1 dal = 10L
Therefor, 2.4 dal = 2.4 x 10 = 24
Our answer: 24
Answer:
The correct option is D. <u>Helicase</u> is the enzyme that unwinds that double stranded DNA at the beginning of DNA replication.
Explanation:
DNA replication can be described as a process by which the strand of DNA gets duplicated into two new daughter strands.
Helicases can be described as the enzymes whose main function is to unwind the double helix structure of the DNA. The helicases break the hydrogen bonds present between the two strands of DNA. The helicases are the first machinery for DNA replication as they form the replication fork and begin unwinding of the DNA from the origin of replication
Answer;
Examples of selectively permeable materials or items are;
1. Vacuum filter- it traps dust but allows air through
2. Coffee filter- allows water to pass through, but does not allow the coffee grounds.
3. Colander -used to strain liquid food from foods. it drains water but is not permeable to larger food particles.
Explanation;
Selective permeability means that a material or a substance allows the passage of some molecules or ions and other materials to pass through it, and inhibits the passage of others.
For example a cell membrane is semi-permeable and thus regulates the transport of materials such that some particles pass through while others can not cross. A semi-permeable membrane allows selective movement of materials, it acts like a filter that allows particles to pass or not according to their size, solubility, electrical charge, or other chemical and physical properties.
Wear gloves ,goggles,clean every thing after you leave , also use specific clothes
<span>a.
</span>The frequency of the dominant allele: ___0.4______
<span>b.
</span>The frequency of the recessive allele: ____0.6_____
<span>c.
</span>The percentage of mice that are homozygous
dominant: __16%_______
<span>d.
</span> The
percentage of mice that are heterozygous: _48%________
<span>e.
</span>The percentage of mice that are homozygous
recessive: __36%_______
Let us assign the dominant allele (that of brown hair) letter R
while
We assign the recessive
allele (that of white hair) letter r
We then note down the
Hardy-Weinburg equation p2 + 2pq + q2 = 1
Brown fur population (p2
+ 2pq) = 64% = 0.64
White fur population (q2) = 36% = 0.36
Then we also remember that the frequencies of both allele
add up to 1 (p + q = 1);
Therefore q = ü.36
= 0.6
P = 1 – q = 1 – 0.6 = 0.4
The heterozygous population will be 2*0.6*0.4 = 0.48 = 48%
Homozygous domain population will therefore be (64% - 48%) =
16%
