Taxis is often characterized by light, heat, moisture, sound, or chemicals. Kinesis is another type of movement that involves orientation. Kinesis is a movement that is random and doesn’t involve a stimulus.
They probably don’t have very good eyesight. If you suddenly turned a rock over and found isopods under it, what would you expect them to be doing? If you watched the isopods for a few minutes, how would you expect to see their behavior change? – I would expect them to scurry around, probably find and tunnel and go down it to keep out of the sunlight. If you found one that wasn’t already underground, I would probably see it walking out in the grass, looking for another object to go under. Is the isopod’s response to moisture best classified as kinesis, or taxis? Explain your response.
This is a supplement to the first printing of the AP Biology Lab Manual. Updates to the AP Biology Investigative Lab Manual. Fermentation Lab Review. Faculty Advisors for Biology. In the experiment you performed in lab. Cellular Respiration AP Biology Lab 5 Introduction. Explain the results shown in the sample graph in your lab manual. As the temperature increased.
Which of the following gels produced by electrophoresis would represent the separation and identity of these fragments? 194 Name: Date: Period: LabBench Activity 7: Genetics of Organisms Analysis of Results In the laboratory you breed your flies and analyze the results of the breeding through the F 2 generation. The exercises below are designed to help you understand the patterns of inheritance in your fly populations. Reversing the Procedure One way to discover patterns of inheritance is by working backward. In other words, you determine the genotype of the original parental generation by careful analysis of the F 1 and F 2 generations. Let's examine two sample cases that trace eye color. For each, look at the data chart with the number of male and female flies exhibiting each eye color.
Plates I and II each contain a plasmid that is resistant to that antibiotic. Plate III has antibiotic agar, but E. Coli that has been transformed to be resistant to tetracycline can grow. Plate IV has no antibiotic. There are no tetracycline-resistant cells on Plate II. Analysis of Results II Each fragment of DNA is a particular number of nucleotides, or base pairs, long. When researchers want to determine the size of DNA fragments produced with particular restriction enzymes, they run the unknown DNA alongside DNA with known fragment sizes.
Using a standard to estimate an unknown is sometimes called 'interpolation'; you will interpolate the size of the unknown fragments. You begin by making a standard curve for the known sample, DNA plus HindIII. Measure the distance each HindIII fragment migrated on the gel and then complete the chart.
It serves as an electron donor and blocks the formation of NADPH. Keygen for windows 10 home. It is an electron acceptor and is reduced by electrons from chlorophyll. It is bleached in the presence of light, and can be used to measure light levels. Some students were not able to get many data points in this experiment because the solution went from blue to colorless in only 5 minutes for the unboiled chloroplasts exposed to light.
The curve in graph C would rise more steeply and level off sooner. The curve in graph A would have the same general shape as the curve in graph C. The chloroplasts would absorb more light energy, so there would be no change. What is the role of DPIP in this experiment? It mimics the action of chlorophyll by absorbing light energy.
A group of asci formed from crossing light-spored Sordaria with dark-spored produced the following results: Number of Asci Counted Spore Arrangement 7 4 light/4 dark spores 8 4 dark/4 light spores 3 2 light/2 dark/2 light/2 dark spores 4 2 dark/2 light/2 dark/2 light spores 1 2 dark/4 light/2 dark spores 2 2 light/4 dark/2 light spores 5. How many of these asci contain a spore arrangement that resulted from crossing over? 10 e From this sample, calculate the map distance between the gene and centromere. 10 map units b. 20 map units c. 30 map units d. 40 map units 182 183 Name: Date: Period: LabBench Activity 4: Plant Pigments & Photosynthesis Analysis of Results I If you did a number of chromatographic separations, each for a different length of time, the pigments would migrate a different distance on each run.
Watch hindi serial laagi tujhse lagan online. Note: The average time for onion root tip cells to complete the cell cycle is 24 hours = 1440 minutes. To calculate the time for each stage, do the following:% of cells in the stage 1440 minutes = number of minutes in the stage Phases: A = B = C = D = E = 179 180 Data Table Number of Cells% of Total Cells Counted Time in each stage Interphase Prophase Metaphase Anaphase Telophase Total: Questions 1. Select the phase of the cell cycle depicted. Select the phase of the cell cycle depicted.
Which of the following statements is correct? Crossing over occurs in prophase I of meiosis and metaphase of mitosis. DNA replication occurs once prior to mitosis and twice prior to meiosis.
What is the percent of crossovers? Take the number of asci with crossovers divided by total number of asci multiplied by 100. For the sample shown here, what is the map distance between the gene for spore color and the centromere? Take the percent of crossovers divided by 2.
D Base your answers to questions 5 & 6 on the following information. A group of asci formed from crossing light-spored Sordaria with dark-spored produced the following results: Number of Asci Counted Spore Arrangement 7 4 light/4 dark spores 8 4 dark/4 light spores 3 2 light/2 dark/2 light/2 dark spores 4 2 dark/2 light/2 dark/2 light spores 1 2 dark/4 light/2 dark spores 2 2 light/4 dark/2 light spores 5. How many of these asci contain a spore arrangement that resulted from crossing over? 10 e From this sample, calculate the map distance between the gene and centromere. 10 map units b. 20 map units c. 30 map units d.
Synapsis occurs in prophase of mitosis. The cell cycle in a certain cell type has a duration of 16 hours.
Which of the following conclusions is supported by the data? The rate of respiration is higher in nongerminating seeds than in germinating seeds. Nongerminating peas are not alive, and show no difference in rate of respiration at different temperatures. The rate of respiration in the germinating seeds would have been higher if the experiment were conducted in sunlight. The rate of respiration increases as the temperature increases in both germinating and nongerminating seeds. The amount of oxygen consumed could be increased if pea seeds were substituted for corn seeds.
Positively charged DNA migrates more rapidly than negatively charged DNA. Uncut DNA migrates farther than DNA cut with restriction enzymes 2. How many base pairs is the fragment circled in red below? A ml/min b ml/min c. 0.8 ml/min d ml/min 191 192 3. An instructor had her students perform this laboratory beginning with setting up their own restriction enzyme digests.
You begin by making a standard curve for the known sample, DNA plus HindIII. Measure the distance each HindIII fragment migrated on the gel and then complete the chart. It is very difficult to get exact numbers as you read this graph. If your response is in a close range, that is acceptable.
One team of students had results that looked like those at the left. What is the most likely explanation for these results?
LB agar without ampicillin, +amp R cells b. LB agar without ampicillin, amp R cells c. LB agar with ampicillin, +amp R cells d. LB agar with ampicillin, amp R cells Choice: Choice: Choice: Choice: 188 189 Questions Refer to the following information and images of Plates I, II, III, and IV to answer questions 1 4. In a molecular biology laboratory, a student obtained competent E. Coli cells and used a common transformation procedure to induce the uptake of plasmid DNA with a gene for resistance to the antibiotic kanamycin.
Let's examine two sample cases that trace eye color. For each, look at the data chart with the number of male and female flies exhibiting each eye color. Then answer the questions. Case 1 Case 2 Based on the data obtained, this cross is a. Monohybrid b. Dihybrid This cross is: a.
The rate of oxygen consumption is higher for nongerminating corn at 12 C than at 22 C. If the experiment were run for 30 minutes, the rate of oxygen consumption would decrease 2. What is the rate of oxygen consumption in germinating corn at 12 C? A ml/min b ml/min c. 0.8 ml/min d ml/min 3.
Analysis of Results II Each fragment of DNA is a particular number of nucleotides, or base pairs, long. When researchers want to determine the size of DNA fragments produced with particular restriction enzymes, they run the unknown DNA alongside DNA with known fragment sizes. The known DNA acts as a marker. In your laboratory, the DNA that has been cut with HindIII is the marker; you will use it to help you determine the fragment sizes in the EcoRI digest. On the next pages we go through the procedure using HindIII and two generalized DNA samples. Making a Standard Curve for HindIII DNA Fragments If you know the fragment sizes in the HindIII digest, how do you determine the fragment sizes in an unknown sample? You use data from the marker to prepare a standard curve, which will provide a standard for comparison to the unknown fragment sizes.
X w X w X w Y 194.
So an animal would respond to light by moving everywhere in random directions. Agonistic behavior is when animals respond to each other in aggressive or submissive movements. Like the hair on dogs backs when they get ready to fight. Another excellent example is the Betta fish, which is sometimes studied in labs. Mating behaviors are activities that involve finding, courting, and mating with a member of the same species. An example would be a peacock fluffing up its feathers to attract females.
On the next pages we go through the procedure using HindIII and two generalized DNA samples. Making a Standard Curve for HindIII DNA Fragments If you know the fragment sizes in the HindIII digest, how do you determine the fragment sizes in an unknown sample? You use data from the marker to prepare a standard curve, which will provide a standard for comparison to the unknown fragment sizes. Using a standard to estimate an unknown is sometimes called 'interpolation'; you will interpolate the size of the unknown fragments. You begin by making a standard curve for the known sample, DNA plus HindIII.
Biotic and abiotic factors are limiting factors that control the maximum size of a given population. Favorable conditions are desired by an organism of its home environment. Because of this, an animal must search for the environment to fit its structure and lifestyle. This is called habitat selection. An animal can display many different types of behaviors, two being taxis and kinesis.
If you used the same experimental design to compare the rates of respiration of a 35g mammal at 100 degrees Celsius, what results would you expect? Explain your reasoning.
It binds with carbon dioxide to form a solid, preventing CO 2 production from affecting gas volume. Its attraction for water will cause water to enter the respirometer. 187 188 Name: Date: Period: LabBench Activity 6: Molecular Biology Analysis of Results I If there is no ampicillin in the agar, E. Coli will cover the plate with so many cells it is called a 'lawn' of cells. Only transformed cells can grow on agar with ampicillin.
Select the phase of the cell cycle depicted. Interphase b. Select the phase of the cell cycle depicted. Interphase b.
189 190 Refer to the following information and images of Plates I, II, III, and IV to answer questions 5 & 6. A student has forgotten which antibiotic plasmid she used in her E. Coli transformation. It could have been kanamycin, ampicillin, or tetracycline. She decides to make up a special set of plates to determine the type of antibiotic used. The plates below show the results of the test.
Positively charged DNA migrates more rapidly than negatively charged DNA. Uncut DNA migrates farther than DNA cut with restriction enzymes 2. How many base pairs is the fragment circled in red below? A ml/min b ml/min c. 0.8 ml/min d ml/min 191 192 3. An instructor had her students perform this laboratory beginning with setting up their own restriction enzyme digests.
Which of the following conclusions is supported by the data? The rate of respiration is higher in nongerminating seeds than in germinating seeds. Nongerminating peas are not alive, and show no difference in rate of respiration at different temperatures.
The exercises below are designed to help you understand the patterns of inheritance in your fly populations. Reversing the Procedure One way to discover patterns of inheritance is by working backward. In other words, you determine the genotype of the original parental generation by careful analysis of the F 1 and F 2 generations. Let's examine two sample cases that trace eye color.
The amount of oxygen consumed by germinating corn at 22 C is approximately twice the amount of oxygen consumed by germinating corn at 12 C. The rate of oxygen consumption is the same in both germinating and nongerminating corn during the initial time period from 0 to 5 minutes. The rate of oxygen consumption in the germinating corn at 12 C at 10 minutes is 0.4 ml O 2 /minute.
Rate = slope of the line,. In this case, Δ y is the change in volume, and Δ x is the change in time (10 min).
Describe and explain the relationship between the amount of oxygen consumed and time. The amount of oxygen that was consumed was the greatest in the warmer water. The oxygen consumption increased over time in the germinating peas. Why is it necessary to correct the readings from the peas with the readings from the beads? This is necessary to show the actual rate at which cellular respiration occurs in peas.
Alcoholic fermentation uses enzymes to convert glucose into ATP (which the cell can use to do work) in the absence of oxygen. Alcohol and carbon dioxide are also produced in this reaction. In the experiment you performed in lab, you measured the amount of CO 2 produced in alcoholic fermentation under different conditions. In the lab experiment, you added water, glucose (or dextrose), and either 0, 1, or 3 ml of yeast to the reaction tubes. You placed the tubes containing each reaction into the apparati shown at the left. This apparatus provided an oxygen free chamber in which alcoholic fermentation occurred.
The students did not allow enough time for the electrophoresis separation. The agarose prepartion was faulty. The methylene blue did not stain the DNA evenly.
The results below were obtained. On which petri dish do only transformed cells grow? Which of the plates is used as a control to show that nontransformed E. Coli will not grow in the presence of kanamycin? If a student wants to verify that transformation has occurred, which of the following procedures should she use?
However, the migration of each pigment relative to the migration of the solvent would not change. This migration of pigment relative to migration of solvent is expressed as a constant, R f (Reference front). It can be calculated by using the formula: R f = Look at the black ink chromatogram to the left. Calculate the R f value for green. Show your work. Answer: Questions 1.
What is the role of DPIP in this experiment? It mimics the action of chlorophyll by absorbing light energy. It serves as an electron donor and blocks the formation of NADPH. It is an electron acceptor and is reduced by electrons from chlorophyll. It is bleached in the presence of light, and can be used to measure light levels. Some students were not able to get many data points in this experiment because the solution went from blue to colorless in only 5 minutes for the unboiled chloroplasts exposed to light. What modification to the experiment do you think would be most likely to provide better results?
Both mitosis and meiosis result in daughter cells identical to the parent cells. Karyokinesis occurs once in mitosis and twice in meiosis.
The nuclei of 660 cells showed 13 cells in anaphase. What is the approximate duration of anaphase in these cells? 13 minutes c. 19 minutes d. 32 minutes e. 647 minutes 181 182 Base your answers to questions 3 & 4 on the following figure: 3. For an organism with a diploid number of 6, how are the chromosomes arranged during metaphase I of meiosis?
194 Name: Date: Period: LabBench Activity 7: Genetics of Organisms Analysis of Results In the laboratory you breed your flies and analyze the results of the breeding through the F 2 generation. The exercises below are designed to help you understand the patterns of inheritance in your fly populations. Reversing the Procedure One way to discover patterns of inheritance is by working backward. In other words, you determine the genotype of the original parental generation by careful analysis of the F 1 and F 2 generations.
Plate IV has no antibiotic. There are no tetracycline-resistant cells on Plate II.
Which graph would be the most likely result of performing the photosynthesis experiment using fresh chloroplasts placed in light and DPIP? What is the best explanation for graph B?
Sex-linked b. Autosomal From the data presented, determine the genotype of the parental generation (before the F1 generation; not shown here). + = wild type (red eyes) w = white eyes a. X + X + X + Y b. X + X w X + Y c.
(Keep track of your counts with paper and pencil.) In this exercise, we are interested only in asci that form when mating occurs between the black-spore strain and the tan-spore strain, so ignore any asci that have all black spores or all tan spores. Occasionally the asci rupture and spores escape. You can see them here as individual spores not in one of the possible arrangements, so don't include them in your count. In the photo, how many asci marked with an X show no evidence of crossing over? In the photo, how many asci marked with an X show evidence of crossing over? In the photo, what is the total number of asci marked with an X? What is the percent of crossovers?
Select the phase of the cell cycle depicted. Interphase b. Interphase b. Telophase 180 181 Analysis of Results II Study this small section of a slide of Sordaria to determine if crossing over has occurred in the asci designated by an X. If the ascospores are arranged 4 dark/4 light, count the ascus as 'No crossing over.' If the arrangement of ascospores is in any other combination, count it as 'Crossing over.' (Keep track of your counts with paper and pencil.) In this exercise, we are interested only in asci that form when mating occurs between the black-spore strain and the tan-spore strain, so ignore any asci that have all black spores or all tan spores.
Making a Standard Curve for HindIII DNA Fragments If you know the fragment sizes in the HindIII digest, how do you determine the fragment sizes in an unknown sample? You use data from the marker to prepare a standard curve, which will provide a standard for comparison to the unknown fragment sizes. Using a standard to estimate an unknown is sometimes called 'interpolation'; you will interpolate the size of the unknown fragments.
What would be the rate of oxygen consumption if the respirometer readings were as shown here? Answer: Questions Refer to the following figure for questions 1, 2, & 187 1. Which is the following is a true statement based on the data?