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Veseljchak [2.6K]
3 years ago
15

Two DNA sequences, each 48 kb in size, are treated with the same restriction enzyme. There is only one site for this restriction

enzyme on each of these sequences. Digests are separated by gel electrophoresis. Sequence A shows ONE band that is 48 kb in size. Sequence B shows TWO bands, one 30 kb and the other 18 kb in size What can you conclude from these data?
A. Sequence A is linear, sequence B is circular
B. Both sequences are linear. >X
C. They are identical sequences
D. Sequence A is circular, sequence B is linear
Chemistry
1 answer:
morpeh [17]3 years ago
7 0

Answer:

Sequence A is circular, sequence B is linear.

Explanation:

Restriction enzymes is also known as the molecular scissors that cuts the specific DNA fragments. Two main types of restriction enzymes are restriction exonuclease and restriction endonuclease.

The DNA fragments that are digested by the restriction endonuclease can be used for the preparation of the maps. The circular DNA ganaretes the equal number of fragments as it has the restriction site and linear DNA produce one extra fragments.

Thus, the correct answer is option (D).

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The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0 °c is 4.48. what is the value of ka for hbro?
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____________________________________________________

Answer:

Your answer would be a). 2.0 × 10-9

____________________________________________________

Work:

In your question the "ph" of a 0.55 m aqueous solution of hypobromous acid temperature is at 25 degrees C, and it's "ph" is 4.48.

You would use the ph (4.48) to find the ka for "hbro"

[H+]

=

10^-4.48

=

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=

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or: [H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-]

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____________________________________________________

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