Introduction

A bioassay is defined as an experiment that helps in the determination of the potency as well as concentration of the compound my calculating the biological responses. The bioassays measures drug concentration, drug potency and calculate pharmacological activity of other drugs (Stypuła-Trębas et al. 2017). The necessary requirements of bioassays. In the following practical, bioassays are done to quantify drug concentration in biological activity. Biological substances activities of an unknown are compared with international standard and the results are expressed as clotting factor, hormones or other mediator (Cutignano et al. 2015) 

Methods

The isolation of guinea-pig ileum was prepared like previous experiment. Two assays followed are

Single point assay:

Increased concentration of 10µl of 10µg/ml ACh of standard ACh was added to tissues till maximum response was reached to plot log dose-response curve. The linear part of the curve was noticed and estimated 50% of maximum response. Unknown was added to the organ bath so that to obtain a volume that stimulates the 50% of responses and (V_U) was read the exact response (R_U) to the bulk from the curve as well as the organ bath concentration was obtained from log dose dependant response for the standard that resembles to the reaction.

Unknown of ‘Y’ ml produced a response of R, then the organ bath concentration would be B (ng/ml). The ‘X’ ml of unknown have (B x organ bath) ng. Hence, the concentration of unknown obtained was (B x organ bath vol)/Y) ng/ml. The following formula was used in determining the stock concentration of unknown

 Organ bath vol. (ml)                X Concentration of unknown read off (ng/ml)

  Vol. of unknown added (ml)

Bracketing assay:

Two concentration were selected for standard which comes under the linear portion of the log concentration-response curve, among 30% plus 70% of concentrated response towards the standard (S1 and S2). The volume of the unknown was determined that produces a reaction roughly midway amongst the above two standard responses (U). The responses were repeated as per the bracketing assay such as S1-U-S2-U-S1-U-S2-U-S1. Each of the ‘brackets’ were plotted on distinct log concentration standard-response graphs. The calculation for the original concentration of the unknown by the technique designated for single point assay were performed. Four estimations were provided for the concentration of the original unknown solution shown in µg/ml or ng/ml. The mean ±standard deviation was obtained for the value received. The concentration was converted to Molar concentration, such as (g/L)/ MWt (Molar weight of Ach = 181.6).

Result

Two concentrations such as 1.114 and 1.1431 of the standard falls in the linear portion of the log concentration-response curve, among 30% and 70% of maximum response produced to the S1 and S2 standard. A 40ul volume of unknown produced a response that is approximately halfway between the two standard responses (U). Table for acetylcholine graph Log Conc. Ng/ml 0.477 0.845 1.114 1.431 1.724 2.068. Response 0.1589 0.2351 0.2874 0.762 0.8011 0.8926. etc.

Trace of result

The result obtained suggested that Unknown result U1 and U2 have a response rate between 20% to 80% and by doing graphical analysis it was obtained that U1 have a response of 0.60 and 0.40 respectively.  All the data interpretation was done from the data obtained. The activity of unknown sample is between two known samples.

Calculation for final drug organ bath concentration

Organ bath concentration =

Vol of ACh added (µl)    x Conc of stock ACh (µg/ml)

Organ bath volume (µl)

10 (µl)    x 10 (µg/ml) = 0.033 ug/ml= 0.33 x 1000= 33.3 ng/ml

3000 (µl)

20 (µl)    x 10 (µg/ml) = 0.0286 ug/ml= 0.0286×1000 = 28.61 ng/ml

6990 (µl)

30 (µl)    x 10 (µg/ml) = 0.0230 ug/ml = 0.0230×1000 = 23.07 ng/ml

13000 (µl)

50 (µl)    x 10 (µg/ml) = 0.0094 ug/ml = 0.0094x 1000 = 9.45 ng/ml

52900 (µl)

100(µl)    x 10 (µg/ml) = 0.00855 ug/ml = 0.00855 x1000= 8.55 ng/ml

116900 (µl)

Table of responses against the organ bath concentration

Amount added	Organ bath conc (ng/ml)	[Organ bath] ml	Log [organ bath]Ng/ml	Response (g)
10μl of 10μg/ml ACh	33.3	3	0.477	0.1589
20μl of 10μg/ml ACh	28.61	6.99	0.845	0.2351
30μl of 10μg/ml ACh	23.07	13	1.114	0.2874
40μl of 10μg/ml ACh	14.8	26.9	1.431	0.762
50μl of 10μg/ml ACh	9.4	52.9	1.724	0.8011
100μl of 10μg/ml ACh	8.55	116.9	2.068	0.8926

Log concentration curve

Fig 1: Log Concentration-response curve

Calculation of Unknown concentration

a) single point bioassay

A 40ul volume of unknown produced a response that is approximately halfway between the two standard responses

Concentration of unknown = (organ bath conc x organ bath volume)/ unknown vol (ml)

= (14.8 x 26.9)/ 0.04 (40ul)

= 9953 ng/ml

= 9.95 ug/ml

b) bracketing bioassay

Standard (S1) = 0.28

Standard (S2) = 0.76

Unknown (U1) = 0.60

Unknown (U2) = 0.40

S1-S2-U1-U2   

S2-U1-U2-S1

U1-U2-S1-S2

U2-S1-S2-U1

Bracket 1	Bracket 2	Bracket 3	Bracket 4
Log dose	response	Log dose	response	Log dose	response	Log dose	response
1.114	0.28	1.431	0.76	1.221	0.40	1.432	0.60
1.431	0.76	1.221	0.40	1.432	0.60	1.114	0.28
1.221	0.40	1.432	0.60	1.114	0.28	1.431	0.76
1.432	0.60	1.114	0.28	1.431	0.76	1.221	0.40

Fig 2: 4 bracket graphs

Concentration of unknown received (approx.)

Concentration of unknown = (organ bath conc x organ bath volume)/ unknown vol (ml)

= (14.8 x 26.9)/ 0.04 (40ul)

= 9953 ng/ml

= 9.95 ug/ml (1)

(23.07x 13)/0.04 =7.49 ug/ml (2)

(9.4 x 52.9)/ 0.04= 12.4 ug/ml (3)

(14 x 26)/ 0.04= 9.1 ug/ml (4)

Four unknown concentration: 9.95 ug/ml, 7.49 ug/ml, 12.4 ug/ml and 9.1 ug/ml

Unknown concentration	Mean	Standard deviation
9.95	9.735	2.048715
7.49
12.4
9.1

Average concentration: 9.735 ug/ml

Ach molar weight 181.6

1M= 181.6 g/L

Therefore, 1M will be (1ml/1000ml) x 181.6g = 0.1816 g.ml

Conversion from g to ng,

1 g = 1000000 ng

Hence, 0.1816x 1000000 ng= 181600000 ng

So, 181600000 ng/ml

9735ng/ml/181600000

5.36 x10^-7 M (Molar concentration)

Discussion

The most accurate concentration of unknown would be 9.1ug/ ml as the concentration is close to the standard drug concentration as the 40ul ranges between 30 to 70% of the drug activity and the drug shows its highest activity when it reaches 50% mark (Nagai 2018). The potential error in the experiment is the addition of drug volume which is done manually and hence there are chances of minor error. A very minor change would affect the further concentration. Bioassay experimental design need to be improved as chances of error are more. The addition of drug can be done with digital apparatus so that human error can be avoided.

Bioassay is mainly used in analysing biological threats and it also gives a quality valuation of a mixture. It is frequently used to assess the ecological influence as well as safety towards new technologies. It helps is detecting the concertation of the substance that brings biological changes (Lu et al. 2016).

References

Cutignano, A., Nuzzo, G., Ianora, A., Luongo, E., Romano, G., Gallo, C., Sansone, C., Aprea, S., Mancini, F., D’Oro, U. and Fontana, A., 2015. Development and application of a novel SPE-method for bioassay-guided fractionation of marine extracts. Marine drugs, 13(9), pp.5736-5749.

Lu, Y., Lu, H., Kim, K., Sun, Y., Fu, J. and Guo, Y., 2016. A New Bioassay Method for Evaluation Allelopathic Potential of Rice Germplasm. Journal of Life Sciences, 10, pp.128-133.

Nagai, T., 2018. A novel, efficient, and ecologically relevant bioassay method using aquatic fungi and fungus‐like organisms for fungicide ecological effect assessment. Environmental toxicology and chemistry, 37(7), pp.1980-1989.

Stypuła-Trębas, S., Minta, M., Radko, L., Jedziniak, P. and Posyniak, A., 2017. Nonsteroidal mycotoxin alternariol is a full androgen agonist in the yeast reporter androgen bioassay. Environmental toxicology and pharmacology, 55, pp.208-211.