Answer:
The correct answer is d. suggests that the time since divergence may not be sufficient for complete reproductive isolation to have occurred.
Explanation:
Polar bears are evolved from brown bear and grizzly bears are subspecies of brown bear so polar bears and brown bears are different species. These two species habitat is different so they are geographically isolated but due to the melting of ice by global warming, their niche started overlapping.
Commonly they do no reproduce with each other because of the reproductive isolation but they can produce viable hybrids called growlers and pizzlies if interspecies reproduction takes place which shows that the time interval since divergence is not enough that can make these two species completely reproductive isolated. So the right answer is d.
Answer:
A burning candle is extinguished when it is covered with a jar.
Explanation:
the candle extinguishes because there is no oxygen in the empty jar
the candle burns longer in a jar with a plant because thye plant gives out o2 and o2 is a supporter of combustion.
the plant also lives longer because the candle gives out co2 after combustion which the plant uses co2 for photosynthesis.
Answer:
Answer is 3/16.
Explanation:
If the F1 progeny has red axial flowers this shows us that red and axial genes are dominant. If we say that R is for red dominant gene, r is for white recessive gene.
If we say A is axial dominant gene, a is for terminal recessive gene.
All F1 progeny has AaRr phenotype.
When we cross them, Aa x Aa can have AA Aa Aa aa
When Rr x Rr crossed, RR Rr Rr rr
The F2 progeny can have white axial flowers by having a and R in the phenotype with the possibility having aa= 1/4 , R in the phenotype , the possibility is 3/4.
1/4 x 3/4 = 3/16 in all F2 progeny
Answer:
a. 29, 29, 27, 27.
Explanation:
Separation of homologous chromosomes to the opposite poles during anaphase-I would have produced the haploid chromosome number of the daughter cells. This means that the parent cell with 56 chromosomes would produce four daughter cells each of which would have 56/2= 28 chromosomes. If one of the chromosome pairs did not segregate during anaphase-I, one daughter cells formed by the end of meiosis-I would have "n+1" chromosomes while the other would have "n-1" chromosomes. This chromosome number is maintained by meiosis-II.
Therefore, nondisjunction at anaphase-I in the parent cell with 56 chromosomes would produce a total of four daughter cells. The two daughter cells would have "n+1= 29" chromosomes and rest two would have "n-1=27" chromosomes.
