Answer:
4 ul Loading Buffer + 19.70 ul dH2O + 0.30 ul DNA Ladder
Load 12 ul on the gel.
Explanation:
DNA Ladder concentration = 1000 ug/ml
1000 ug DNA in 1 ml DNA Ladder solution → 150 ng DNA = 0.15 ug DNA in..... 0.00015 ml = 0.15 ul DNA Ladder solution
6x DNA Loading Buffer → it has to be diluted by an equal volume 6 times (1 ul LB + 1 ul distilled H2O)
An appropriate volume to load on an average agarose gel is 12 ul, so:
2 ul Loading Buffer + 9.85 ul dH2O + 0.15 ul DNA Ladder = 12 ul
But since 0.15 ul is a very small volume and mistakes could be made while measuring it, let's make double:
4 ul Loading Buffer + 19.70 ul dH2O + 0.30 ul DNA Ladder = 24 ul
And load half of that solution (12 ul) on the gel.
Answer: The answer is C.
C. They're examples of cell organelles.
Explanation: I just looked up the question and I got that answer.
The step in translation initiation that is unique to the eukaryotes is:
<span>formation of the preinitiation complex ribosome assembly
</span>
Here are the processes involved in the Translation Initiation of Eukaryotes
1) 5'cap is used to position the mRNA on the 40S ribosomal subunit
2) ribosome scans down the mRNA looking for an AUG.
3) There is an initiator methionine-tRNA
4) The initiating AUG codon is often within a consensus sequence called the Kozak sequence (5'-ACCAUGG-3')
5) After binding the cap, ribosomes scan down the mRNA until the Kozak sequence is reached and translation begins
<span>6)The poly (A) tail and 5'-cap binding proteins help the initiation complex form
</span>
Answer:
Like a prokaryotic cell, a eukaryotic cell has a plasma membrane, cytoplasm, and ribosomes. However, unlike prokaryotic cells, eukaryotic cells have: a membrane-bound nucleus. numerous membrane-bound organelles (including the endoplasmic reticulum, Golgi apparatus, chloroplasts, and mitochondria)
Explanation:
Given: The population density of frogs in a portion of a pond is 12 frogs per square meter. The size of the entire pond is 370 square meters.
To find: The total population size of the frogs in the pond.
Method: Using unitary method of proportions
Solution:
Let p be the density of the frogs present per square meter.
Let A denote the total area of the pond.
∴ p = 12 frogs/m² and A = 370 m²
The total number of frogs in the pond is given by,
Therefore, there are 4440 frogs in the pond.
