We use Chargaff's rule to get the answer.
[A] + [G] = [C]+ [T].
[A] + [G] + [C] + [T] = 100%
Where is A is Adenine, G is Gaunine, T is Thymine and C is Cytosine.
In DNA, Adenine always pairs with Thymine, and Guanine always pairs with Cytosine.
Therefore if Thymine is 35%, then Adenine will also be 35% to make 70% in total.
The remaining percentage will be 100% - 70% = 30%.
The 30% will be shared equally among Cytosine and Guanine, at 15% each. Therefore Cytosine will be 15%
Chlorella is a genus of about thirteen species of single-celled green algae belonging to the division Chlorophyta. The cells are spherical in shape, about 2 to 10 μm in diameter, and are without flagella. Their chloroplasts contain the green photosynthetic pigments chlorophyll-a and -b. In ideal conditions cells of Chlorella multiply rapidly, requiring only carbon dioxide, water, sunlight, and a small amount of minerals to reproduce.
Answer:
e. secretion, absorption and chemical protection
Explanation:
Epithelial tissues is responsible for the protection of the skin, absorption of nutrients during digestion and secretion of waste materials. These tissues form covering to all body surfaces. They also perform a variety of other functions such as excretion of waste substances, filtration of air from dirt and particles and clean the air that is inhaled, diffusion, and sensory reception.
Hardy-Weinberg Equation (HW) states that following certain biological tenets or requirements, the total frequency of all homozygous dominant alleles (p) and the total frequency of all homozygous recessive alleles (q) for a gene, account for the total # of alleles for that gene in that HW population, which is 100% or 1.00 as a decimel. So in short: p + q = 1, and additionally (p+q)^2 = 1^2, or 1
So (p+q)(p+q) algebraically works out to p^2 + 2pq + q^2 = 1, where p^2 = genotype frequency of homozygous dominant individuals, 2pq = genotype frequency of heterozygous individuals, and q^2 = genotype frequency of homozygous recessive individuals.
The problem states that Ptotal = 150 individuals, H frequency (p) = 0.2, and h frequency (q) = 0.8.
So homozygous dominant individuals (HH) = p^2 = (0.2)^2 = 0.04 or 4% of 150 --> 6 people
Heterozygous individuals (Hh) = 2pq = 2(0.2)(0.8) = 0.32 or 32% of 150
--> 48 people
And homozygous recessive individuals (hh) = q^2 = (0.8)^2 = 0.64 = 64% of 150 --> 96 people
Hope that helps you to understand how to solve these types of population genetics problems!
