Answer: polyunsaturated fatty acid residues
Explanation: Unsaturated fatty acids have a lower melting point so arctic animals have these fatty acids in their membranes.
Answer:
C. They all use a cut and paste mechanism.
Explanation:
DNA transposons can go through a replicative or nonreplicative transposition.
The replicative transposition uses a "copy and paste" mechanism that consists of the introduction of a new copy of the transposable element in a new position, meanwhile <u>the old copy remains in the original position</u>. This determines an increase in the number of copies.
The nonreplicative transposition uses a "cut and paste" mechanism that consists of the cleavage of the transposable element from its position and its <u>insertion in a new position</u> without increasing the number of copies.
Retrotransposons, on the other side, move through RNA intermediates generated by the reverse transcriptase.
Codominance is when the two parent phenotypes are expressed together in the offspring. An example is a white flower and a red flower producing offspring with red and white patches.
Answer:
TACGTACGTTAC (so the second one)
Answer:
E) Either anaphase I or II
Explanation:
Failure of segregation of homologous chromosomes during anaphase I or failure of segregation of sister chromatids during anaphase II leads to the presence of the abnormal number of chromosomes in resultant gametes. In the given example, the egg mother cell with 48 chromosomes (24 pairs) would enter meiosis I but the failure of one pair of homologous chromosomes to segregate from each other followed by normal meiosis II would result in the formation of two gametes with one extra chromosome and two gametes with one less chromosome.
On the other hand, if the nondisjunction occurs at anaphase II of meiosis II, two normal gametes, one gamete with one extra chromosome and one gamete with one less chromosome will be formed. Therefore, nondisjunction at anaphase I or anaphase II would have resulted in the production of eggs with one extra chromosome.
