Condensation of water droplets causes the formation ofA. rain
B. lakes
C. clouds
D. water vapor

Even though the light dependent reactions of photosynthesis are not affected by a change in temperature, the light independent reactions of photosynthesis can be dependent on temperature. They are reactions of enzymes. As the enzymes approach their optimum temperatures the rate increases.

While mapping the motor cortex, researchers Foerster and Penfield found that body areas requiring the greatest control occupied the greatest amount of cortical space

The majority of the afferent fibres that enter each region of the cerebral cortex come from a single thalamic projection nucleus. This thalamocortical projection is topographically organised so that discrete thalamic volumes project to discrete fields in the cortex, while continuous thalamic volumes project to a continuous region in the cortex. A point-to-point projection could be thought of in this respect, however this oversimplification ignores the mutual overlap of thalamic fibres inside their "modular" cortical arborization zones . Because the cerebral cortex is a two-dimensional surface and the thalamus is a three-dimensional body, the topographical depiction of the thalamic projection nuclei over the cerebral cortex is a topological problem.

Hence, Foerster and Penfield found that body areas requiring the greatest control occupied the greatest amount of cortical space.

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Answer:

with no doubt I think the answer is diploid itdoes not access meiosis at all

Answer:

haploid cells

Explanation:

Answer:

B. False

Explanation:

Life history characteristics and survivorship patterns within a species are pre-set but may be changed according to the changes in environmental conditions needed for growth and reproduction. Allocation of energy to reproduction or other processes varies according to food availability, temperature, rainfall, etc.  

Abundant resource availability favors the production of a large number of smaller offspring while limited resources make a population to produce a small number of larger progeny to increase the survival rate.  

Example: Clutch size in birds is adjusted to obtain the maximum survival of progeny.  

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.