Refer to the chart below. Which organism has the greatest fitness?Select one:
a. Organism C
b. Organism D
c. Organism A
d. Organism B

Aristotle illustrates virtue in a way distinct from what one is usually taught in school, but it is much closer to how one thinks on a regular basis. One is usually taught that courage is the contrary of cowardice, and generosity is the reverse of miserliness and others.  

Although, Aristotle illustrates virtue as the mean between the two extremes, which requires to be avoided. For Aristotle, virtue is the golden mean between the two extremes. Though the mean is not a strict arithmetic mean. Virtue comes in between the two extremes, but where it actually comes depends on a very large extent to a particular situation.  

Answer:

Aristotle defines a virtue as a habit that requires practice.

Explanation:

Moreover, as with any habit, becoming virtuous requires practice, repeatedly doing similar kinds of things until it becomes second nature.

Answer:

a. p=0.725, q = 0.275: P=0.51, H=0.43, Q=0.06

Explanation:

Let state our given parameters from the question:

Frequencies of coat color genes at the C locus for population 1 are .85 for C

This implies that the Allelic frequency C for population p1 =0.85

Frequencies of coat color genes at the c locus for population 1 are .15 for c

This implies that the Allelic frequency c for population q1 = 0.15

Frequencies for Care .6 i.e p2= 0.6

Frequencies for care .4 i.e, let that be q2= 0.4

The table below shows a diagrammatic representation of the above expression:

Alllelic Frequency                      C                                          c

Population 1                        (p1)   0.85                              (q1)   0.15

Population 2                       (p2)   0.6                               (q2)   0.4

Now, from above: let think of the table as a punnet square and then cross it together;

                                            (p1)  = 0.85                              (q1) =  0.15

p2 = 0.6                               p1p2                                       p2q1

                                            = 0.6 × 0.85                           = 0.15 × 0.6

                                            = 0.51 (P)                                = 0.09 (H)              

q2 = 0.4                               p1q2                                       q1p2

                                            = 0.85 × 0.4                           = 0.4 × 0.15

                                            =0.34 (H)                                = 0.06 (Q)

From the above table, the heterozygous are represented by (H)

∴ Frequency of heterozygous can be calculated as:

= 0.09 + 0.34

= 0.43

Thus, we can conclude that the progenyF1 genotypic frequencies are:

P= 0.51

H= 0.43

Q= 0.06

Now, let us calculate the allelic frequencies, p and q in F1

p = P + 1/2 × (H)

= 0.51 + (1/2 × 0.43)

= 0.51 + 0.215

= 0.725

q =Q + 1/2× (H)

= 0.06 + (1/2 × 0.43)

= 0.06 × 0.215

= 0.275

Hence, p=0.725, q = 0.275: P=0.51, H=0.43, Q=0.06 , This makes option a the correct answer.

The correct answer is B. Tests experimental and control groups in parallel

Explanation:

In science, a controlled experiment is a type of experiment in which scientist intervene or manipulate variables. This often implies scientists use a control group in which the variable is constant and an experimental group that has a different treatment or changes in the variable studied. This allows scientists to evaluate the effect of the variable by comparing the results of both groups and thus confirm or discard a hypothesis. Therefore, a controlled experiment can be defined as on that "tests experimental and control groups in parallel".

Final answer:

A controlled experiment is one that tests experimental and control groups in parallel, with all other variables kept constant. This enables the scientist to observe the impact of a single variable.

Explanation:

A controlled experiment is one that can be defined as an experimental set up in which the scientist keeps all variables constant, except for the one being studied. The correct option here would be option 'b'. Such an experiment tests experimental and control groups in parallel.

This methodology allows the researcher to observe the effect of one variable on the study subject while ensuring that all other conditions remain the same. For example, if you were to experiment with plant growth, you might keep factors like sunlight, water, and type of soil constant while changing the type of fertilizer used.

Learn more about Controlled Experiment here:

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D. or B.because algae is cabable of producing oxygen through photosynthesis

Answer:

Option A

Percent of the population that has skin pigment but also carries the recessive allele is percent

Explanation:

As per hardy-weinberg equation,

------------- Eq(A)

where

frequency of the homozygous genotype (dominant)

frequency of the homozygous genotype (recessive)

frequency of the heterozygous genotype (recessive)

Also, sum of the allele frequencies at the locus is equal to one.

Thus,   ------------- Eq(B)

Here frequency of recessive allele i.e %

Substituting this in equation B, we get

frequency of dominant allele i.e %

Substituting the value of frequency of both dominant and recessive allele in equation A, we get

Thus, percent of the population that has skin pigment but also carries the recessive allele is percent