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Fantom [35]
3 years ago
15

Given a template strand of 3' ATGCTTGGACA 5' and partially made complementary strand containing only 5' TAC 3', what would be th

e sequence of the new strand of DNA (including the 5' TAC 3') if the only additional nucleotides available to DNA polymerase were those containing the bases G, A, and C?
Biology
1 answer:
madreJ [45]3 years ago
3 0

<span>The enzyme DNA polymerase will  form hydrogen bonds with the nucleotides of the template DNA and add new nucleotides, adenine will bond with thymine and guanine will bond with cytosine, forming the complementary strand.
</span>
<span>3' ATGCTTGGACA 5 is the DNA template
</span><span>5' TACGAACC .......3' is the complementary strand
you just need to join the right nucleotides. If there is no T, the DNA polymerase has nothing appropriate to connect to the A, so it stops there</span>
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If A/A ⋅ B/B is crossed with a/a ⋅ b/b and the F1 is testcrossed, what percentage of the testcross progeny will be a/a ⋅ b/b if
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a) 25%; b) 50%; c) 45%; d) 38%

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<h3 /><h3>a) Unlinked genes</h3>

A/A B/B X a/a b/b

F1: A/a B/b

Testcross A/a B/b X a/a b/b

The homozygous recessive individual only produces <em>ab</em><em> </em>gametes.

The F1 produces four types of gametes (each of them with a frequency of 1/4):  <em>AB</em>, <em>Ab</em>, <em>aB </em>and <em>ab</em>.

25% of the progeny will be a/a b/b.

<h3>b) Completely linked genes</h3>

AB/AB X ab/ab

F1: AB/ab

Testcross AB/ab X ab/ab

The F1 produces only two types of gametes, the parentals (each of them with a frequency of 1/2):  <em>AB</em> and <em>ab</em>.

50% of the progeny will be ab/ab.

<h3>c) 10 m.u. apart</h3>

AB/AB X ab/ab

F1: AB/ab

Testcross AB/ab X ab/ab

The F1 produces four types of gametes, the parentals <em>AB</em> and <em>ab </em>and the recombinants <em>Ab</em> and <em>aB</em>.

Since the genes are 10mu apart, 10% of the produced gametes will be recombinant and 90% will be parentals. Since there are two types of parental gametes, each of them has a frequency of 45%.

45% of the progeny will be ab/ab.

<h3>d) 24 m.u. apart</h3>

This is very similar to c).

Since the genes are 24mu apart, 24% of the produced gametes will be recombinant and 76% will be parentals. Since there are two types of parental gametes, each of them has a frequency of 38%.

38% of the progeny will be ab/ab.

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