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vampirchik [111]
2 years ago
8

student prepared a test tube containing yeast, glucose, and water. After 24 hours, the test tube was analyzed for the presence o

f several substances.What substance would the student expect to find if respiration occurred in the test tube?
Biology
1 answer:
expeople1 [14]2 years ago
5 0

Answer:

In the presence of oxygen, respiration of yeast produces carbon dioxide and water. In the absence of oxygen, respiration of yeast produces carbon dioxide and alcohol.

Under the assumption that the yeast has insufficient exposure to oxygen in the surrounding air in the testtube, alcohol and carbon dioxide would be expected to be found in the test tube

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When enzymes are elongating (building) a newly synthesized dna strand in pcr, new nucleotides are being added to the ____ of the
Tamiku [17]
The correct answer is c. 3 ' hydroxyl of deoxyribose.  
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During the DNA synthesis via PCR, polymerization occurs in 5’-3’ direction. This means that enzyme DNA polymerase which synthesizes a new DNA strand is complementary to the DNA template strand (3’-<span>5’ direction). 5’ phosphate group of the free nucleotides which are added is attached to the 3'-hydroxyl group at the end of the elongating DNA strand.</span></span>
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Answer:

B. CO2

Explanation:

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Which of the following correctly identifies the decomposers in the forest food web shown below? (1 point)
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4 0
2 years ago
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Three small bones connect the tympanic membrane to the oval window. what is the function of these bones??
lesantik [10]
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6. Delete any base (between the READING FRAME, excluding the start and stop codons) and show the change in the amino acid sequen
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This is the DNA. I'm going to only use the upper strand to demonstrate what this strand would code for before and after a single bp deletion (so write it as mRNA). I will also write it how it's easier to see this which is to split them up into the 3 base codon system. Note that you don't need to know the amino acid code - you use a table to find these.

ORIGINAL (mRNA on top, Amino Acid (AA) on bottom:

5'-AGC GGG AUG AGC GCA UGU GGC GCA UAA CUG-3'
    SER  GLY  MET SER ALA  CYS GLY ALA  STOP LEU 

Note that the protein would stop being made at the stop codon and the LEU wouldn't matter at the end...

Now, I will remove one bp...(I bolded it up top). Rewrite the mRNA and find the corresponding AA...

NEW
5'-AGC GGG AUG GCG CAU GTG GCG CAU AAC UG-3'
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Completely different amino acid sequence after the methionine (MET). The stop codon is gone...the protein would continue being translated until it reaches another stop codon...so not what was supposed to be made! 
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1 year ago
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