Answer: A) The resident populations are more likely affected by these threats because of their proximity to shore.
Explanation:
All of these threats are higher in coastal regions.
As contaminants usually reach the ocean through rivers or because of coastal cities, the highest degree of pollution is there. All chemicals will accumulate in this region first, before they disperse into the offshore region. This is also caused by the fact that the coastal waters have a lower volume and surface area than the offshore which results in a higher degree of contamination.
Answer:
a gene present on the Y chromosome that triggers male development
Explanation:
Answer:
Competition for resources like food and space cause the growth rate to stop increasing, so the population levels off. ... The carrying capacity (K) is the maximum population size that can be supported in a particular area without destroying the habitat. Limiting factors determine the carrying capacity of a population. Population Growth Limits | CK-12 Foundation
<span>UGAGCC
There
are three principles to keep in mind when predicting the sequence of
the mRNA produced by transcription of a particular DNA sequence.
The RNA polymerase reads the sequence of DNA bases from only one of the two strands of DNA: the template strand.The
RNA polymerase reads the code from the template strand in the 3' to 5'
direction and thus produces the mRNA strand in the 5' to 3' direction.In
RNA, the base uracil (U) replaces the DNA base thymine (T). Thus the
base-pairing rules in transcription are A→U, T→A, C→G, and G→C, where
the first base is the coding base in the template strand of the DNA and
the second base is the base that is added to the growing mRNA strand.</span>
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.
