Active sites in Enzymes can be divided in subsides (S) which consist of single amino acids. The subsides are flanking the catalytic site [C] and are numbered starting at the catalytic site. An Enzyme with 4 subsites on the one side and 3 subsides on the other side of the catalytic site has the following nomenclature S4-S3-S2-S1-[C]-S1'-S2'-S3'
The corresponding positions (P) of the substrate have the same numbering as the subsides they occupy. The positions are counted from the point of cleavage. Note that not in all cases every subside in occupied by a aminoacid of the substrate. Thus finally the P residue describes the aminoacid or molecule which can be found at that specific location of substrate.
P4-P3-P2-P1------P1'-P2' ------ = point of cleavage
S4-S3-S2-S1-[C]-S1'-S2'-S3'
Reference: Schechter and Berger (1967) On the size of active sites in proteases (Biochem. Biophys. Res. Commun. 27, 157-162
Let p1 and p2 be the two prime numbers. Because they are prime, their divisors are div(p1) = {1,p1} and div(p2) = {1,p2}. So GCD(p1,p2) = Greatest Common Divisor of p1 and p2 = p1 if p1 equals p2 1 if p1 is different from p2
P1 or parental
P1-e is an expression, not a formula.
In genetics, in a pure-breeding population, the parental generation is the P1 generation. The off-spring of the P1 Generation is called the F1 Generation
If the old population is P1, the new population is P2, and the growth rate is G, G = (P2 - P1) ÷ P1 x 100%
P1: tt F2: tt
#include#include#define n 100void del_char(char str1[n],char str2[]){char str[n],*p1,*q,i,j;p1=str1;q=str;*q=*p1;i=0;while(*p1!='\0'){{if(*p1==str2[i]){p1++;i++;if(*p1==str2[i]){p1++;i++;if(*p1==str2[i]){p1++;i++;}else{q++;}}else{q++;}}else{p1++;q++;}*q=*p1;}i=0;}printf("%s\n",str);}main(){char str1[n],str2[]={"to "};clrscr();printf("please enter a text and includ 'to':\n");gets(str1);printf("remove 'to',remaining :\n");del_char(str1,str2);getch();return 0;}Output:please enter a text and includ 'the':Write a c program to simulate the calculator using functionremove 'the',remaining :Write a c program to simulate calculator using function
P1 Material is the name given to Carbon Manganese Steel in the engineering industry
ed=(q1-q2)/q1/(p1-p2)/p1
P1" is the highest short-term rating category for Moody's Investor Service. P1 rating are considered to be of high credit quality
(p1/v1) = (p2/v2)For Apex (P1 N1)= (P2N2 )
#include<stdio.h> #include<conio.h> int main() { char p[10][5],temp[5]; int i,j,pt[10],wt[10],totwt=0,pr[10],temp1,n; float avgwt; clrscr(); printf("enter no of processes:"); scanf("%d",&n); for(i=0;i<n;i++) { printf("enter process%d name:",i+1); scanf("%s",&p[i]); printf("enter process time:"); scanf("%d",&pt[i]); printf("enter priority:"); scanf("%d",&pr[i]); } for(i=0;i<n-1;i++) { for(j=i+1;j<n;j++) { if(pr[i]>pr[j]) { temp1=pr[i]; pr[i]=pr[j]; pr[j]=temp1; temp1=pt[i]; pt[i]=pt[j]; pt[j]=temp1; strcpy(temp,p[i]); strcpy(p[i],p[j]); strcpy(p[j],temp); } } } wt[0]=0; for(i=1;i<n;i++) { wt[i]=wt[i-1]+et[i-1]; totwt=totwt+wt[i]; } avgwt=(float)totwt/n; printf("p_name\t p_time\t priority\t w_time\n"); for(i=0;i<n;i++) { printf(" %s\t %d\t %d\t %d\n" ,p[i],pt[i],pr[i],wt[i]); } printf("total waiting time=%d\n avg waiting time=%f",tot,avg); getch(); }