Answer:
it condenses the DNA so that it can fit inside the cell.
Explanation:
Supercoiling is a level of genome packaging and is important for the packaging of the DNA which is about 2m in length inside a nucleus that is about 6microm. Supercoiling refers to the over (positive supercoiling) or under winding (negative supercoiling) of a DNA strand, and is an expression of the strain on that strand. Supercoiling is in compacting DNA, and also allows for the regulation of access of transcriptional machinery and other biological activities to the DNA and in turn gene expression. Certain enzymes such as topoisomerases are able to change DNA topology to facilitate access of DNA replication or transcription.
The molecule of water is different from the molecules of oxygen and hydrogen in the following ways:
1. Combustion: Oxygen supports combustion, hydrogen burns in the air to form water. Water is not combustible. It rather inhibits combustion when poured over burning substances.
2. State of matter: Water is a liquid in room temperature whereas oxygen and hydrogen are gases. The polar nature of water contributes to hydrogen bonding between the hydrogen and oxygen atoms of water.
3. Density: Water is denser than both oxygen and hydrogen. Hydrogen bonding helps pack more molecules of water in less space.
4. Freezing point: Water freezes to ice at 0°C whereas oxygen freezes at -218°C and hydrogen freezes at -259°C.
<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%
Answer:A,B,E
Explanation:
That’s what the answer is
