<span>False. E.coli is generally about 2 micrometers in size compared to white blood cells which are around 13 micrometers in size. Also, white blood cells have a characteristic segmented nucleus with two to five lobes joined by fine strands of chromatin.</span>
Answer:a mature women's cell, especially a human or other animal that may split into an embryo generally only after male cell fertilization.
Explanation:
Answer:Our ancestors’ perception of taste was important for survival and thriving. Now researchers are trying to produce food with reduced calories or salt that remains palatable
Explanation:
Evolution of taste
Our ancestors had to seek out their food from the environment, and their perception of taste was important for survival and thriving. ... They work with foods that are being produced in a manner that reduces calories or salt or adds ingredients with potential health benefits from plants
Answer:
The correct answer is
- Carbon dioxide, ATP, and NADPH,
- In the stroma of the chloroplast.
Explanation:
The product of the light-dependent reaction that uses in the Calvin cycle or the light independent cycle. Calvin cycle is the light independent cycle that involves the production of the glucose by converting carbon dioxide and other products of light reaction.
The light-independent reaction takes place in the stroma a fluid-filled region of the chloroplast in the photosynthetic organisms.
Thus, the correct answer is -
- Carbon dioxide, ATP, and NADPH,
- In the stroma of the chloroplast.
<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%