Answer:
D
Explanation:
because mintochodria is the power house hence it provides energy to the cell
Answer:
48 amino acids
Explanation:
The wild type gene codes for a protein with 100 amino acids. One amino acid is encoded by one triplet code of the gene. This means that the wild type gene has a total 100 triplets or 300 nucleotides to code for a protein of 100 amino acid. Mutation in this protein has introduced the code "UAA" at the 49th codon. The code "UAA" is a stop codon. Therefore, the mRNA transcribed from the mutant allele would code for a protein having 48 amino acids as the protein synthesis will be stopped once the stop codon at the 49th position is read.
Muscular exercise presents a dramatic test of the body's homeostatic control systems because it results in large amounts of heat production.
Homeostatic control systems- A body's physiological ability to maintain a steady internal environment in response to changes in the external environment is known as homeostasis.
Heat Production- The term "thermogenesis" refers to the process through which energy is lost by producing heat with specialization.
Energy- In biology, cells frequently store energy in macromolecules, especially lipids and carbohydrates (sugars). When chemical bonds are formed, such as during the redox reactions of cellular aerobic respiration, energy is released.
Redox Reactions- A reaction that happens when an oxidizing material and a reducing substance come into contact.
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Answer:
The correct answer would be 2 in 4.
According to the question Xo and XO show codominance and express themselves completely when present in heterozygous condition. Cats bearing XoXO show patchwork of black and orange fur and are called tortoiseshell cats.
Xo codes for orange color fur and XO codes for black color fur. In addition, Y chromosome does not contain any gene associated with fur color.
Now, genotype of mother cat is XOXO (orange fur). So, the gametes formed would be XO only.
The genotype of father cat is XoY(black fur). So, the gametes would be Xo and Y.
The cross would lead to the formation of two male cats each having XOY as their genotype and two female cats each with XOXo as their genotype.
Hence, both the male cats would show orange fur and both the female cats would show patchwork of orange and black fur.
Therefore, we can conclude that 2 out of 4 would exhibit tortoiseshell coloring.
The <span>hardy-weinberg equation is
p2 + 2pq + q2 = 1
where are already given with
c = 0.1 and C = 0.9
SO
p2 = 0.9^2
p2 = 0.81
q2 = 0.01
0.81 + 2pq - 0.01 = 1
2pq = 0.18
The answers are
CC = 0.81
cc = 0.01
Cc = 0.18</span><span />
