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Damm [24]
3 years ago
15

The brilliant research scientist, Professor "Iron-Mike" Scrawll is studying a new pore-forming toxin from the pathogenic bacteri

um Minnesotious goferious, the causative agent of Rusty Rodent disease, which transforms normally docile copper gophers to aggressively attacking the ankles of golfers and stealing their golf balls on the U of I blue course. Professor Scrawll has isolated the virulence factor BurD, which he has discovered to be a very stable, large, pore-forming exotoxin. His research indicates that, when BurD is applied to cultured human ankle cells, the cells are rapidly lysed and die. Excitedly, Professor Scrawll sends a sample of his purified BurD to his colleague, Dr. D. Bogey, who attempts to repeat the same experiment. Disappointedly, Dr. Bogey reports back that BurD did not lyse and kill cultured ankle cells in her lab. After an emergency meeting at the 19th hole, Professor Scrawll and Dr. Bogey hold a press conference to explain the discrepancies in their data. Based solely on what you know about BurD, which of the following explanations most likely explains experimentally why Professor Scrawl and Dr. Bogey obtained different results?
a. Professor Scrawll applied 100 nM BurD to the cells for 12 hours, while Dr. Bogey applied 100 nM BurD to the cells for 24 hours
b. Dr. Bogey accidentally contaminated her ankle cells with bleach
c. Professor Scrawll applied 100 nM BurD to the cells for 24 hours, while Dr. Bogey applied 1 nM BurD to the cells for 12 hours
d. Professor Scrawll applied 1 nM BurD to the cells for 24 hours, while Dr. Bogey applied 100 nM BurD to the cells for 24 hours
e. Dr. Bogey applied 100 nM BurD to the cells for 24 hours, while Professor Scrawll applied 1 nM BurD to the cells for 12 hours
Biology
1 answer:
Dafna11 [192]3 years ago
5 0

Answer:

The correct option is C - Professor Scrawll applied 100 nM BurD to the cells for 24 hours, while Dr. Bogey applied 1 nM BurD to the cells for 12 hours.

Explanation:

The correct option is C - Professor Scrawll applied 100 nM BurD to the cells for 24 hours, while Dr. Bogey applied 1 nM BurD to the cells for 12 hours.

As the exotoxin, BurD is very stable and lyse the ankle cells very quickly, more concentration and more time of action should only lyse the cells. Perhaps Dr.Bogey's ankle cells were not lysed because the concentration she used was only 1nM compared to the 100nM concentration used by Dr. Scrawll, and the time period of incubation was only 12 hours compared to the 24 hours used by Dr. Scrawll.

Considering the other explanations given in the remaining options, the concentration and time of incubation used by Dr. Bogey are more than that used by Dr. Scrawll which should only have possibly lysed the cells. Moreover, contamination with bleach also should have only lysed the cells.

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3 years ago
Which of the following statements about stability is NOT true?
ahrayia [7]
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8 0
3 years ago
Read 2 more answers
Match the following:Part A1. Transverse foramina present.2. No canals or foramen present. It articulates superiorly with the sac
Stels [109]

Answer:

1. Transverse foramina present: Cervical

2. No canals or foramen present. It articulates superiorly with the sacrum: Coccyx

3. Receive the most stress: Lumbar

4. Attach to ribs: Thoracic

5. Articulates with hip bones of the pelvis: Sacrum

Explanation:

The vertebral column is a <u>series of 33 bones called vertebrae</u> that play a key role in organ protection, movement of body, and overall support. The column has been divided into <u>5 different regions</u> with the number of bones.

<u><em>1. Cervical Vertebrate</em></u><u>:</u> These are the group of <u>seven vertebrae of the neck</u>, start immediately below the skull. Two cervical bones C1 and C2 are unique in function. They are responsible for the movement of the head.  They have <u>transverse foramina</u> which <u>gives passage to vertebral artery and vein</u>.

<em><u>2. Thoracic Vertebrae:</u></em> They are a group of twelve small bones that form the vertebral spine in the upper trunk. The function of the thoracic vertebrae is to articulate with ribs to produce the bony thorax.

<em><u>3. Lumbar Vertebrae: </u></em>This is the largest segment of the vertebral column that consists of 5 bones between the rib cage and pelvis. They <u>carry all of the upper body weight</u> providing flexibility and movement to the trunk region.  This is why it can <u>receive the most stress easily</u>.

<em><u>4. Sacrum Vertebrae: </u></em>There are 5 sacral vertebral fused bones. It <u>connects to the hip bones and play role in forming a strong pelvis</u>.

<em><u>5. coccyx Vertebrae:</u></em> These are a group of 4 fused bone. There is <u>no vertebral canal due to a lack of vertebral arches</u>. They <u>provide </u>an <u>attachment site for muscles</u> ligament and tendons. They also <u>play a role in stabilization and support</u> while sitting.

5 0
3 years ago
Which statement concerning osmosis and active transport is correct?
marshall27 [118]

The question is incomplete, however, the statements associated with this question is given in the comments and here as well:

Neither facilitated diffusion nor osmosis requires cell energy.

Diffusion of gases and other small molecules requires no energy on the part of the cell.

Active transport requires cell energy and osmosis doesn't.

Both endocytosis and active transport require cell energy.

Answer:

The correct answer is - Active transport requires cell energy and osmosis doesn't.

Explanation:

Osmosis is an example of passive transport as it does not require energy to facilitate the movement of solvent In the process of osmosis,. It moves from high concentration to low concentration through the semipermeable membrane which is along the gradient so no requirement of energy.

In the case of Active transport, it requires energy to facilitate the movement of transport as it is the movement of a substance from low concentration to a high concentration area that is against the concentration gradient.

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Answer:

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Explanation:

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DNA replicatese by semi-conservative mode of replication in which one the daughter DNA consists of one newly synthesized strand and one parental strand. This mode of replication was explained by Messelson and Sthal experiment. The cells from 15N medium are transferred to 14N medium the band of intermediate density was observed. The daughter DNA consists of one 15N DNA strand and one 14N DNA strnd. This strand undergo replication in the 14N medium. This time one low density DNA band ( contains 14N - 14N DNA strand) and one intermediate density band ( 14N-15N DNA).

Thus, the correct answer is option (d).

8 0
3 years ago
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