ATP hydrolysis _______.to ADP and Pi is the most common mechanism for transferring free energy to drive endergonic reactions

would be more exergonic if phosphate did not have resonance structures

results in products that are less solvated than the ATP

is usually in the form of ATP -> ADP Pi or ATP -> AMP PPi

Answer:

immunoglobulins; chemokines

Explanation:

Chemicals known as ____ and ____ guide neuron.

Answer In contrast, neurons in many regions, including the cerebral cortex, cerebellum, hippocampus, and spinal cord, are guided to their final destinations by crawling along a particular type of glial cell, called radial glia, which acts as a cellular guide

Answer:

adaptation through evolution and ability to reproduce

Answer:

A

Explanation:

The vast majority of bacteria that produce gastrointestinal symptoms are transmitted via fecal-oral transmission. This means that when sub-optimal hygiene is present, bacteria from the hands of the infected person that got there from contact with its own feces (after going to the bathroom, for example) is passed to the next person. This happens a lot in the food business, and it's the mode of transmission of the much famous Salmonella.

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.

Answer:C

Explanation:I just had this question