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velikii [3]
3 years ago
8

*NEED HELP?!!!?? ASAP!!! PLEASE HELP!!!****

Biology
1 answer:
Ket [755]3 years ago
6 0
1. A or D
2. D Urbanization
3. <span>D groundwater pollution
4. </span><span>C contamination of beaches and sea vessels with pollutants.</span>
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Identify the independent variable in the experiment described. Identify the plasmid that was used as a negative control for luci
bagirrra123 [75]
<h3>Question details: </h3>

CD3 is a signaling protein that is typically found only in the plasma membrane of immune system T lymphocytes. CD3 is composed of several different polypeptides, including a gamma chain, CD3γ . Scientists analyzed the promoter of the CD3γ chain gene for regulatory sequences that might have positive or negative effects on expression of the gene. The scientists cloned fragments of the CD3γ gene that included the first transcribed nucleotides plus up to 789 nucleotides of upstream regulatory sequences into plasmids in which the gene for the firefly enzyme luciferase immediately follows the fragments. The plasmids were then introduced into a line of T lymphocytes (Figure 1), and the cells were allowed to grow for a short while. Because the regulatory sequences of the CD3γ gene immediately precede the luciferase gene in the plasmids, the activity, either positive or negative, of the regulatory sequences affected the amount of luciferase gene expression by the T lymphocytes. Luciferase catalyzes a reaction that results in the release of light and is responsible for the bioluminescence (light flashes) of fireflies. By quantifying the bioluminescence, or luciferase activity, in the cells, the scientists were able to determine the effects of each CD3γ gene fragment cloned into the plasmids (Figure 2) on expression of the gene.

<h3>Figures uploaded as attachment</h3>

Figure 1. Summary of experimental procedure. A series of plasmids containing fragments of the CD3γ upstream regulatory sequences cloned immediately before the luciferase gene were constructed. Each type of plasmid was introduced into T lymphocytes. The amount of luciferase produced by the lymphocytes was dependent on the regulatory sequences present in each plasmid.

Figure 2. Analysis of the luciferase activity of T lymphocytes containing plasmids with different amounts of the CD3γ regulatory sequence. Names of the plasmids are shown on the left. Included regulatory sequences and the resulting luciferase activity are shown on the right. Error bars represent ±2SEx¯. The transcription start site is indicated by +1.

<h3>Answer:</h3>

Identify the independent variable in the experiment described: the plasmid sequence, which contains different amounts of CD3γ regulatory sequence

Identify the plasmid that was used as a negative control for luciferase activity: Parent plasmid

Justify including the plasmid with the non- CD3γ  active promoter in the experiments: To show that luciferase activity occurs in these cells

<h3>Explanation:</h3>

The aim of the experiment is to analyse regulatory sequences present upstream of the CD3γ  gene that might influence the expression of the gene. To do this, they took a plasmid with a luciferase gene present. They then inserted sequences of varying length in the plasmid upstream of the luciferase genes. They then inserted this plasmid into T cells and performed luciferase assays to assess how highly expressed the luciferase gene was from the different regulatory sequences. This would be a good indication of the regulatory activity of that sequence.

The only variable that changes throughout the experiment is the regulatory sequence that is inserted upstream of the luciferase gene. This means researchers can determine the effect of that regulatory sequence on the expression of a gene, without the influence of other factors.

The plasmid used as a negative control is one that we expect not produce any luciferase activity. In this case, it is the parent plasmid, which does not have an upstream promoter/regulatory sequence, so it will not be able to be expressed, and the assay will detect no (or very little) luciferase activity. This is shown in the attached figure 2. This controls for expression of luciferase not produced by the promoter.

The plasmid with the non-CD3 active promoter shows the "normal" activity of the luciferase gene. This controls for the possibility that the inserted sequences do not positively affect gene expression. If this happened and we did not have our control, we wouldn't know if the assay is working properly or whether this sequence just doesn't act as a promoter. Therefore, the non-CD3 active promoter shows us that the luciferase assay is working as expected, and we can therefore judge any changes in expression caused by our regulatory sequence.

5 0
3 years ago
Scientists have documented that spring seasons starting earlier due to climate change have resulted in some caterpillar species
andriy [413]

Answer:

phenotypic adaptive landscape

Explanation:

Global climate change will not only affect migration patterns, biotic interactions, and local population dynamics, but also the selective pressures experienced by populations. Coupled ecological and evolutionary changes are therefore expected under climate change.

8 0
3 years ago
The cell division process that only occurs in specific cells and at specific times is _____. meiosis mitosis
LUCKY_DIMON [66]

Answer:

Meoises

Explanation:

Cell división

4 0
3 years ago
According to the cladogram shown, which organism is most closely related to Arabidopsis?
ANEK [815]

Cycad is more closely related to the Arabidopsis . It is so as the closest species from the node of cladogram posses the most common traits as each node represents the end of the ancestral taxon and the species that further split from the node are called sister taxa.

7 0
3 years ago
A compound containing Na, C, and O is found to have 1.06 mol Na, 0.528 mol C, and 1.59 mol O. What is the empirical formula of t
dsp73

Answer:

So first understand that empirical formula refers to ratio of different atoms in lowest integer form.

To calculate this let's make mol C as our default ratio of 1* and divide everything by 0.528 mol.

Na 1.06/0.528 = 2.00757575758

C 0.528/0.528 = 1

O 1.59/0.528= 3.01136364

Na2CO3

It's sodium carbonate

5 0
3 years ago
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