The correct answer choice among the statement above about strings of amino acids which can be decoded using single letter amino acid abbreviations to a type of
secondary structure is:
Alanine luecine proline histidine alanine histidine glutamic acid leucine isoleucine cysteine glutamic acid serine. Option a is the correct answer
- The structure of mature messanger RNA encoding the average eukaryotic proteins is: 5' cap, 5' UTR, coding region, 3' UTR, and poly(A) tail. Option a is the correct answer
- The correct answer choice which is extrinsic is: the TATAAT consensus sequence at the -10 posi!on of prokaryotic promoters
- The diagram solution of the coding strand, 3’, 5’, translation on, transcription replication to the last isquestion is attached
<h3>Amino acid</h3>
Amino acid can simply be defined as those organic molecules or substances which combine to form proteins. However, amino acids are generally characterized by some key features or properties; some of these properties of amino acids includes the following:
- They are soluble in water
- They are base units of proteins
- Amino acid are insoluble in organic solvent
- Amino acids are colourless
- They are crystalline solid
- All amino acids however too, have at least one acidic carboxylic acid (-COOH) group and one basic amino (-NH2) group.
So therefore, the correct answer choice among the statement above about strings of amino acids which can be decoded using single letter amino acid abbreviations to a type of
secondary structure is:
Alanine luecine proline histidine alanine histidine glutamic acid leucine isoleucine cysteine glutamic acid serine. Option a is the correct answer
Learn more about amino acids:
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Answer:
It is a beneficial mutation.
Explanation: Mutations are permanent changes in the nucleotide sequence of a DNA. Mutations can beneficial, neutral and harmful or deleterious. When change in the nucleotide sequence of DNA a mutation enhances the effectiveness of a protein or improves the protein function, it is said to be beneficial. When a mutation causes the synthesis of a protein which have the same amino acid as the original protein and performs the same function as the original protein, it is said to be silent or neutral. When a mutation results in the synthesis of a protein with an altered amino acid sequence and a nonfunctional protein, it is said to be harmful.
The study of onion root tips is widely known for the clear appearance of the stages of mitosis since the chromosomes are large and are clearly visible. They are easily stained with the stainer and are visible clearly under the electron microscope. The phase of mitosis are
Interphase
Prophase
Metaphase
Anaphase
Telophase and
Cytokinesis
Each of these phases show the different stages in which the division of one cell into two daughter cells takes place. The cytokinesis is the final phase in the mitosis cell division. This is the part of cell division process during which the cytoplasm of the cell divides into two equal parts and forms into two daughter cells.
During the cytokinesis process, the spindle apparatus partitions and transports the duplicated chromatids and moves into the cytoplasm of the dividing daughter cells. In this process a dividing structure called the cell plate is formed to distinguish the daughter cells. This cell plate later grows into a double layered cell wall which then splits the parent cell into two separate cells.
Hence the option D is the right answer
Answer: To do different kinds of jobs
Explanation:
Answer:
The correct answer is -
A) BbTt x bbTt
B) .5^4 or 1/16 or 0.0625
Explanation:
As it is given that Brachydactylus and PTC tasters both traits are autosomal dominant conditions which mean only one allele would be enough.
For Branchydactylus and For tasting : man and women will be heterogeneous.
Hence,
The Genotype of man = BbTt
The Genotype of wife = bbTt
b. Answer = 0.0625
For Branchydactylus:
Bb X bb
Possible genotypes:
B b
b Bb( Brachydactylus) bb(normal)
b Bb (Brachydactylus) bb (normal)
The probability of a single child being Branchydactylus = 2/4 = 0.5
So, Probability of all 4 child being Branchydactylus = .5 x .5 x .5 x .5 = 0.0625
