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trapecia [35]
3 years ago
13

After drinking a great deal of coffee (pH 5), a human's blood buffering system would need to ________ as the coffee was digested

to lower the level of acid present in the blood stream.
Biology
1 answer:
spayn [35]3 years ago
3 0

Answer:

Hydrogen ion(H+)

Explanation

The human blood has carbonic acid and bicarbonate anion as buffer to maintain the blood pH between 7.35 and 7.45. A virtual value higher than 7.8 and 6.8 can lead to death.

Taking coffee which has a low pH will lower the pH of the blood. In order to raise the pH back to normal and lower the level of acid present in the system. The human buffering system will to take up Hydrogen ion.

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What is the explanation of why avery, macleod, and mccarty's in vitro transformation experiment showed that dna, but not rna or
astraxan [27]

The Griffith's experiment, the Avery-MacLeod-McCarty experiment, and the  Hershey–Chase experiments were the set of experiments that established DNA as the key hereditary molecule. The Avery-MacLeod-McCarty experiment was an extension to the Griffith's experiment. The heat killed virulent S strain cells of the Griffith's experiment were lysed to form a supernatant containing a mix of RNA, DNA, proteins and lipids from the cell. The supernatent was equally divided into 3 parts after the removal of the lipids. The 3 parts were respectively treated with an RNAase to degrade the RNA, DNAase to degrade the DNA and proteinase to degrade the proteins. The treated supernatant was then added into the culture containing the non-virulent R cells. In case of the supernatant treated with the DNAse, no transformation of R cells into S cells occurred. The transformation of R cells to S cells occurred in the proteinase and the RNAse cases. This indicated that DNA was the hereditary molecule and not protein or RNA.

5 0
3 years ago
How many possible open reading frames (frames without stop codons) are there that extend through the following sequence? 5'... C
fenix001 [56]

Answer:

There are 4 reading frames without stop codons.

Explanation:

1) To find the possible reading frames you need to separate your sequence in codons (trios)  for the 5'-> 3' sense, first, excluding the first nucleotide, then excluding 2 nucleotides and then not excluding nucleotides.

5'... C TTA CAG TTT ATT GAT ACG GAG AAG G... 3'  NO STOP CODON

5'... CT TAC AGT TTA TTG ATA CGG AGA AGG... 3'  NO STOP CODON

5'... CTT ACA GTT TAT <u>TGA</u> TAC GGA GAA GG... 3'

2)Then you find the complementary sequence (which you already have)

3'... GAA TGT CAA ATA ACT ATG CCT CTT CC... 5'

3'... GA ATG TCA AAT AAC TAT GCC TCT TCC... 5'

3'... G AAT GTC AAA TAA CTA TGC CTC TTC C... 5'

3) Next, you have to obtain the reverse complementary

CCT TCT CCG TAT CAA TAA ACT GTA AG NO STOP CODON

CC TTC TCC GTA TCA ATA AAC TGT AAG NO STOP CODON

C CTT CTC CGT ATC AAT AAA CTG <u>TAA</u> G

4)Now you need to find start (AGT in terms of DNA) and stop codons(<u>TAA</u>, <u>TAG,</u> <u>TGA</u>, also in terms of DNA), in 5->3 and de reverse complementary.

<em />

<em>As you can see, you have for open reading frames that lack of stop codon, unfortunately, they also lack of AGT, the start codon.</em>

I added a table where you can find thestart and stop codons as well as the proteins that they translate to in terms of RNA.

I hope you find this information interesting and useful, good luck!

8 0
3 years ago
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