Author: Biology Junction Team

Ink Chromatography

Chromatography of Inks

Introduction:

One of the main jobs of biochemists is to unravel the complexities of chemical compounds and reduce them to their individual components.  The term chromatography comes from two Greek words, “chromat” meaning color and the word “graphon” meaning to write.  Separation of the components of chemical compounds can be done by using several methods. Liquids can be separate by High Performance liquid Chromatography (HPLC), while the components of gases are separated by Gas Chromatography.  Chromatography is a method for analyzing complex mixtures (such as ink) by separating them into the chemicals from which they are made. Chromatography is used to separate and identify all sorts of substances in police work. Drugs from narcotics to aspirin can be identified in urine and blood samples, often with the aid of chromatography.

Chromatography was first used to separate pigments (colors) in leaves, berries, and natural dyes. Paper chromatography is a technique used to separate, isolate, and identify chemical components of a compound. In paper chromatography, the solid surface is the cellulose fibers in the chromatography paper.  A solvent or developer (water, alcohol, or acetone) is placed in the bottom of the chromatography chamber. The paper acts as a wick to pull the solvent up the paper. The solvent front will “wick” up the chromatography paper by capillary action.  A minute drop of the ink or chemical mixture to be separated is placed near the bottom of the strip of chromatography paper, but slightly above the level of the solvent in the chamber.  As the solvent passes over the drop of ink, the components of the ink dissolve in the solvent. Because the components of the ink do not all dissolve at the same rate, as the components of the mixture move upward, they show up as colored streaks.  The separated substances on the chromatography paper form a color pattern called a chromatogram.

To determine the rate of migration for each pigment or component of the ink, the Rf value for each pigment must be calculated. The Rf value represents the ratio of the distance a pigment moved on the chromatogram relative to the  distance the solvent front moved. Each pigment or compound will have a unique Rf value that scientists can use to identify the substance. The Rf value is calculated using the following formula:

Rf = distance traveled by the compound / distance traveled by the solvent

Objective:

Use the process of paper chromatography to separate the pigments in various markers and then determine the Rf value for each color on your chromatogram.

Materials:

Plastic vials, paper clips, markers in assorted colors, chromatography paper, scissors, pencil

Procedure:

  1. Obtain chromatography vials and chromatography strips, and different color markers so that each person in the group will have two chromatograms.
  2. Cut one end of the chromatography strip to a point. The bottom of the point will mark the starting point for movement of the solvent (H2O).
  3. About 2.0 centimeters from the bottom of the strip, draw a faint horizontal line with pencil. This will mark the starting point for measuring the migration distance of each color.
  4. Using a different color marker for each strip, drop a dot of ink on the center of the horizontal pencil line.  Let this dry a moment & then add more ink to the dot.
  5. Add a small amount of water to the bottom of the chromatography chamber. (The ink dot should be ABOVE the surface of the water.)
  6. Straighten a paper clip and poke a hole through the top of your chromatography strip
  7. Use the paper clip to hang the strip in your chamber. (The straighten paper clip will lay across the top of the chamber.)
  8. MAKE SURE THE TIP OF THE STRIP BUT NOT THE INK IS IMMERSED IN THE WATER!
  9. Notice the separation of the ink as both the solvent and ink travel up the chromatography strip.
  10. Once the solvent front has neared the top of the strip, remove the strip from the chamber and lay it on a piece of paper towel.
  11. Immediately mark the solvent front with a faint pencil line.
  12. Immediately mark the leading edge of each color with an “x”.
  13. Measure, in millimeters, the distance the solvent migrated from the tip of the strip to your solvent front pencil line.
  14. Measure, in millimeters, the distance each color migrated from the point of origin (pencil line where the ink dot was placed) to the leading edge of the color (marked with an “x”.
  15. Record all data in Data table 1.
  16. Calculate and record the Rf value for each color using the formula below.

Rf = distance traveled by the compound / distance traveled by the solvent

Data Table 1

 

Color pen/marker used:

Separated colors
(list top of strip to bottom) 		Distance each color traveled

(mm)

Distance solvent (H2O)
(mm) 		Rf Value for each color

(Distance color traveled / Distance solvent traveled)

       
       
       
       
       
       
       
       

 

 

 

Color pen/marker used:

Separated colors
(list top of strip to bottom) 		Distance each color traveled

(mm)

Distance solvent (H2O)
(mm) 		Rf Value for each color

(Distance color traveled / Distance solvent traveled)

       
       
       
       
       
       
       
       

 

 

Questions:

1. Which color of marker did you use?

2. which color separated out first from your ink dot?

3. Why did the inks separate?

 

4. What was your solvent?

5. If you had used markers that weren’t water-soluble, how would you have had to change this lab?

 

6. Why did some inks move a greater distance than others?

 

7. How do scientists use paper chromatography in their investigations?

 

 

Identifying Controls and Variables

Identifying Controls and Variables

 

Smithers thinks that a special juice will increase the productivity of workers. He creates two groups of 50 workers each and assigns each group the same task (in this case, they’re supposed to staple a set of papers). Group A is given the special juice to drink while they work. Group B is not given the special juice. After an hour, Smithers counts how many stacks of papers each group has made. Group A made 1,587 stacks, Group B made 2,113 stacks.

 

 Identify the:

1. Control Group

2. Independent Variable

3. Dependent Variable

4. What should Smithers’ conclusion be?

 

5. How could this experiment be improved?
Homer notices that his shower is covered in a strange green slime. His friend Barney tells him that coconut juice will get rid of the green slime. Homer decides to check this out by spraying half of the shower with coconut juice. He sprays the other half of the shower with water. After 3 days of “treatment” there is no change in the appearance of the green slime on either side of the shower.

 

 6. What was the initial observation?

Identify the-
7. Control Group

8. Independent Variable

9. Dependent Variable

10. What should Homer’s conclusion be?

 

 

 

Bart believes that mice exposed to microwaves will become extra strong (maybe he’s been reading too much Radioactive Man). He decides to perform this experiment by placing 10 mice in a microwave for 10 seconds. He compared these 10 mice to another 10 mice that had not been exposed. His test consisted of a heavy block of wood that blocked the mouse food. he found that 8 out of 10 of the micro waved mice were able to push the block away. 7 out of 10 of the non-micro waved mice were able to do the same. Identify the-
11. Control Group12. Independent Variable

13. Dependent Variable

14. What should Bart’s conclusion be?

15. How could Bart’s experiment be improved?
Krusty was told that a certain itching powder was the newest best thing on the market, it even claims to cause 50% longer lasting itches. Interested in this product, he buys the itching powder and compares it to his usual product. One test subject (A) is sprinkled with the original itching powder, and another test subject (B) was sprinkled with the Experimental itching powder. Subject A reported having itches for 30 minutes. Subject B reported to have itches for 45 minutes. Identify the-
16. Control Group17. Independent Variable

18. Dependent Variable

19. Explain whether the data supports the advertisements claims about its product.

Lisa is working on a science project. Her task is to answer the question: “Does Rogooti (which is a commercial hair product) affect the speed of hair growth”. Her family is willing to volunteer for the experiment.

 20. Describe how Lisa would perform this experiment. Identify the control group, and the independent and dependent variables in your description.

 

 

HW casesheets

Case:

 

Initial Class Frequencies: AA __________ Aa __________ aa _________

My In initial Genotype: __________

F1 Genotype __________

F2 genotype __________

F3 Genotype __________

F4 Genotype __________

F5 Genotype __________

 

Final Class Frequencies: AA __________ Aa __________ aa __________

P: __________     q: __________

 

Number of A alleles present at the fifth generation:

Number of offspring with genotype AA __________ x 2 = __________ A alleles

Number of offspring with genotype Aa __________ x 1 = __________ A alleles

Total = __________ A alleles

P =     TOTAL number of A alleles                                           = ____________
TOTAL number of alleles in population

Number of a alleles present at the fifth generation:

Number of offspring with genotype aa __________ x 2 = __________ a alleles

Number of offspring with genotype Aa __________ x 1 = __________ a alleles

Total = __________ a alleles

 

q =     TOTAL number of a alleles                                            = ____________
TOTAL number of alleles in population

 

BACK

HW answers

Hardy-Weinberg Answers

PROBLEM #1.    You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:

  1. The frequency of the “aa” genotype. Answer: 36%, as given in the problem itself.
  2. The frequency of the “a” allele. Answer: The frequency of aa is 36%, which means that q2 = 0.36, by definition. If q2 = 0.36, then q = 0.6, again by definition. Since q equals the frequency of the a allele, then the frequency is 60%.
  3. The frequency of the “A” allele. Answer: Since q = 0.6, and p + q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%.
  4. The frequencies of the genotypes “AA” and “Aa.” Answer: The frequency of AA is equal to p2, and the frequency of Aa is equal to 2pq. So, using the information above, the frequency of AA is 16% (i.e. p2 is 0.4 x 0.4 = 0.16) and Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48).
  5. The frequencies of the two possible phenotypes if “A” is completely dominant over “a.” Answers: Because “A” is totally dominate over “a”, the dominant phenotype will show if either the homozygous “AA” or heterozygous “Aa” genotypes occur. The recessive phenotype is controlled by the homozygous aa genotype. Therefore, the frequency of the dominant phenotype equals the sum of the frequencies of AA and Aa, and the recessive phenotype is simply the frequency of aa. Therefore, the dominant frequency is 64% and, in the first part of this question above, you have already shown that the recessive frequency is 36%.

 

PROBLEM #2.    Sickle-cell anemia is an interesting genetic disease. Normal homozygous individials (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, many of these individuals become very ill from the parasite and many die. Individuals homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect. However, individuals with the heterozygous condition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these “partially defective” red blood cells. Thus, heterozygotes tend to survive better than either of the homozygous conditions. If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene? Answer: 9% =.09 = ss = q2. To find q, simply take the square root of 0.09 to get 0.3. Since p = 1 – 0.3, then p must equal 0.7. 2pq = 2 (0.7 x 0.3) = 0.42 = 42% of the population are heterozygotes (carriers).

 

PROBLEM #3.    There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. Sorry. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and recessive alleles, and if the four (4%) represent the frequency of the homozygous recessive condition, please calculate the following:

  1. The frequency of the recessive allele. Answer: Since we believe that the homozygous recessive for this gene (q2) represents 4% (i.e. = 0.04), the square root (q) is 0.2 (20%).
  2. The frequency of the dominant allele. Answer: Since q = 0.2, and p + q = 1, then p = 0.8 (80%).
  3. The frequency of heterozygous individuals. Answer: The frequency of heterozygous individuals is equal to 2pq. In this case, 2pq equals 0.32, which means that the frequency of individuals heterozygous for this gene is equal to 32% (i.e. 2 (0.8)(0.2) = 0.32).

 

PROBLEM #4.    Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Given this simple information, which is something that is very likely to be on an exam, calculate the following:

  1. The percentage of butterflies in the population that are heterozygous.
  2. The frequency of homozygous dominant individuals. Answers: The first thing you’ll need to do is obtain p and q. So, since white is recessive (i.e. bb), and 40% of the butterflies are white, then bb = q2 = 0.4. To determine q, which is the frequency of the recessive allele in the population, simply take the square root of q2 which works out to be 0.632 (i.e. 0.632 x 0.632 = 0.4). So, q = 0.63. Since p + q = 1, then p must be 1 – 0.63 = 0.37. Now then, to answer our questions. First, what is the percentage of butterflies in the population that are heterozygous? Well, that would be 2pq so the answer is 2 (0.37) (0.63) = 0.47. Second, what is the frequency of homozygous dominant individuals? That would be p2 or (0.37)2 = 0.14.

PROBLEM #5.    A rather large population of Biology instructors have 396 red-sided individuals and 557 tan-sided individuals. Assume that red is totally recessive. Please calculate the following:

  1. The allele frequencies of each allele. Answer: Well, before you start, note that the allelic frequencies are p and q, and be sure to note that we don’t have nice round numbers and the total number of individuals counted is 396 + 557 = 953. So, the recessive individuals are all red (q2) and 396/953 = 0.416. Therefore, q (the square root of q2) is 0.645. Since p + q = 1, then p must equal 1 – 0.645 = 0.355.
  2. The expected genotype frequencies. Answer: Well, AA = p2 = (0.355)2 = 0.126; Aa = 2(p)(q) = 2(0.355)(0.645) = 0.458; and finally aa = q2 = (0.645)2 = 0.416 (you already knew this from part A above).
  3. The number of heterozygous individuals that you would predict to be in this population. Answer: That would be 0.458 x 953 = about 436.
  4. The expected phenotype frequencies. Answer: Well, the “A” phenotype = 0.126 + 0.458 = 0.584 and the “a” phenotype = 0.416 (you already knew this from part A above).
  5. Conditions happen to be really good this year for breeding and next year there are 1,245 young “potential” Biology instructors. Assuming that all of the Hardy-Weinberg conditions are met, how many of these would you expect to be red-sided and how many tan-sided? Answer: Simply put, The “A” phenotype = 0.584 x 1,245 = 727 tan-sided and the “a” phenotype = 0.416 x 1,245 = 518 red-sided ( or 1,245 – 727 = 518).

 

PROBLEM #6.    A very large population of randomly-mating laboratory mice contains 35% white mice. White coloring is caused by the double recessive genotype, “aa”. Calculate allelic and genotypic frequencies for this population. Answer: 35% are white mice, which = 0.35 and represents the frequency of the aa genotype (or q2). The square root of 0.35 is 0.59, which equals q. Since p = 1 – q then 1 – 0.59 = 0.41. Now that we know the frequency of each allele, we can calculate the frequency of the remaining genotypes in the population (AA and Aa individuals). AA = p2 = 0.41 x 0.41 = 0.17; Aa = 2pq = 2 (0.59) (0.41) = 0.48; and as before aa = q2 = 0.59 x 0.59 = 0.35. If you add up all these genotype frequencies, they should equal 1.

 

PROBLEM #7.    After graduation, you and 19 of your closest friends (lets say 10 males and 10 females) charter a plane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a deserted island. No one finds you and you start a new population totally isolated from the rest of the world. Two of your friends carry (i.e. are heterozygous for) the recessive cystic fibrosis allele (c). Assuming that the frequency of this allele does not change as the population grows, what will be the incidence of cystic fibrosis on your island? Answer: There are 40 total alleles in the 20 people of which 2 alleles are for cystic fibrous. So, 2/40 = .05 (5%) of the alleles are for cystic fibrosis. That represents p. Thus, cc or p2 = (.05)2 = 0.0025 or 0.25% of the F1 population will be born with cystic fibrosis.

 

PROBLEM #8.    You sample 1,000 individuals from a large population for the MN blood group, which can easily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). They are typed accordingly:

 

BLOOD TYPE GENOTYPE NUMBER OF INDIVIDUALS RESULTING FREQUENCY
M MM 490 0.49
MN MN 420 0.42
N NN 90 0.09

 

Using the data provide above, calculate the following:

  1. The frequency of each allele in the population. Answer: Since MM = p2, MN = 2pq, and NN = q2, then p (the frequency of the M allele) must be the square root of 0.49, which is 0.7. Since q = 1 – p, then q must equal 0.3.
  2. Supposing the matings are random, the frequencies of the matings. Answer: This is a little harder to figure out. Try setting up a “Punnett square” type arrangement using the 3 genotypes and multiplying the numbers in a manner something like this:

    MM (0.49) MN (0.42) NN (0.09)
    MM (0.49) 0.2401* 0.2058 0.0441
    MN (0.42) 0.2058 0.1764* 0.0378
    NN (0.09) 0.0441 0.0378 0.0081*

     

    Note that three of the six possible crosses are unique (*), but that the other three occur twice (i.e. the probabilities of matings occurring between these genotypes is TWICE that of the other three “unique” combinations. Thus, three of the possibilities must be doubled. MM x MM = 0.49 x 0.49 = 0.2401
    MM x MN = 0.49 x 0.42 = 0.2058 x 2 = 0.4116
    MM x NN = 0.49 x 0.09 = 0.0441 x 2 = 0.0882
    MN x MN = 0.42 x 0.42 = 0.1764
    MN x NN = 0.42 x 0.09 = 0.0378 x 2 = 0.0756
    NN x NN = 0.09 x 0.09 = 0.0081
  3. The probability of each genotype resulting from each potential cross. Answer: You may wish to do a simple Punnett’s square monohybrid cross and, if you do, you’ll come out with the following result: MM x MM = 1.0 MM
    MM x MN = 0.5 MM 0.5 MN
    MM x NN = 1.0 MN
    MN x MN = 0.25 MM 0.5 MN 0.25 NN
    MN x NN = 0.5 MN 0.5 NN
    NN x NN = 1.0 NN

PROBLEM #9.   Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following:

  1. The frequency of the recessive allele in the population. Answer: We know from the above that q2 is 1/2,500 or 0.0004. Therefore, q is the square root, or 0.02. That is the answer to our first question: the frequency of the cystic fibrosis (recessive) allele in the population is 0.02 (or 2%).
  2. The frequency of the dominant allele in the population. Answer: The frequency of the dominant (normal) allele in the population (p) is simply 1 – 0.02 = 0.98 (or 98%).
  3. The percentage of heterozygous individuals (carriers) in the population. Answer: Since 2pq equals the frequency of heterozygotes or carriers, then the equation will be as follows: 2pq = (2)(.98)(.02) = 0.04 or 1 in 25 are carriers.

PROBLEM #10.    In a given population, only the “A” and “B” alleles are present in the ABO system; there are no individuals with type “O” blood or with O alleles in this particular population. If 200 people have type A blood, 75 have type AB blood, and 25 have type B blood, what are the alleleic frequencies of this population (i.e., what are p and q)? Answer: To calculate the allele frequencies for A and B, we need to remember that the individuals with type A blood are homozygous AA, individuals with type AB blood are heterozygous AB, and individuals with type B blood are homozygous BB. The frequency of A equals the following: 2 x (number of AA) + (number of AB) divided by 2 x (total number of individuals). Thus 2 x (200) + (75) divided by 2 (200 + 75 + 25). This is 475/600 = 0.792 = p. Since q is simply 1 – p, then q = 1 – 0.792 or 0.208.
PROBLEM #11.    The ability to taste PTC is due to a single dominate allele “T”. You sampled 215 individuals in biology, and determined that 150 could detect the bitter taste of PTC and 65 could not. Calculate all of the potential frequencies. Answer: First, lets go after the recessives (tt) or q2. That is easy since q2 = 65/215 = 0.302. Taking the square root of q2, you get 0.55, which is q. To get p, simple subtract q from 1 so that 1 – 0.55 = 0.45 = p. Now then, you want to find out what TT, Tt, and tt represent. You already know that q2 = 0.302, which is tt. TT = p2 = 0.45 x 0.45 = 0.2025. Tt is 2pq = 2 x 0.45 x 0.55 = 0.495. To check your own work, add 0.302, 0.2025, and 0.495 and these should equal 1.0 or very close to it.

Human Hand Adaptations

 

Human Hand Adaptation

Introduction:        Living things have bodies that are adapted for the places they live and the things they do. Fish have gills so that they can remove oxygen that is dissolved in water. Most plants have green leaves which contain chlorophyll so that they can make food. Jellyfish have stinging cells to capture prey. Birds have hollow spongy bones so that they will be light enough to fly. Arctic animals have layers of fat and thick coats of fur to keep warm in the frigid Arctic climate. There are hundreds of examples of ways that organisms are adapted for a successful lifestyle.       Humans, too, are adapted for the things they do. One of our adaptations is our hand. Humans, as well as monkeys, gorillas, and other primates, have a hand that can grasp objects. We are able to grasp objects because of our opposable thumb. When students first hear or read about the opposable thumb during discussions of human evolution, they may perceive it as an anatomical fact with little seeming importance. In this activity, students will discover which of their simplest daily activities are possible only because of their opposable thumbs, which activities take longer without the use of an opposable thumb, and what sort of human activities would not be likely in the absence of an opposable thumb.   In this lab exercise, you will perform several common actions. Then you will change your hand so that it resembles that of a non-primate animal. You will determine whether or not you can successfully perform the same actions. This will demonstrate how the human hand is adapted for the actions it performs. You will work with a partner to do this exercise.   Materials: (per group)

  • masking tape
  • scissors
  • paper clips
  • zip-lock storage bag
  • plastic fork and knife
  • small amounts of food items to be cut
  • pencil
  • jar with screw-on lid
  • paper
  • roll of tape
  • balloons
  • comb
  • book
  • lace-up shoe
  • clock with a second hand
  • Piece of yarn or string
  • balloon
  • clothes with zippers & buttons

Procedure: Using masking tape, have your partner tightly tape each of your thumbs to the palm of the hand. Then, try to complete the tasks that are listed below. Be careful not to use your thumbs. Have your partner record on your data table how long it takes to do each task with your thumb taped and then with your thumb free. If an activity takes longer than 2 minutes, record the event as unsuccessful . After completing each item, write out the answers to the following questions:

  • Is the task more difficult with or without an opposable thumb?
  • How did you have to change your usual technique in order to complete this task?
  • Do you think organisms without opposable thumbs would carry out this task on a regular basis? Why or why not?

Tasks:

  1. Pick up a single piece of paper. Put it down on your desk.
  2. Pick up a pen or pencil from the table top. Use it to write your name on the piece of paper.
  3. Open a book. Turn single pages in the book.
  4. Unscrew a bottle cap or jar cover.
  5. Use a fork and knife to cut a food item into small pieces.
  6. Tear off a small piece of tape.
  7. Turn on the water faucet. (Complete activity #8!) Turn it off.
  8. Moisten a paper towel and wash and dry the desktop.
  9. Sharpen a pencil.
  10. Cut a circle out of a piece of paper using scissors.
  11. Pick up all the scraps from activity #10 and throw them into the recycling box.
  12. Comb your hair.
  13. Open a door.
  14. Pick up one paper clip. Clip a pile of papers together.
  15. Tie your shoelaces.
  16. Button several buttons.
  17. Zip up your jacket.
  18. Blow up a balloon and tie it.
  19. Tie a knot in a piece of string.
  20. Close a zip-lock bag.

Data:

Table 1 – Time It Took To Perform Various Tasks

 

Task Time Taken for Event: Task Difficulty With Taped Thumb
(More/Less)		 Modification Made to complete Task
Thumb Free Thumb Taped
Pick up paper
Write name
Turn book pages
Open jar
Use knife & fork
Tear off tape
Turn faucet on & off
Clean desk top
Sharpen a pencil
Cut out a circle
Pick up the scraps of paper
Comb hair
Open door
Clip papers together
Tie shoelaces
Button & unbutton garment
Use zipper
Blow up & tie balloon
Knot string
Close zip-lock bag

Conclusion:   1. Explain why dog and cat paws are not adapted for doing the six actions you tested.     2. What are cat and dog paws adapted for?     3. Describe how your hand is adapted for doing the actions you tested.       4. You have an opposable thumb. Explain what this means.     5. Why do you feel that human hand adaptations have helped to make humans such a successful species on earth?