Experiment exercises will deal with Diffusion and Osmosis

This unit’s Experiment exercises will deal with Diffusion and Osmosis. Read through the introductory material located below and complete the questions found in the Unit 3 Experiment Answer Sheet.

How to Proceed

  • Read through the Experiment Exercise Introductions below.
  • Open the Unit 3 Experiment Answer Sheet and complete the following Experiment exercises this unit:
    • Experiment 3 Exercise 1 – Diffusion (~1 hr)
    • Experiment 3 Exercise 2 – Osmosis (~1.5 hrs)
  • Save your completed Unit 3 Experiment Answer Sheet and submit it no later than Sunday midnight (CT).

Diffusion – Introduction

This unit we are learning about the structure and function of cells. The plasma membrane, for example, is an important structure of all cells and it is responsible for regulating the passage of materials into and out of the cell. Plasma membranes are differentially (selectively) permeable, meaning some substances are allowed to enter and exit the cell, while the movement of other materials is either carefully regulated or blocked. Two ways in which materials can move freely across the cell membrane are diffusion and osmosis.

Diffusion is the movement of solutes (material dissolved in liquid) from an area of high concentration to an area of low concentration. If these areas are separated by a membrane, that membrane may or may not be permeable to the solute. The membrane is always permeable to water though and the movement of water across a membrane is a special form of diffusion called osmosis.

In our first exercise, we will examine diffusion of solutes through a semipermeable membrane and the factors that affect their movement. You’ll want to be sure to review our online lecture this unit on Cell Structure and pp 83 – 86 in your book. View the following two animations BEFORE starting this exercise:

McGraw-Hill. 2006. How Diffusion Works
http://highered.mheducation.com/sites/0072495855/student_view0/chapter2/animation__how_diffusion_works.html (Links to an external site.)

McGraw-Hill. 2006. How Osmosis Works
http://highered.mheducation.com/sites/0072495855/student_view0/chapter2/animation__how_osmosis_works.html (Links to an external site.)

When you are ready to begin, open the Unit 3 Experiment Answer Sheet and answer the questions associated with the first exercise.

Osmosis – Introduction

In our second exercise this unit, we will to take a closer look at osmosis; the movement of water across a membrane. The direction water moves depends on the relative concentration of solute molecules on either side of the membrane (in this case, these solutes are not able to cross the membrane). Furthermore, the presence or absence of cell walls (e.g., in plant cells) influences how cells respond to osmotic fluctuations in their environment. This exercise will examine the forces that determine whether water moves into or out of a cell.

We will be using the following website in this exercise. Be sure you are able to access and use this website before starting.

The Biology Place. No Date. Osmosis: Movement of Water across Membranes
http://www.phschool.com/science/biology_place/biocoach/biomembrane1/osmosis.html  (Links to an external site.)

Open the Unit 3 Experiment Answer Sheet and complete the questions for this exercise.

WEEK 3 EXPERIMENT ANSWER SHEET Please submit to the Week 3 Experiment dropbox no later than Sunday midnight.

SUMMARY OF ACTIVITIES FOR WEEK 1 EXPERIMENT ASSIGNMENT

· Experiment 3 Exercise 1 – Diffusion: Movement of Solutes across a Membrane

· Experiment 3 Exercise 2 – Osmosis: Movement of Water across a Membrane

Experiment 3 Exercise 1: Diffusion – Movement of Solutes across a Membrane

We will be using dialysis tubing to simulate a semipermeable membrane. This tubing allows small molecules (e.g., water, ions, glucose) to pass while preventing large molecules (e.g., macromolecules like proteins, starch, glycogen) from moving across. Be sure you have read over the suggested material before starting this exercise and that you have reviewed the following animations:

McGraw-Hill. 2006. How Diffusion Works https://highered.mcgraw-hill.com/sites/0072495855/student_view0/chapter2/animation__how_diffusion_works.html

McGraw-Hill. 2006. How Osmosis Works https://highered.mcgraw-hill.com/sites/0072495855/student_view0/chapter2/animation__how_osmosis_works.html

Experimental Design

A. The dialysis bag we will use is permeable to water and small molecules (e.g., less than 500 g/mol) and impermeable to large molecules (e.g., more than 500 g/mol).

B. The dialysis bag is filled with a mixture of glucose (molecular weight = 180 g/mol) and protein (molecular weight = 10,000 g/mol) dissolved in water. A small subsample of the dialysis bag contents is saved and will be used in Step 4.

C. The dialysis bag is then placed into a beaker of water. A small subsample of beaker water is also saved and is to be used in Step 4 as well.

image1.png

The presence or absence of glucose and protein will be determined using indicators. Indicators change colors in the presence certain materials. The two tests that we’ll use are the Benedict’s test for simple sugars (e.g., glucose) and the Biuret test for the presence of proteins.

· If glucose is present, the Benedict’s indicator will turn green. If no glucose is present, the solution will be blue.

· If protein is present, the Biuret indicator will turn violet. If the solution remains clear, then no protein is present.

D. The subsample of dialysis bag solution and the beaker water are tested for the presence of glucose and protein. See Table 1 below for the results.

E. The dialysis bag is then left in the beaker of water for 60 minutes.

F. At the end of 60 minutes, the dialysis bag solution and the beaker water are again tested for the presence of glucose and protein. See Table 1 below for the results.

Table 1. Results of testing of the dialysis bag and beaker contents at the beginning and end of the Experiment.

  Test for Glucose Test for Protein
  Beginning End Beginning End
Dialysis Bag Green Green Violet Violet
Beaker Blue Green Clear Clear

Questions

1. Summarize the results regarding the presence (+) or absence (-) of glucose and protein in the dialysis bag and beaker in Table 2 below (4 pts):

Table 2.

  Glucose Protein
  Beginning End Beginning End
Dialysis Bag        
Beaker        

2. Explain the movement or lack of movement of protein and glucose across the dialysis bag membrane (4 pts)

3. Which solution, that in the bag or that in the beaker, is hypotonic compared with the protein solution (2 pts)?

4. What factors affect the movement of molecules across a semipermeable membrane? Which factor plays the greatest role in biological systems (4 pts)?

5. Briefly explain what active transport is and how it differs from passive transport, especially in terms of concentration gradients (4 pts).

 

Experiment 3 Exercise 2: Osmosis – The Movement of Water across a Membrane

Before starting, let’s see what you know about the terms hypotonicisotonic and hypertonic. Examine the diagrams below. Note that the small green circles represent dissolved solutes like salt, glucose, and amino acids. You can assume that the additional space surrounding the solutes is water and that the tan area is INSIDE the cell.

image2.png

Question

1. Define each term below in terms of solute concentration outside compared to the inside of the cell. You do not need to explain which direction water will move (3 pts).

a. Hypotonic –

b. Isotonic –

c. Hypertonic –

Procedure

A. Open the following website to get started:

The Biology Place. No Date. Osmosis: Movement of Water across Membranes http://www.phschool.com/science/biology_place/biocoach/biomembrane1/osmosis.html

B. Read over the information presented and then Click on image3.png

C. Then, Click on image4.png. Read through the information presented and be sure to click on Animate beneath the illustration.

Questions

2. What concentration of salt is isotonic to animal cells (1 pts)?

3. When cells are in isotonic solution, is there movement of water into or out of the cell? If so, describe this movement (3 pts).

 

Procedure (continued)

D. Click on image5.png.

E. Read through the information presented and be sure to click on Animate beneath the illustration. When ready, answer the following question.

Question

4. Describe the net movement of water molecules when cells are placed in a hypotonic solution. Explain why water moves this way (3 pts).

 

Procedure (continued)

F. Click on image6.png

G. Read through the information presented and be sure to click on Animate beneath each of the illustrations. Answer the following questions. Your answers should incorporate the terminology used in the animations.

Questions

5. What happens to an animal cell when placed in a hypotonic solution (2 pts)?

6. What happens to plant cells when placed in a hypotonic solution? What accounts for the difference in outcomes between animal cells and plant cells (3 pts)?

 

Procedure (continued)

H. Click on image7.png

I. Then, Click on image8.png. Read through the information presented and be sure to click on Animate beneath the illustration. Answer the following question.

Question

7. Describe the net movement of water molecules when cells are placed in a hypertonic solution. Explain why water moves this way (3 pts).

 

Procedure (continued)

J. Click on image9.png

K. Read through the information presented and be sure to click on Animate beneath the illustration. Answer the following questions.

Questions

8. Compare and contrast what happens to plant and animal cells when placed in a hypertonic solution. Be sure to use proper terminology (4 pts).

9. Based on what you learned in this exercise, explain why salt might make a good weed killer (3 pts).

 

Week 3 Experiment Grading Rubric

Component Expectation Points
Experiment 3 Exercise 1 Interpretation of results and demonstrated understanding of diffusion (Table 2 and Questions 1-5). 18 pts
Experiment 3 Exercise 2 Demonstrates understanding of isotonic solutions and no net movement of water (Questions 1-3). 7 pts
  Demonstrates understanding of hypotonic solutions and the movement of water (Questions 4-6). 8 pts
  Demonstrates understanding of hypertonic solutions and the movement of water (Questions 7-9). 10 pts
TOTAL  
 
"Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!"

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

 
"Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!"

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.

 
"Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!"

Monohybrid crosses

How to Proceed

  • Read through the introductory materials below.
  • Open the Unit 6 Experiment Answer Sheet and complete the following Experiment exercises this unit:
    • Experiment 6 Exercise 1 – Monohybrid crosses (~45 min)
    • Experiment 6 Exercise 2 – Dihybrid crosses (~30 min)
    • Experiment 6 Exercise 3 – Inheritance of Human Traits (~30 min)
  • Save your completed Unit 6 Experiment Answer Sheet and submit it no later than Sunday midnight (CT).

Monhybrid and Dihybrid Crosses – Introduction

Mendel crossed true-breeding pea plants in order to develop and understanding of how traits are inherited. True-breeding means that if a plant was crossed with itself, it always generated offspring that looked like the parent. Although Mendel didn’t know this at the time, it meant that the parent plant was homozygous or had two copies of the same allele that controlled the appearance of the trait.

Mendel noticed that when he crossed two true-breeding plants exhibiting different versions of a trait (e.g., green and yellow); the offspring (F1) always looked like only one of the parent plants. We know now that the F1 individuals looked like the parent that carried the dominant trait. But what surprised Mendel, was that when he crossed the F1 individuals with each other, the F2 offspring exhibited BOTH traits! Based on this observation, he concluded that the F1 individuals were hybrids, meaning they carried both alleles for a given trait. Only the dominant trait was expressed in the F1 individuals and the recessive trait, although present, was masked.

monohybrid cross is when you are interested in crossing individuals that vary in only a single trait (e.g., flower color, seed color, stem length). In a dihybrid cross, we are crossing individuals that differ at two traits (e.g., flower color and seed color, flower color and stem length). Obviously, the more traits that vary, the more complex the crosses become!

By examining the distribution of the various traits obtained following different types of crosses, Mendel was able to describe the general pattern of genetic inheritance. Be sure to review the online lecture this unit on Genetics and pp 146-153 in your book before starting these first two exercises.

We will be using the following website for the first exercise. Be sure that you can access it and use it before beginning:

Glencoe-McGraw Hill. No date. Punnett Squares
http://glencoe.mcgraw-hill.com/sites/dl/free/0078759864/383934/BL_05.html (Links to an external site.)

You will need to complete the Tables and answer the questions in the Unit 6 Experiment Answer Sheet for Exercises 1 and 2.

Inheritance of Human Traits – Introduction

Some human traits are controlled by a single gene that has only two alternative alleles. If a characteristic is determined by the dominant allele, one or both parents express that trait and many of the children will as well. Dominant characteristics will most likely be present in every generation, since the expression of these traits requires only one of the dominant alleles in order to be expressed. If the characteristic is determined by the recessive allele, then neither parent may express the trait nor few of the children. This is because two copies of the recessive allele must be present in order for the recessive trait to be expressed. If a trait is X-linked recessive; meaning the gene for the trait is found on the X chromosome, it will be expressed primarily in males.

The application of human genotypes in medicine and genetic counseling is becoming more and more necessary as we discover more about the human genome. Despite our increasing ability to decipher the chromosomes and their genes, an accurate family history remains one of the best sources of information concerning the individual. In this exercise you will determine your genotype for certain characteristics that are controlled by a single gene with two alleles based on your phenotype. We will not be looking at any X-linked traits in this exercise.

Use the information about the traits of interest in the Unit 6 Experiment Answer Sheet to answer the questions found there.

WEEK 6 EXPERIMENT ANSWER SHEET Please submit to the Week 6 Experiment dropbox no later than Sunday midnight.

SUMMARY OF ACTIVITIES FOR WEEK 6 EXPERIMENT ASSIGNMENT

· Experiment 6 Exercise 1 – Monhybrid Crosses

· Experiment 6 Exercise 2 – Dihybrid Cross

· Experiment 6 Exercise 3 – Inheritance of Human Traits

Experiment 6 Exercise 1: Monohybrid Crosses

You will be conducting monohybrid crosses using fruit flies. Open in the following website:

Glencoe-McGraw Hill. No date. Punnett Squares http://glencoe.mcgraw-hill.com/sites/dl/free/0078759864/383934/BL_05.html

Procedure

A. Open the above website and click on the VCR to listen to the introduction. Close the window when done.

B. Click on the Lab Notebook on the lab bench. A breeding scenario will be presented to you that you will need to carry out. Here are the possible phenotypes and genotypes you will using:

a. Normal wings (LL or Ll) or vestigial wings (ll)

b. Gray body (GG or Gg) or black body (gg)

C. Enter the Scenario number in Table 1 below.

D. Based on the Scenario, use the down arrows beneath the Parent 1 and Parent 2 boxes to select the appropriate parents. Look carefully at the flies so that you know which ones to select.

E. Before proceeding, click on the Check Parents button. If necessary, make corrections. If you are correct, the maternal and paternal alleles will be added to the Punnett Square.

F. Next, drag the correct allele combinations and the corresponding fly types to the boxes in the Punnett Square.

G. When you are done, click on Check Offspring. If necessary, make corrections.

H. Record your data in Table 1 below. An example has been given, but note that the example is not using the genotypes and phenotypes used in this exercise.

I. Click on Reset. Repeat steps B – H four more times for a total of five crosses. If a scenario is presented that has already been completed, click Reset again. DO NOT REPEAT a given scenario!

Note that the scenario number you need to record in the Table below is the number associated with the specific scenario you completed.

Table 1. Results of crosses.

 

Parent Genotypes Offspring Genotypes Offspring Phenotype
Scenario # Parent 1 Parent 2 # % # %
Example Rr rr 2 Rr

2 rr

50% Rr

50% rr

2 red

2 white

50% red

50% white

             
             
             
             
             

Questions

1. Which type of cross gave you the greatest number of genotypes? Was the number of phenotypes the same as the number of genotypes or different? If different, why (2 pts)?

2. Can the genotype for a gray-bodied fly be determined? If so, how? (3 pts)?

Experiment 6 Exercise 2: Dihybrid Cross

We will continue to use flies for our crosses, but this time we will examine the inheritance of TWO different traits: body color (gray or black) and wing type (long or vestigial). As with our first crosses, the gray body color is dominant (GG or Gg) over the black body color (gg). And the long wing type is dominant (LL or Ll) over vestigial (ll). Be sure you have reviewed our online Genetics lecture and this week’s reading before proceeding. An example of a dihybrid cross is shown on p 150 in your book.

Recall our flies from the previous exercise. We have the following traits:

· Gray body (GG or Gg) is dominant over black body (gg)

· Long wings (LL or Li) is dominant over vestigial wings (ll)

We will cross a gray bodied fly with long wings which has the genotype GGLl with a gray bodied fly with long wings with a genotype of GgLl . Note that even though the phenotypes are the same, the genotypes of the two parents are different.

Identify the four possible gametes produced by these two individuals. Note that each gamete must consist of two alleles (G or g and L or l):

Parent 1 Parent 2

image1.png image2.png

GGLl GgLl

Parent 1 (GGLl) Gametes: _______ ______ _______ _______ (1 pts)

Parent 2 (GgLl) Gametes: _______ ______ _______ _______ (1 pts)

Create a Punnett square to show the outcome of a cross between these two individuals (GGLl and GgLl) using the gametes you identified above (3 pts).

    Parent 1
           
Parent 2          
           
           
           

Questions

1. What are the possible F1 genotypes (these must now have four alleles) and their percentages (4 pts)?

2. Recall that GG and Gg individuals are gray bodied, while gg individuals are black bodied and that LL and Li individuals have long wings, while ll individuals have vestigial wings. What are the phenotypes of the resulting offspring and what are the percentages of these phenotypes (2 pts)?

Experiment 6 Exercise 3: Inheritance of Human Traits

Read over the Inheritance of Human Traits Introduction under the Week 6 Experiment link in our course before beginning.

Procedure

A. For each of the heritable traits describe below, determine which form you have (dominant form or recessive form). This is your phenotype.

B. Record your phenotype information in Table 2 below. Then, enter the possible genotype(s) you have based on your phenotype.

C. Answer the questions found following Table 2 below.

Description of Heritable Traits

Trait Possible Alleles Dominant Form Recessive

Form

Examples
Ear lobes E or e Detached (Free) Attached  

image3.png 

Hairline W or w Widows peak Straight image4.jpg Widow’s peak Straight
Tongue rolling T or t Able to roll Unable to roll image5.jpg
Hand folding R or r Right thumb on top Left thumb on top image6.jpg
Chin C or c Cleft chin No cleft chin image7.jpg
Tongue folding F or f Can fold tongue backwards Cannot fold tongue backwards image8.jpg
Thumb H or h Straight thumb (cannot bend backwards) Hitchhiker’s thumb (can bend it backwards) image9.jpg
Little Finger B or b Bent inwards Straight image10.jpg
Mid-digital hair M or m Hair on fingers No hair on fingers  

image11.png 

An example is shown as to what should be entered in RED. Please correct the entry for “Ear lobes” based on your personal data. For the Genotypes, please use the letters provided above (8 pts).

Table 2. Your phenotypes and genotypes.

Trait Phenotype Genotype
Ear lobes Unattached OR Detached EE/Ee OR

ee

Hairline    
Tongue Rolling    
Hand Folding    
Chin    
Tongue Folding    
Thumb    
Little Finger    
Mid-digital Hair    

Questions

1.  Which traits did you have that were dominant (1 pts)?

2.  Which traits did you have that were recessive (1 pts)?

3. What does it mean to be homozygous for a trait? Cite source(s) used (1 pts).

4.  What does it mean to be heterozygous for a trait? Cite source(s) used (1 pts).

5.  Define genotype and phenotypeCite source(s) used (1 pts).

6.  Which traits do you know for sure that you were homozygous (1 pts)?

Week 6 Experiment Grading Rubric

Component Expectation Points
Experiment 6 Exercise 1 Correctly perform and record the outcome of five monohybrid crosses (Table 1).

5
 

Demonstrate an understanding of the possible outcomes of monohybrid crosses with respect to genotypes and phenotypes (Questions 1-2).

5
Experiment 6 Exercise 2 Determine the correct parental gametes and conduct a dihybrid cross.

5
 

Correctly evaluate the outcome of a dihybrid cross (Questions 1-2).

6
Experiment 6 Exercise 3

Correctly recognize one’s phenotype and assigns the correct genotype (Table 2).

8
 

Demonstrate an understanding of dominant and recessive traits, genotype vs phenotype and homozygous vs heterozygous (Questions 1-6).

6
TOTAL  

35 pts

Updated October 2013

 
"Looking for a Similar Assignment? Get Expert Help at an Amazing Discount!"