Category: Curriculum Map

Cell Division Study Guide BI

Cell Division Study Guide

What molecule contains the information needed to direct all the activities of a cell?
Where in a cell are prokaryotic chromosomes found? eukaryotic chromosomes?
A human somatic cell contains how many homologous chromosomes?
How many chromosomes are in an human egg cell? sperm cell?
What is a karyotype?
Are gametes diploid or haploid?
Zygotes will have what chromosome number?
Does cell division in bacteria take place in the same way as it does in eukaryotes? Explain.
In what stage do cells spend most of their life cycle?
Is mitosis asexual or sexual reproduction?
A new nuclear envelope develops during cell division in what stage?
In what stage do chromatids separate from each other?
How does the number of chromosomes in newly divided cells compare with the number of chromosomes in the original cell?
During what type of cell division do haploid cells develop from diploid cells?
In order for DNA to fit into a cell, what must be done to compact it?
What is a centromere?
How many chromosomes are in a human skin cell? a human ovum?
Bacteria reproduce by a method known as _____________  ______________.
What is the shape of a bacterial chromosome?
Chromosomes are arranged along the equator of a cell during which stage of cell division?
Spindle fibers are made of ________________.
Be able to recognize sketches of the stages of mitosis.
What happens during cytokinesis in a plant cell?
Homologs separate during ________________.
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Calorimetry lab

Calorimetry – Measuring the energy in Foods

Introduction:
There are two processes that organisms use to make usable energy. The process by which autotrophs convert sunlight to a usable form of energy is called photosynthesis. Photosynthesis supports all life on earth. Products from photosynthesis include food, textiles, fuel, wood, oils, and rubber. During photosynthesis, light energy is used to make organic compounds from inorganic water and carbon dioxide. Photosynthesis goes through light dependent reactions and the light independent reactions which include the Calvin cycle.
The process where heterotrophs break down food molecules to release energy for work is called cellular respiration. Cellular respiration is the reverse of photosynthesis; the reactants of one are the products of the other. The reactants of cellular respiration are glucose and oxygen, and the products are carbon dioxide, water, and energy.  Cellular respiration breaks down glucose to form carbon dioxide and water, while releasing energy usable by the cells. The first step, glycolysis is the process  that converts glucose to pyruvate and releases a small amount of cellular energy.  The second step may be aerobic or anaerobic depending on the amount of oxygen available.  Aerobic respiration is the breakdown of pyruvate in the presence of oxygen.  A larger amount of cellular energy or ATP is produced during the Kreb’s cycle and electron transport chain. Anaerobic respiration is the breakdown of food molecules in the absence of oxygen. Less ATP is produced by anaerobic respiration or fermentation.

Hypothesis:
If the heat given off by a burning pecan is measured by how much the temperature increases in a given amount of water, then the number of calories of energy stored in the nut during photosynthesis can be determined.

Materials:
Items needed for the lab included a large paper clip, a 100 ml graduated cylinder, thermometer, 2 soft drink cans, electronic balance, butane lighter, plastic tray, scissors, paper, and pencil.

Procedure:
Use a graduated cylinder to measure 100 ml of water and add this to an empty soft drink can. Cut holes on two sides of a second soft drink can so there is room to place a large bent paper clip.  Measure and record the mass of one pecan using the electronic balance. Bend a large paper clip to make a “nut stand” and measure and record  the mass of this clip. Place the pecan on the nut stand and put the stand inside the cut-out drink can.  Use a thermometer to measure and record the temperature of the water in the second can.  Place this can on top of the can with the nut. Use a butane lighter to ignite the nut. Record the temperature of the water when the nut is completely burned. Complete the data table by calculating the  the total calories in the pecan.

Data:    

Data Table 1
Before Burning After Burning Difference
Mass of Nut 1.7 g 0.1g 1.6g
Temperature of Water 20 40.1 20.1
Mass of Paper Clip 1.4g 1.4g 0g

 

Data Table 2
Mass of pecan 0.1 g
Temperature change of 100 ml of water 20.1
Calories required to produce temperature change in 100 ml water 2010
Calories per gram contained in the pecan 1182.4

Error Analysis:
Errors may have occurred in several ways during this experiment. One error that may have occurred is that some of the energy may have been lost during the burning. Some of the pecan’s energy was lost as light instead of heat energy. Also some of the heat measured in the water could have been due to the butane lighter used to ignite the pecan.

Conclusion:
The temperature of the 100 ml of water in the can above the burning pecan was changed by the energy given off by the pecan when it was burned.  The energy given off by the burning pecan was great enough to increase the water temperature by 20.1 degrees Celsius. The mass of the unburned pecan was 1.7g. It takes 100 calories to raise the temperature of 1 ml of water by 1 degree Celsius. The temperature of 100 ml of water was recorded to have increased by 20.1 degrees Celsius; therefore, the total number of calories in the pecan equals 20.1 x 100 or 2010 calories. Since the nut had a mass of 1.7g, the number of calories per gram equals 2010 divided by 1.7 or 1182.4 calories per gram.
The increase of temperature in the water showed that energy had been stored in the pecan. In this experiment, the amount of calories of heat energy stored in a pecan during photosynthesis was measured by a process known as calorimetry.

 

Campbell Problem 7

Molecular Genetics Problem 7
7. Using the information from problem 6, a further testcross was done using a heterozygote for height and nose morphology. The offspring were tall-upturned nose, 40; dwarf-upturned nose, 9; dwarf-downturned nose, 42; tall-downturned nose, 9. Calculate the recombination frequency from these data; then use your answer from problem 6 to determine the correct sequence of the three linked genes.

Experiment 3. (Frequency/Distance between T and S)

Determine the recombination frequency for the genes controlling Tallness and Snout:

40 tall-upturned snout = 40% expected
42 dwarf-downturned snout = 42% expected
9 dwarf-upturned snout = 9% recombinant
9 tall-downturned snout = 9% recombinant

Total = 100%

Therefore this recombination frequency between genes T and S is 18%

One can determine the relative frequency between genes using the percent frequencies as distances.

The Recombinant relationships from experiments 1-3 are:

Exp. 1 T-A = 12 map units Exp. 2 A-S = 5 map units Exp. 3 T-S = 18 map units

An arrangement that fits the data would be:

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Catalase Bi Sample Lab 2

 

 

Enzyme Rate of Reaction for Catalase

 

Introduction:
Life would not be possible without chemical reactions. Chemical reactions are responsible for speeding up the process. A chemical reaction is the process of breaking chemical bonds, forming new bonds or both. The four things that can speed up a chemical reaction is heat, increasing the concentration of reactants, decreasing the concentration of products, and enzymes. Enzyme is a catalase, most the time a protein. Enzymes can control the rate of a reaction, and they also lower activation energy. Enzymes are important in regulating chemical pathways, synthesizing materials needed by cells, releasing energy, and transferring information. Enzymes are involved in digestion, respiration, vision, movement, and thought. There are several things that can affect the function of enzymes like temperature, the pH, and the amount of reactant or product. Simple cells may have as many as 2000 different enzymes, each one catalyzing a different reaction. In this particular lab, your hands act as the enzyme “Catalase”. This enzyme, which is found in your cells, splits hydrogen peroxide, a byproduct made by your cells during cellular respiration,  into water and oxygen.

 

Hypothesis:
If  time is increased, then more hydrogen peroxide molecules will be split into water and oxygen

 

Materials:
The materials used in this lab were pencils, scissors, envelope, 100 paper hydrogen peroxide molecules, and a watch with a second hand so that a person would be able to keep time for the person tearing the strips.

 

Methods:
Take a paper template and cut out 100 hydrogen peroxide molecules. Place the cut out pieces into an envelope. Then have a person act as a catalase and take one piece of the paper molecules out of the envelope at a time and rip it in two and  place the pieces back into the envelope. Have a person hold the envelope person, while another student keeps track of the “tearing” time intervals (10, 20, 30 ,60, and 60 seconds). Count how many molecules are  ripped at the end of each time interval and record this number in your data table. When all time intervals and counts are completed, use the formula below to figure the reaction rate for catalase. Record this rate in  your data table.
M2 – M1 = reaction rate
T2 – T1

Results:

 

 

Time in seconds

  

Ripped Hydrogen Peroxide Molecules

  

Rate of reaction
 

0-10

 3 .3
 

10-30

 10 .35
 

30-60

 24 .47
 

60-120

 63 .65
 

120-180

 124 1.02

 

1. What is an enzyme? What are its functions in living things?
Enzymes are proteins in living systems. Enzymes can control the rate of a reaction, and they lower activation energy.

 

2. Name several things things that can affect the function of an enzyme?
Temperature, the amount of reactant or product and the pH.

3. Write the chemical equation for the breakdown of hydrogen peroxide by the enzyme catalase.
hydrogen peroxide + catalase yields water + oxygen

4. An enzyme’s efficiency increases with greater substrate concentration, but only up to a point. Why?
all of the active sites of the enzymes become filled with hydrogen peroxide molecules

 

5. If you were allowed to continue this lab and rip hydrogen peroxide molecules for 240 and 300 seconds. What would happen to the rate of reaction and why would this happen?
It would increase.

 

6. What can you say about the length of time and the rate of the reaction?
The less time, the more the reaction rate is lowered, and the more time, the more the reaction rate is higher.

 

7. What would happen to the rate of reaction if you remove the water  and oxygen molecules as soon as they are produced?
It would be faster.

 

Error Analysis:
All pieces must be returned to the envelope each time interval to correctly simulate what occurs within a cell.

 

Discussion and Conclusion:
As the time intervals increased, the reaction rate of catalase increased also. In a living cell, more hydrogen peroxide would be broken down by catalase over a longer period of time.

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Carbon Dioxide Use in Plants

 

 

Do Plants Consume or Release CO2?

 

Introduction

The rate of photosynthesis can be determined by measuring the rate of production of sugar or oxygen or by measuring the rate of decrease in carbon dioxide concentration. A common aquarium plant called, Elodea,  can be used to show fast carbon dioxide is being removed from the water in which the Elodea is submerged.

6CO2 + 12H2O + light energy —> C6H12O6 + 6O2 + 6H2O

 

In this lab, you will use phenol red as an indicator to show whether CO2 is being consumed or produced in a reaction. It is well known that in the presence of light, plants perform photosynthesis. At the same time, plants are also performing cell respiration. To demonstrate this, we will determine whether CO2 is consumed or produced as Elodea is placed in either a light or dark environment. The change in CO2 will be detected by the pH indicator phenol red. Phenol red is yellow under acidic conditions (high H+ ion concentration), pink to magenta under basic or alkaline conditions (low H+ ion concentration), and orange under neutral conditions. A change in the amount of CO2 will cause a directly proportional change in H+ ion.

If the CO2 concentration decreases, the H+ ion concentration will also decrease, and the solution will change to pink, becoming basic.

If the CO2 concentration increases, the H+ ion concentration will also increase, and the solution will change to yellow, becoming acidic.

Neutral solutions of phenol red will be orange.

Materials:

phenol red solution,  4 sprigs of Elodea, soda straw, 4 test tubes, labeling marker, 100 ml graduated cylinder, beaker, aluminum foil

Procedure:

  1. Create a solution of phenol red by adding concentrated phenol red to about 100 ml of water in a beaker. The phenol red may change color as a result of adding water (depending on how acidic your tap water is). Your goal is to make your solution a neutral orange color. You can do this by gently blowing into the solution with a straw.
  2. Label 4 test tubes 1, 2, 3, and 4.
  3. Once you have the solution at an orange color, transfer it to 4 test tubes (they should be filled about 2/3 full with your orange solution).
  4. Place a cut piece of Elodea (cut end up) into tubes 2 and 4 and tightly cap.
  5. Test tubes 3 and 4 will not have Elodea. Cap and then cover these tubes with aluminum foil so no light can enter.
  6. Place tubes 1 and 2 in bright light.
  7. Place tubes 2 and 4 in the dark.
  8. After 24 hours, uncover and examine all 4 test tubes and record the results.

Data:

 

Test Tube # Contents of Tube Initial Color Final Color
1
2
3
4

 

Conclusion:

1. What test tubes served as the controls in this experiment. Why?

 

 

2. What was the dependent variable?

 

3. Do you think there would have been any change in any of the test tubes if they were left for 48 or 72 hours? Explain.

 

 

4. Describe and explain what happened in the test tubes.

 

 

 

5. Why did the color change occur?

 

6. Where does the carbon dioxide that is removed from the solution go?

 

7. What other process goes on in plant cells that requires oxygen and produces carbon dioxide?

 

8. What was the purpose in tightly capping all four test tubes?