The graph shows, in a period of time between 1978 and 2010. Therefore, it's related to the last 50 years, not the last 100 years, meaning that alternative D is wrong. The amount of sea ice shown in the graph is shown in the scale of million square miles. So, as the graph shows an initial value around 3.2 (3 millions and 2 hundred miles) and it ends up in a value around 2.2 (2 millions and 2 hundred miles), still the real values are around 2.8 (2 million and 8 hundred thousands initially) and 2.0 (2 millions at the end), having a lower value of 1.8 (1 million and 8 hundred thousand); So, in both scenarios, arctic sea ice has shrunk by 1,000,000 (1.0) square miles, being under 2,000,000, not over. Therefore, the highest number to be inside this amount is 2,000, so the correct answer is A: Arctic sea ice has shrunk by over 2,000 square miles in the last 50 years.
Answer:
18%
Explanation:
Chargaff's rule states that A = T and C = G
The equation for this must be A + C = T + G
A and T should both be .32 since they're equal
--> .32 + C = .32 + G
Both sides must equal each other by .50
.50 - .32 = .18
.18 × 100 = 18% Guanine and also 18% Cytosine
Answer:
a) Jar C, b) Jar A
Explanation:
a) The amount of CO2 is the highest in jar C because it only contains CO2-producing organisms: mice. A plant in the same jar, like in jar B, would counteract this CO2 production by using it to produce oxygen, which would balance it out.
b) Jar A has the lowest amount of CO2 because the plant is by itself only producing oxygen. There is no life form in jar b that produces CO2.
Options for the question have not been given. They are as follows:
A) 1/8
B) 3/4
C) 1/16
D) 1/2
Answer:
D) 1/2
Explanation:
Let the allele for extra digits be represented by D. Since it is a dominant trait, DD and Dd will result in extra digits whereas dd will result in normal number of digits. The woman has normal number of digits so she has dd genotype. The man has extra digits so he can be either DD or Dd.
The couple's first child has normal number of digits so she has dd genotype. She has obtained one d allele from her mother and another d allele from her father. Hence, the man's genotype is Dd.
Man = Dd
Woman = dd
Their children :
D d
d Dd dd
d Dd dd
Half of the progeny will have Dd genotype and will have extra digits. Other half will have dd genotype and will have normal number of digits. So, the probability of their next child to have extra digits is 1/2.
Answer:
Sexual reproduction produces offspring which are distinct from the parent.This strategy is successful because this reproduction increases genetic variation, which entailsthe odds of some progeny, which are suited to new and challenging conditions.
Explanation:
Sexual reproduction produces genetic variability, as two nucleic unites to produce a new offspring this allows the fungus to adapt to new environments. Sexual reproduction in the fungi consists of three sequential stages plasmogamy, karyogamy, and meiosis.
It involves diploid chromosomes pulled into two daughter cells, each containing an haploid set of chromosome.
Plasmogamy is the fusion of two protoplasts which brings together two compatible haploid nuclei this nuclear types are present in the same cell, but are not yet fused. Karyogamy then leads to the fusion of the haploid nuclei and the formation of a diploid nucleus. The new cell formed by karyogamy is the zygote.
