ATP is considered as nature dehydrating agent.what does it means?show with the help of examples

Industrial melanism refers to the dark pigmentation that evolved in some insects giving them protective coloration on vegetation

Natalka [10]

Answer:

The frequency of the dominant allele, p = 0.542

The proportion of black moths that are heterozygous 2pq = 0.496

Explanation:

According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the allelic dominant or recessive forms. The genotypic frequencies after one generation are p² (Homozygous dominant), 2pq (Heterozygous), q² (Homozygous recessive). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.

In the same way, the sum of genotypic frequencies equals 1, this is

p² + 2pq + q² = 1

Being

p the dominant allelic frequency,

q the recessive allelic frequency,

p² the homozygous dominant genotypic frequency

q² the homozygous recessive genotypic frequency

2pq the heterozygous genotypic frequency

In the exposed example, 79% of the moths of the species Biston betularia were black due to the presence of a dominant gene for melanism.

If the genotypic frequency of back moths is 0.79, then, by performing the following equation we can get the not-black moths genotypic frequency:

p² + 2pq + q² = 1

where p² is the homozygous dominant genotypic frequency, q² the homozygous recessive genotypic frequency, and 2pq is the heterozygous genotypic frequency.

As 0.79 is the phenotypic frequency of black moths, then this frequency equals p²+2pq.

Clearing the equation:

p² + 2pq + q² = 1

0.79 + q² = 1

q² = 1 - 0.79

q² = 0.21

The genotypic frequency of non-black moths is 0.21. So, from here we can calculate the allelic frequency:

q² = 0.21

q= v 0.21

q = 0.458

If 0.46 is the allelic frequency of non-black moths, then by clearing the equation p + q = 1, we can get the p allelic frequency:

p + q = 1

p + 0.458 = 1

p = 1 - 0.458

p = 0.542

The genotypic frequency p² = (0.542)² = 0.294

The heterozygote genotypic frequency

2 x p x q = 2 x 0.542 x 0.458 = 0.496

Finally, we can check this answer by clearing the following equation:

p² + 2pq + q² = 1

0.294 + 0.496 + 0.21 = 1

4 0

3 years ago

What substance makes up 75 to 90% of every cell in the human body?

elixir [45]

I believe that substance would be water.

7 0

3 years ago

Read 2 more answers

Where does the primary source of energy needed for life come from

tresset_1 [31]

Answer:

the sun provides all energy

4 0

3 years ago

Read 2 more answers

PLEASE solve number 1 and 2 because i don’t get it , don’t mind my answers

Readme [11.4K]

Answer:

1) RNA strand = UCA GCA AAU

2) Amino acids: Asparagine Arginine Stop

(ASN) (ARG) (STOP)

Explanation:

Some theory that you need to know to answer the question:

Each tRNA has two important sites. One of them that couples with the codon of the mRNA molecule, named anticodon. The other site couples with an amino acid through the action of the aminoacyl-tRNA synthase enzyme. The whole complex, amino acid + enzyme + tRNA is named aminoacyl-tRNA. Considering that there are twenty amino acids available, there are also twenty complexes of aminoacyl-tRNA, one for each amino acid.

tRNA allows amino acids to align according to the nucleotidic sequence in the mRNA molecule.

Once the new amino acid joins the growing peptidic chain, the binding between the amino acid and the tRNA molecule breaks. The tRNA is now free to join another amino acid and repeat the cycle.

Now, to fill in the boxes, you need to pay attention to two things:

• Anticodon: In this box, you need to place the sequence of bases that pairs with the mRNA codon.

• Amino Acid: In this box, you need to place the amino acid that is codified by the mRNA codon.

For instance, let us say that the codon of mRNA is CUC. The anticodon for this sequence is GAG (because guanine joins cytosine and adenine joins uracyl).

The amino acid places on the 3´extreme of the tRNA molecule will be the one that is codified by the codon, NOT the anticodon. So, as the codon is CUC, the amino acid will be Leucine, Leu. Do not use the anticodon GAG to fill these boxes. Remember that tRNA is translating information carried by the mRNA, so we need to use this information to grow the protein.

1)

Strand of DNI → AGT CGT TTA

Each of these letters are bases → Adenine, A - Guanine, G - Thymine, T - Cytocine, C
Each of these three-lettered groups are codons.

So, we need to pair these bases, knowing that:

Adenine (A) and guanine (G) derive from purines, while Thymine (T) Uracile (U), and Cytosine (C) derive from Pyrimidines.

In the DNA molecule, Adenine (Purine) always pairs with Timine (Pyrimidine), while Guanine (Purin) forms pairs with Cytosine.

In the RNA molecule, Adenine is paired with Uracil, U. So, whenever there is an Adenine in the DNI strand, it pairs with Uracil of RNA strand. If there is a Thymine in the DNI strand, it pairs Adenine of the RNA strand. And, as always, Guanine pairs with Cytosine.

A ⇒ U
T ⇒ A
G ⇔ C

Knowing how to pair bases, we can fill the RNA molecule boxes.

Strand of DNI → AGT CGT TTA
Strand of RNA → UCA GCA AAU

A ⇒ U C ⇒ G T ⇒ A

G ⇒ C G ⇒ C T ⇒ A

T ⇒ A T ⇒ A A ⇒ U

2)

Now we need to translate these new codons of mRNA to amino acids.

To fill the boxes you should have a genetic code chart to see the codons that codify for each amino acid.

Codon: AAU CGA UAG

Amino acids: Asparagine Arginine Stop

(ASN) (ARG) (STOP)

The new protein is synthesized from the amino terminus to the carboxy terminus of the mRNA molecule. The added amino acids to the chain are coded by a codon formed by three bases in the mRNA. mARNs also have a start and end codon that are the signals of the synthesis initiation and finish. When the ribosome reaches a stop codon, protein synthesis is over.

8 0

3 years ago

How do frogs utilize extensor and flexor muscles in order to swim, jump, and do other frog things efficiently; include specific

ahrayia [7]

The structure of the feet and legs varies greatly among frog species, depending in part on whether they live primarily on the ground, in water, in trees or in burrows. Frogs must be able to move quickly through their environment to catch prey and escape predators, and numerous adaptations help them to do so. Most frogs are either proficient at jumping or are descended from ancestors that were, with much of the musculoskeletal morphology modified for this purpose. The tibia, fibula, and tarsals have been fused into a single, strong bone, as have the radius and ulna in the fore limbs (which must absorb the impact on landing). The metatarsals have become elongated to add to the leg length and allow the frog to push against the ground for a longer period on take-off. The illium has elongated and formed a mobile joint with the sacrum which, in specialist jumpers such as ranids and hylids, functions as an additional limb joint to further power the leaps. The tail vertebrae have fused into a urostyle which is retracted inside the pelvis. This enables the force to be transferred from the legs to the body during a leap 
The muscular system has been similarly modified. The hind limbs of ancestral frogs presumably contained pairs of muscles which would act in opposition (one muscle to flex the knee, a different muscle to extend it), as is seen in most other limbed animals. However, in modern frogs, almost all muscles have been modified to contribute to the action of jumping, with only a few small muscles remaining to bring the limb back to the starting position and maintain posture. The muscles have also been greatly enlarged, with the main leg muscles accounting for over 17% of the total mass of the frog.

7 0

3 years ago