the allele “a”. The genotype of such an organisms is then Aa, and it is described as heterozygous for

this chromosomal locus.

The genotype of an organism is the listing of the two alleles for each trait that it possesses. The phenotype of an

organism is a description of the way a trait is displayed in the structure, behavior, or physiology of the organism.

Some alleles are dominant to others and mask the presence of other alleles. The dominant condition is indicated

by uppercase letters (e.g., “A”). The alleles that are masked are called recessive alleles. The recessive condition

is indicated by lowercase letters (e.g., “a”). When both dominants are present in the genotype (AA), the organism

is said to be homozygous dominant for the trait, and the organisms will show the dominant phenotype (trait

expression A). When both recessives are present in the genotype (aa), the organism is said to be homozygous

recessive for the trait, and the organisms will show the recessive phenotype (trait expression a). In the case of

complete dominance, the dominant allele completely masks the recessive allele, and an organism with a

heterozygous genotype (Aa) will show the dominant phenotype (trait expression A).

 

Evolutionary Agents

Evolution is a process resulting in changes in gene frequencies (= the genetic make-up) of a population over

time. The mechanisms of evolution include selection (which can cause change over time & adaptation), and

forces that provide variation and cause change over time (but not adaptation). Factors that change gene

frequencies over time are referred to as evolutionary agents.

A powerful way to detect the presence of evolutionary agents is the use of the Hardy–Weinberg model. This

model can be applied to traits that are influenced by several loci; the simplest case is for a trait that is regulated

by one locus with two alleles.

With the Hardy–Weinberg model, the frequency of genotypes in the population can be predicted from the

probability of encounters between gametes bearing the different alleles. With alleles R and B occurring at

frequencies p and q, respectively, the frequency of genotypes in the population is described by the formula:

 

p 2

+ 2pq + q 2

= 1

Hardy-Weinberg

equilibrium

If p is the frequency of one allele, and q

is the frequency of the other allele, then:

 

p + q = 1

 

 

Foundations of Biological Sciences I Evolutionary Agents – 2

If certain conditions are met, the proportions of genotypes that make

up a population remain constant from generation to generation, and

can be predicted from the Hardy-Weinberg equilibrium.

For example, if flower color is controlled by two alleles (R & B),

and the allele for red color is present in the population 80% of the

time, than the other allele for blue color must be present 20% of the

time. Consequently, the allele frequencies in the population are p =

0.8 and q = 0.2. (0.8 + 0.2 = 1).

From this we can calculate the expected genotype frequencies in the

population. Since p = 0.8, we would expect 64% of the flowers in

the populations to be homozygous for red flower color (expected

genotype frequency for p 2

= 0.8 x 0.8 =0.64). 32% of the

populations would be heterozygous for flower color. They would

have one R allele (p = 0.8) and one B allele (q = 0.2), and if neither

allele were dominant they would appear purple. The expected

genotype frequency of these purple individuals is 2pq = 2 x 0.8 x 0.2 = 0.32. Finally, 4% of the population would be homozygous for the

blue flower color (q 2

= 0.2 x 0.2 = 0.04).

The Hardy–Weinberg model applies

when the following conditions are met:

1) No genetic drift

2) No selection

3) No mutation

4) No migration

 

By contrast, there will be change in gene

frequencies in a population when at least

one of these conditions occur:

1) Genetic drift

2) Selection

3) Mutation

4) Migration

 

In today’s lab, you will do a series of exercises that illustrate the effect of the different evolutionary agents on the

genetic structure of a model population. You will work with populations composed of individuals that are

represented by colored beads. White beads represent individuals that are homozygous for the white allele (WW);

red beads are individuals that are homozygous for the red allele (RR), and pink beads are heterozygous (WR).

These beads live in a habitat – a plastic dishpan filled with smaller beads. The larger beads of our population will

be retained by the mesh, while the smaller beads pass through the mesh.

When the individuals are recovered with the help of the mesh, the frequencies of the color alleles are determined

using the Hardy-Weinberg model. The alleles in our populations are codominant – each white individual

possesses two white alleles, each red individual two red alleles, and the pink individuals have one red and one

white allele. Consequently, the total number of color alleles in a population of 40 individuals is 80. If such a

population contains 10 white individuals, 20 pink individuals, and 10 red individuals, the frequency of white

alleles (p) is

(2 x number white beads) + number pink beads

p = —————————————————————-

(2 x number of beads total)

(2 x 10) + 20

p = ———————– = 0.5

80

Because p + q = 1, the frequency of the red allele (q) must also be 0.5.

1. NAT URAL S E L E CT I O N

Natural selection disturbs the Hardy-Weinberg equilibrium by discriminating between individuals with respect to

their ability to survive and reproduce. Individuals that are better at surviving to produce young will contribute

more genes to the next generation; they are said to have greater fitness than those individuals that leave no or

fewer offspring.

In this experiment you will test the hypothesis that individuals are more likely to survive and reproduce when

their coloration makes it easier to hide from predators in the environment.

 

 

Foundations of Biological Sciences I Evolutionary Agents – 3

1. Work in groups of four. Each group member assumes one of the following roles

Predator: Search for prey (large beads)

Data Recorder / Timer: record numerical results and time the predation sessions

Calculator: use a calculator to calculate the allele frequencies

Caretaker: look after and manipulate the experimental setup

2. Create a white habitat by filling the dishpan with small white beads. Establish an initial population by

adding 10 large white beads, 20 large pink beads and 10 large red beads into the habitat. “Hide” the

individuals in the habitat by mixing the large and small beads. The predator will prey on the large beads,

removing as many individuals as possible in a set amount of time. The survivors will reproduce a new

generation, upon which the predator will prey again. This cycle will be repeated several times. Make a

prediction as to how the frequency of red alleles in the populations will change over time.

Prediction:

 

3. The predator hunts for prey (large beads) in the habitat, and uses the pair of forceps to catch as many prey

items as possible in 30 seconds.

4. After the predation (selection) episode, strain the habitat with the sieve and count the remaining red, pink

and red individuals. Record the numbers in the second row in Table 1. Calculate the frequencies of the white

(p) and red (q) alleles remaining in the population, and record them in Table 2 (under First generation). For

example, if 6 white, 8 pink and 8 red individuals remain, the frequency of the white alleles is

To calculate p, use the observed numbers of each color within the formula:

(2 x number white beads) + number pink beads

p = ——————————————————–

(2 x number of beads total)

(2 x 6) + 8

p = ———————– = 0.45

44

 

Table 1: Large- Bead counts before and after four rounds of Natural Selection (Predation)

 

Population

White Beads

Pink Beads

Red Beads

Total Beads

 

Initial

Before 10 20 10 40

After

 

Second Generation

Before

After

 

Third Generation

Before

After

 

Fourth Generation

Before

After

 

 

Foundations of Biological Sciences I Evolutionary Agents – 4

Table 2: Allele and genotype frequencies due to Natural Selection (Predation)

Population p q p 2

2pq q 2

Initial 0.5 0.5 0.25 0.5 0.25

 

First Generation

 

Second Generation

 

Third Generation

 

Fourth Generation

 

5. Based on the new values (after selection) for allele frequencies, calculate the genotype frequencies for the

homozygous white (p 2 ) and red (q

2 ) individuals, and for the heterozygous pink individuals (2pq). Record the

new allele frequencies in Table 2. For example, if p now equals 0.43, the frequency of the homozygous

white individuals is

p 2

= (0.48) 2

= 0.23

6. Assuming that 40 individuals comprise the next and all succeeding generations, calculate the number of

white, red and pink individuals to create the next generation, and record the numbers in the Before row

under Second generation in Table 1. For this, and all future calculations, round up or down to the nearest

whole number. For example, if p 2 =0.23, the number of white individuals for the next generation is

p 2

x 40 = 0.23 x 40 = 9.2 or 9 white individuals

7. Calculate the numbers of white, red and pink individuals you need to construct the new generation, and

introduce them into the habitat for a new round of selection.

8. Repeat the selection and reproduction steps for three more rounds, filling in the remaining rows in Tables 1

and 2. When you are done, use the frequencies of the red allele from Table 2 to construct a histogram in the

appropriate space in Figure 1 below. Remember to label your axes and complete the figure caption.

Figure 1: Changes in frequency of the red allele (q) due to selection. ……………………………………………………………………..

……………………………………………………………………………………………………………………………………………………………………………….

……………………………………………………………………………………………………………………………………………………………………………….

 

 

 

 

 

 

 

 

 

 

0 1 2 3 4

1.0

0.5

0.0

 

 

Foundations of Biological Sciences I Evolutionary Agents – 5

What is your conclusion as to the prediction you made in point 2?

There are 3 different patterns of selection. Directional selection favors one extreme phenotype over the other and

causes allele frequencies to change in a predictable direction. Stabilizing selection favors an intermediate

phenotype, rather than one at the extremes. Disruptive selection disfavors intermediate phenotypes, and favors

the extreme ones. Which kind of selection is illustrated by predation of white, pink and red individuals in a white

habitat?

 

If two identical populations inhabited different environments (e.g. white and red habitats), how would the

frequency of the color genes in each habitat compare after a large number of generations?

 

When two populations become genetically different through time (divergence), individuals can lose the ability

to interbreed, and two new species are formed. This process is called speciation.

 

2. EFFECT OF GENE FLOW ON NATURAL SELECTION

New members may join populations (immigrations) or leave the population (emigration). As they do, the

frequencies of alleles in the population change. This gene flow due to migration can be a powerful force in

evolution.

1. Establish a new population as described in the previous section.

2. Begin the selection process as before, but this time 5 red individuals will immigrate into the population

before the new allele frequencies are determined. Write down a prediction of the hypothesis that gene flow

resulting from migration of individuals into a population undergoing predation affects the change in allele

frequencies expected from selection alone. Focus your prediction on the change in the frequency of red

alleles in the population.

3. Conduct 4 cycles of predation with migration. For each generation, write down the number of surviving

individuals in Table 3, and the allele frequencies in Table 4. When you are done, use the frequencies of the

red allele from Table 4 to construct a histogram as your homework.

Homework: Write a hypothesis and prediction for this evolutionary model based on your understanding of

gene flow. Create a histogram that displays the change through four generations of natural selection with

migration. Remember to include a figure caption and axis labels (10 pts). *Hint: see Figure 1

 

 

Foundations of Biological Sciences I Evolutionary Agents – 6

 

Table 3:

Large- Bead counts before and after four rounds of simulated Natural Selection and Gene Flow

 

Population

White Beads

Pink Beads

Red Beads

Total Beads

 

Initial

Before 10 20 10 40

After (First Gen) (survivors + 5)

 

Second Generation

Before

After (survivors + 5)

 

Third Generation

Before _________ _________ __________ _________

After _________ _________ (survivors + 5)

 

Fourth Generation

Before _________ _________ __________ _________

After _________ _________ (survivors + 5)

Table 4: Allele and Genotype frequencies due to Natural Selection and Gene Flow

Population p q p 2

2pq q 2

Initial

First Generation

Second Generation

Third Generation

Fourth Generation

 

How did migration influence the effectiveness of selection in this example?

If white individuals would have immigrated into the population instead of the red ones – how would this have

influenced the change in gene frequencies?

 

Through immigration, new genetic information is introduced into the population. Gene flow thus maintains

genetic variation in a population. Barriers to gene flow can decrease genetic variation within populations, and

also accelerate divergence between populations.

 

 

Foundations of Biological Sciences I Evolutionary Agents – 7

3. MUTATION

New genetic information can also be introduced into a population through mutation.

1. Establish a new population by placing 10 large white beads, 10 large red beads and 20 large pink beads in

the bowl. Do not add the small beads this time.

2. Designate one group member to pick 20 large beads from the bowl (without looking!). Use these 20 beads to establish a new generation. Then replace one white bead with a silver bead. This represents a mutation in the

gamete that one parent contributes to this generation.

3. Calculate the allele frequencies of the new generation, including the frequency of the new color allele (r),

and write them down in Table 5.

 

Table 5: Change in Allele Frequencies due to Mutation

Population p

q

r

 

Initial

 

New Generation

 

After the mutation, three alleles are present in the population (p + q + r = 1). Consequently, the Hardy-Weinberg

equation must be expanded to

p 2 + 2pq + q

2 + 2pr + 2qr + r

2 = 1.0.

In addition to white, pink and red phenotypes, there are now silver, silver-red, and in subsequent generation

potentially black phenotypes. If the next generation contains 50 individuals, how many offspring of each

phenotype would be found in the population? Use Table 6 to calculate the genotype and phenotype frequencies.

 

Table 6: Phenotypes two generations after a mutation

 

Phenotype Genotype Frequency

x 50

 

# Individuals

White p 2

 

 

Pink 2pq

Red q 2

 

 

Silver 2pr

Silver-Red 2qr

Black r 2

 

 

Imagine a population made up of individuals of the color phenotypes in these proportions. What effect will

natural selection have on the phenotypes in a white habitat?

 

Under which conditions would a rare black allele be favored?

 

 

Foundations of Biological Sciences I Evolutionary Agents – 8

4. GENETIC DRIFT

Gene frequencies can change over generations as a result of chance (Genetic Drift). Genetic drift is often a

problem for small populations, because it can result in a loss of genetic variability. In very small populations,

chance can even eliminate an allele from a population.

1. Establish a new population by placing 10 large white beads, 10 large red beads and 20 large pink beads in

the bowl. Do not add the small beads this time.

2. If all 40 members of this population have an equal chance of getting to reproduce, the allele frequencies for

the colors in the population are p = 0.5 and q = 0.5 (see first column of Table 7).

3. Now let’s see what happens when only a subset of the population gets a chance to reproduce. Choose,

without looking, 10 beads from the bowl. They will make up a small group of individuals that reproduce.

Record the allele frequencies in this cluster in the second column of Table 7.

4. Replace the 10 beads to the population and mix well. Then pick 30 new beads. They represent a larger group

of individuals that get to reproduce. Calculate the allele frequencies in this group, and add them to the third

column in Table 7.

 

 

Table 7: Allele frequencies resulting from Genetic Drift

Observed Frequency in

Expected Frequency Small Group Large Group

n

40

 

p

0.5

 

q

0.5

 

Compare the allele frequencies of the three groups of reproducing individuals. How does group size affect the

makeup of the next generation?

 

When members of an old population emigrate to establish a new population, the allele frequencies in the new

population can be heavily influenced by chance.

5. Reestablish the old 10/20/10 beads population you have worked with before. Then pick (without looking) 6

individuals which will represent the 6 founding members of a new population.

6. Move these individuals to a new habitat. Calculate the allele frequencies and record them in Table 8.

 

Table 8: Allele Frequencies in a Founder Population

 

Allele Frequency

Initial population Founder Population

 

p

0.5

 

q

0.5

 

How do the allele frequencies of the small founder population compare to the larger original population?

 
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