13-9 = 4
4-4 = 0
0 divided by 2 = 0
2x = 0
x = 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.
Answer:
40
Step-by-step explanation:
PEMDAS
Do what is in the parentheses first.
-8 * -9 = 72
-9 - 8 + (72) - 15
-9 - 8 - 15 = -32
-32 + 72 = 40
Let the lengths of the east and west sides be x and the lengths of the north and south sides be y. the dimensions you want are therefore x and y.
The cost of the east and west fencing is $4*2*x; the cost of the north and south fencing is $2*2*y. We have to put in that "2" because there are 2 sides that run from east to west and 2 sides that run from north to south.
The total cost of all this fencing is $4(2)(x) + $2(2)(y) = $128. Let's reduce this by dividing all three terms by 4: 2x + y = 32.
Now we are to maximize the area of the vegetable patch, subject to the constraint that 2x + y = 32. The formula for area is A = L * W. Solving 2x + y = 32 for y, we get y = -2x + 32.
We can now eliminate y. The area of the patch is (x)(-2x+32) = A. We want to maximize A.
If you're in algebra, find the x-coordinate of the vertex of this quadratic equation. Remember the formula x = -b/(2a)? Once you have calculated this x, subst. your value into the formula for y: y= -2x + 32.
Now multiply together your x and y values to obtain the max area of the patch.
If you're in calculus, differentiate A = x(-2x+32) with respect to x and set the derivative equal to zero. This approach should give you the same x value as before; the corresponding y value will be the same; y=-2x+32.
Multiply x and y together. That'll give you the maximum possible area of the garden patch.
Answer:
72 pieces
Step-by-step explanation:
since 1/8 pieces would mean 8 per pie then
8*9 = 72
So 72 pieces
