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kap26 [50]
4 years ago
11

Plzzzzzzz give me notes of dna replication​

Mathematics
1 answer:
miskamm [114]4 years ago
4 0
- A DNA strand can act as a template for synthesis of a new nucleic acid strand in which each base forms a hydrogen-bonded pair with one on the template strand (G with C, A with T, or A with U for RNA molecules). The new sequence is thus complementary to the template strand. The copying of DNA molecules to produce more DNA is known as DNA Replication.

-DNA replication takes place at a Y-shaped structure called a replication fork. A self-correcting DNA polymerase enzyme catalyzes nucleotide polymerization in a 5ʹ-to-3ʹ direction, copying a DNA template strand with remarkable fidelity. Since the two strands of a DNA double helix are antiparallel, this 5ʹ-to-3ʹ DNA synthesis can take place continuously on only one of the strands at a replication fork (the leading strand).

-On the lagging strand, short DNA fragments must be made by a “backstitching” process. Because the self-correcting DNA polymerase cannot start a new chain, these lagging-strand DNA fragments are primed by short RNA primer molecules that are subsequently erased and replaced with DNA.

-DNA replication requires the cooperation of many proteins. These include:

*DNA polymerase and DNA primase to catalyze nucleoside triphosphate polymerization;
*DNA helicases and single-strand DNA-binding (SSB) proteins to help in opening up the DNA helix so that it can be copied;
*DNA ligase and an enzyme that degrades *RNA primers to seal together the discontinuously synthesized laggingstrand DNA fragments;
*DNA topoisomerases to help to relieve helical winding and DNA tangling problems. *Many of these proteins associate with each
other at a replication fork to form a highly efficient “replication machine,” through which the activities and spatial movements of the individual components are coordinated.

Major steps involved in DNA replication are as follows:

*Each strand in a parental duplex DNA acts as a template for synthesis of a daughter strand and remains basepaired to the new strand, forming a daughter duplex (semiconservative mechanism).
*New strands are formed in the 5′ to 3′ direction.
*Replication begins at a sequence called an origin.
*Each eukaryotic chromosomal DNA molecule contains multiple replication origins.
*DNA polymerases, unlike RNA polymerases, cannot unwind the strands of duplex DNA and cannot initiate synthesis of new strands complementary to the template strands.
*Helicases use energy from ATP hydrolysis to separate the parental (template) DNA strands.
*Primase synthesizes a short RNA primer, which remains base-paired to the template DNA.
*This initially is extended at the 3′ end by DNA polymerase α (Pol α), resulting in a short (5′ )RNA- (3′)DNA daughter strand.
*Most of the DNA in eukaryotic cells is synthesized by Pol ẟ, which takes over from Pol α and continues elongation of the daughter strand in the 5′ to 3’direction.
*Pol ẟ remains stably associated with the template by binding to Rfc protein, which in turn binds to PCNA, a trimeric protein that
encircles the daughter duplex DNA.
*DNA replication generally occurs by a bidirectional mechanism in which two replication forks form at an origin and move in opposite directions, with both template strands being copied at each fork.
*Synthesis of eukaryotic DNA in vivo is regulated by controlling the activity of the MCM helicases that initiate DNA replication at multiple origins spaced along chromosomal DNA.
*At a replication fork, one daughter strand (the leading strand) is elongated continuously.
*The other daughter strand (the lagging strand) is formed as a series of discontinuous Okazaki fragments from primers synthesized every few hundred nucleotides.
*The ribonucleotides at the 5′ end of each Okazaki fragment are removed and replaced by elongation of the 3′ end of the next Okazaki fragment.
*Finally, adjacent Okazaki fragments are joined by DNA ligase.
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Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai
marta [7]

f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1

Lets check with every option

(a) [-4,-3]

We plug in -4  for x  and -3 for x

f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3  for x  and -2 for x

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.  

(c) [-2,-1]

We plug in -2  for x  and -1 for x

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1  for x  and 0 for x

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0  for x  and 1 for x

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.  

(f) [1,2]

We plug in 1  for x  and 2 for x

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1],  (d) [-1,0], (f) [1,2]

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Answer:

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