Virtual Lab Report
Virtual Lab 1: Virtual Microscopy
A. Gauge the size of various biological components and organisms. The Virtual Microscope (copy and paste this address into a new browser window: http://learn.genetics.utah.edu/content/cells/scale/ ) can be used to make these observations. Estimate the size (length and width in microns) of
An E.Coli cell is around 0.5µm in width and around 2µm in length. A mitochondrion is around 0.5µm in width and approximately 10µm in length. A Red blood cell is around 5.5µm in width and 7.5µm in length. A virus is approximately 0.25µm in width and around 1µm in length. A water molecule is the smallest and is around 0.0001µm.
1. Observe and describe three differences between prokaryotic ad eukaryotic cells.
The First difference between a prokaryotic and eukaryotic cell is the absence of a well-defined nucleus in a prokaryotic cell. The second difference is the presence of a cellular membrane in eukaryotic cells while in prokaryotic cells it is absent. The third difference is the absence of Mitochondria in prokaryotic cells while it is present in eukaryotic cells.
2. Observe and describe three differences and three similarities between plant and animal cells.
A plant cell has a distinct cell wall while a cell wall is absent in animal cell. Animal cells have centrioles while plant cells lack centrioles. The third difference is that plant cells have chlorophyll while animal cells do not have chlorophyll. There are 3 similarities between plant and animal cells. Both have Golgi bodies. Both have a nucleus and thirdly both have cytoplasm.
C. Form a hypothesis
1. Hypothesize about how you might be able to sort a mixed population of cells into prokaryotes and eukaryotes. Try to be practical, build on your understanding of the differences between the two cell classes.
There are many distinctive characteristics of prokaryotic and eukaryotic cells. Using an optical microscope could help in distinguishing eukaryotic and prokaryotic cells. As we know, there are many differences between eukaryotic and prokaryotic cells. If a cell does not have a well-defined nucleus it would most likely be a prokaryotic cell. Under microscopic analysis, if a cell lacks a mitochondria it would be categorized as a prokaryotic cell. Another way of sorting out cells would be on determining the cellular membrane. If a cellular membrane exist, it is a eukaryotic cell. Prokaryotic cells also have smaller ribosomes. Size of ribosomes could be compared and then cells would be sorted.
2. Hypothesize about a means to separate out plant cells from a mixed population of eukaryotic cells.
A plant cell is different from other eukaryotic cells. They have a larger vacuole. In plants the cell wall is made up of cellulose and hemicellulose. Other eukaryotic cell walls ar made up of chitin. Plants contain plastids that are not present in other eukaryotic cells. Plasmodesmata, specialized communication pathways are only observed in plant cells. In this manner, plant cells can be sorted out from other eukaryotic cells.
Virtual Lab 2: Cellular Processes
A. Bacterial Growth. Observing the growth of the bacteria Streptococcus pneumoniae
These Streptococcus bacteria have been placed on a nutrient rich agar medium and their growth visualized. You can monitor their growth by watching the middle frame and moving through time with the time step buttons.
• Estimate how long it takes for this population of bacteria to double. Hint- this population doubles multiple times during the duration of this recording.
According to the time frame, it would take nearly 24 hours for the bacterial population to double. This was based on the time duration of the initial and intermediate frame.
B. Cellular Reproduction: The Cell Cycle (1, 2, 3), Mitosis (1, 2), Meiosis (1, 2), and Binary fission (1, 2).
1. Estimate the percentage of time that a constantly developing cell spends in interphase.
A developing cell spends nearly 12 to 24 hours in interphase. It is nearly 90% of the total time taken for a complete cell cycle. The interphase consist of 4 major steps of the cell cycle: The Gap 0 (G0), the Gap 1 (G1), the S (synthesis) phase and the Gap 2 (G2).
2. In a random selection of 100 such cells, estimate the number that would be undergoing mitosis at any given time.
Since mitosis consist of nearly 10% of the time in the cell cycle. An estimate of 90 out of 100 cells would be undergoing mitosis at a given time. This estimate is based on the time taken during interphase (which is long) and the time taken for mitosis (shorter).
3. Understand the basic differences between mitosis, meiosis, and binary fission. Is mitosis more similar to meiosis or to binary fission? Explain your reasoning.
Mitosis is a process of asexual reproduction while Meiosis is process of sexual reproduction. In mitosis there is division of cells into two, having equal chromosomes. On the other hand, Meiosis involves reduction of chromosomal number to half. Mitosis usually takes place in all animals whereas Meiosis takes place in Humans and animals. Another difference is that there is only one division in mitosis while two divisions take pace in Meiosis. Mixing of chromosomes can only occur in mitosis. Mitosis is similar to binary fission. Binary fission includes division of cells into two identical daughter cells. A similar phenomenon takes place during mitosis, where a cell divides into two having equal number of chromosomes.
C. Cellular Metabolism: Cellular Respiration (1, 2), Photosynthesis (1, 2), and the Carbon Cycle (1, 2, 3) 1. In a paragraph or two compare and contrast photosynthesis and cellular respiration.
The main difference between respiration and photosynthesis is that photosynthesis can take place only in the presence of sunlight while respiration is a continuous process. The second major difference is that respiration involves the absorption of oxygen while photosynthesis involves the evolution of oxygen. There is release of energy during respiration. Energy is required during photosynthesis.
Photosynthesis takes place in chloroplasts while respiration takes place in mitochondria. A catalyst (chlorophyll) is required for photosynthesis while respiration does not involve a catalyst.
2. Describe the ecological relationship between photosynthesis and cellular respiration.
Photosynthesis forms the basis of life form. Photosynthesis involves the release of oxygen in the atmosphere. Without oxygen, no life form would exist today. Because of photosynthesis there is oxygen and because of oxygen life form exists. Photosynthesis also involves the uptake of atmospheric carbon dioxide (C02). Respiration involves intake of oxygen and the release of CO2. Thus both are opposite of each other but important for cellular life form.
3. Hypothesize about what might happen if a large number of producers were suddenly removed from the biosphere. Where might carbon accumulate if the ratio of number of producers to consumers was markedly reduced?
Virtual Lab 3 Genetics 1
A. Phenotype and genotype of Dragons For fun, you can use this web lab to answer these questions: What genotype(s) result in wings? What genotype(s) result about a brown skinned Dragon? If necessary use your own research to answer the following questions.
Define genotype and phenotype.
Genotype is an organism’s inherited genetic code. It is not visible and it causes effects in an organism’s phenotype (visible traits). Examples involve Sickle cell trait in humans.
Phenotype is an organism’s visible trait. It can be observed on basis of distinctive features. Examples involve tall and short heighted individuals.
What is an allele?
An allele is a sequence present on the chromosome that determines an individual’s genotype. The point at which the allele is present is called the locus.
B. Drosophila Lab Enter the lab as a guest. You need to purchase a breeding pair of flies. Purchase a female mutant that has a small (vestigial) wing size and a male wild type fly. Breed them and notice the resulting distribution of phenotypes.
Describe and explain the characteristic of the first generation (F1) of flies. Is the vestigial wing characteristic dominant or recessive?
The F1 generation consisted of only wild type. This illustrates that the entire F1 generation has genotype or traits from the male fly. This simple depicts that the vestigial wing characteristic was recessive for F1 generation. Of the total 1158 flies, 602 were females (wild type) and 556 were male (wild type).
What percentage of the F1 generation would show the recessive characteristic phenotypically according to your table? Are your experimental breeding results consistent with what you expect from this assumption and the logic of the Punnet square?
There was not a single fly with the recessive characteristic. Hence the percentage of F1 generation showing recessive characteristics would be Null. According to the Punnet square, the F1 generation should consist of 50% of wild type and 50% of recessive trait. This was not found in the experiment carried out in the virtual lab.
Breed two of these F1 flies. To do this select a male and a female from the results of your first cross and put them in the breeding jar. Describe and explain the characteristics of the second generation (F2) flies.
The entire population was of the wild type. Since both the parents were of the wild type, there was no chance of mutation or cross-over. This resulted in the F12 generation to have only wild type trait. This result would have been observed if there was a recessive trait present in the wild type. Nature has its own way of dealing with genetic mutations.
C. Genetic Disorders Library Describe the three main classes of genetic disorders and give an example of each.
The three main classes of genetic disorders are Monogenic, Multifactorial inheritance and chromosomal. Cri-du chat syndrome, Down syndrome and Klinefelter syndrome are examples of each disorder respectively.
Virtual Lab 4: Genetics II
A. Learn how electrophoresis works and answer these questions:
On what basis is electrophoresis able to separate molecules? What are the lengths of the three DNA bands that you produce in this lab?
Electrophoresis works on the principles of charged particles. In case of DNA, shorter strands of DNA travel further along the gel while longer strands travel the least. Since DNA is negative in charge, DNA moves from negative to positive end of the gel. This allows distinctive bands to form on the gel after an hour. Three bands were observed in the virtual lab. The length of the first and the longest band is 6000 base pairs (bp). The second band is approximately 3500 bp. The third band was approximately 1500 bp. The length of bands were determined on comparison with a standard.
B. Electrophoresis.
**If the simulation does not work on your computer, open the "ELECTROPHORESIS ALTERNATIVE" below under "Supplemental" to view the screen shots of the activity. Then, answer the questions in the Lab Report.
• Load each lane as follows: lane 1 with Bgl 1; lane 2 with EcoR 1; lane 3 with Hinc II; lane 4 with Ple I; lane 5 with predetermined molecular weight markers.
• Run the gel and describe and explain the number of bands in lanes 2 and 4.
There were 3 bands in lane 2 and 2 bands in lane 4. The bands were compared to the standard. Some of the bands appeared to be closer to each other. Bands were observed under UV. Bands that appeared to be closer might be of different lengths but may differ to a size not more than 100kb.
C. Human blood types and the immune system: Emergency Transfusion!
You can skip the video introduction, but will probably want to read the guide before you attempt the emergency simulation. Repeat the simulation until you can get hired on by the virtual hospital staff. Use your mouse to: 1. Take a blood sample from the patient. 2. Place the blood sample in the antibody solutions. 3. Hang the correct blood transfusion bags, so as to start a blood transfusion.
• Identify the different blood types, the antibodies associated with each, and the types of blood that each type can receive and donate too.
There are 4 types of blood groups. A, B AB and O. Another factor to be taken into consideration is the Rh. Rh is either negative or positive. An individual with blood group A would have B antibodies. An individual with blood group B would have A antibodies. In case of an individual with AB blood group, he would have antibodies against O group only. Hence individuals having AB blood group are said to be Universal receivers. On the other hand, people with blood group O have antibodies against A and B antigens. Hence people with O blood group can receive blood only from are sad to be Universal donors.
