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almond37 [142]
3 years ago
8

In the first step of cellular respiration, also known as glycolysis, a six carbon _________ molecule is converted into _________

.
a. glucose ; pyruvic acid
b. glucose ; ATP
c. ATP ; glucose
d. pyruvic acid ; glucose
Biology
2 answers:
Flauer [41]3 years ago
8 0
The first step of cellular respiration is glycolysis, where a six carbon glucose is broken down into a three carbon, this is called pyruvic acid. You're welcome

If you want the other steps in cellular respiration or cytology, just ask, I'm an A+.
Zigmanuir [339]3 years ago
5 0

Correct answer: A). Glucose ; pyruvic acid  

Glycolysis is the metabolic pathway that is known to convert glucose into pyruvate. The end product that is obtained after glycolysis is called as pyruvate, it can be further used in either anaerobic respiration or aerobic respiration. At the end of glycolysis, one glucose mole is broken into two molecules of pyruvate, ATP.

Hence, the correct answer would be option A.




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mRNA (Messenger RNA) is translated by ribosomes and contains the code that specifies the sequence of amino acids in polypeptide chain.

A single-stranded ribonucleic acid molecule is known as messenger RNA(mRNA) plays a major role in <u>translation</u>.

Translation is the method by which an mRNA codes for a certain protein. mRNA provides the template for<u> protein synthesis</u>.

The ribosome translates the mRNA that is produced from the DNA into a chain of certain amino acids and<u> protein synthesis</u> is facilitated by this <u>amino acid</u> sequence.

<u>The genetic code</u>, which connects the DNA sequence to the amino acid sequence of proteins, is used to "read" the mRNA. Each group of three nucleotides in mRNA forms a codon, and each codon corresponds to a particular amino acid (triplet code).

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2 years ago
The domestic cat genome contains 2.9×109 base pairs. The length of linker DNA in mammals is 50 base pairs. Approximately how man
IgorLugansk [536]

Answer:

Number of nucleosomes in 2.9 * 10^9bp is equal to 1.47 * 10^7

Explanation:

For wounding one nucleosome, total length of DNA required is equal to 146 bp

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Thus , the total length of DNA that confides between two nucleosome is equal to the sum of wounding length of DNA and the linker length

= 146 + 50\\= 196bp

Thus, in 196bp length of DNA, the total number of nucleosomes is equal to 1

Thus, number of nucleosomes in 2.9 * 10^9bp is equal to

\frac{2.9* 10^9}{196} \\1.47 * 10^7

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