Scientists repeat the experiment to further prove it being correct, and they can replicate is with data, models, and other forms of proof of a correct hypothesis
Answer:
Heterogeneous consists of the structure with various components or elements appearing to be irregular or variegated.
Explanation:
An example of this definition is a dermoid cyst in which has the components of a heterogeneous attenuation on CT.
Answer:
A molecule of mRNA is formed.
Explanation:
Translation is the second process that occurs in gene expression. It is the process by which the information encoded in the mRNA transcript is used to synthesize a protein.
The mRNA nucleotide sequence is read in a group of three nucleotides called CODON. Each codon specifies an amino acid. Translation, which occurs in the ribosomes (cytoplasm), reads the codon with an anticodon using the complementary base pairing rule i.e. A-U, G-C. This means that a CODON-ANTICODON pairs.
The anticodon carries a corresponding amino acid to the polypeptide sequence. A peptide bond is formed when two amino acids joins together in a condensation reaction.
Note: A molecule of mRNA is formed during transcription
Answer:
Active transport requires energy from ATP while facilitated diffusion does not
Explanation:
Active transport and facilitated diffusion with the use of channel and carrier proteins are both ways by which ions, polar and large molecules cross a selectively permeable membrane.
The major difference is that; Active transport transport these particles from a low to high concentration, which is against concentration gradient and hence, energy is required to perform the task
Facilitated diffusion transport from a high to low concentration, which is through a concentration gradient and hence, no energy is required to perform the task.
Answer:
The recombination frequency between two genes exhibits a positive correlation with the distance between them, that is, farther they are, and more will be the chance of recombination. Thus, recombination frequency is used to signify distance among the two genes, for example, 1 percent recombination frequency demonstrates distance of 1 map unit.
Let us consider that the heterozygous female of genotype AaBb can generate four kinds of gametes, that is, AB, Ab, aB and ab. Of these, the two gametes are the outcomes of recombination, or it can be said that 50 percent are recombinants. Thus, it can be concluded that in case of two linked genes, the maximum probable recombination frequency is 50 percent.
This shows that any genes, which are distant than 50 map units will function as unlinked and will function as if they were on distinct chromosomes, and the frequency of recombinant frequency will be 50 percent.
In the given question, it is given that the map distance between the two genes is 80 map units, that is, more than 50 map units. The maximum probable recombinant offspring will be 50 percent of the entire offspring.
