Gct tag tcg
Ooh..., tough one. You could say it is the user. But that is lame. You could also say processor, since no computer could work without one. My first computer worked with a 5¼" floppy drive, no hard drive, 256 kb of RAM and a 4.7 Mhz processor. No mouse either. The graphics were CGA and EGA.
GW-BASIC (short for Gee-Whiz Beginner's All-purpose Symbolic Instruction Code) is a version of the BASIC programming language released by Microsoft in 1983 for the DOS operating system. GW-BASIC evolved from Microsoft BASIC-80, also known as MBASIC. The last version was released in 1988. GW-BASIC was superseded by QBASIC in 1991. Note: please ask ONE question at a time.
Before we look at the complimentary mRNA sequence of the given DNA sequence, let us remember that RNA contains uracil (U) in place of Thiamine (T) The querry sequence is: t-a-c-c-t-c-g-c-a-a-c-t So the mRNA sequence would be: A U G G A G C G U U G A
#include <iostream.h> #include <conio.h> # include <process.h> //exit(0) # include <dos.h> # include <stdlib.h> # include <graphics.h> # include <stdio.h> int MaxX, MaxY, MidX, MidY ; int bri[5][20] ; bricks(); //to display bricks and paddle and bulb rounded form delbrick(int,int); int graphmode= CGAHI, graphdriver = CGA; //CGAHI = 640 x 200 (col x row) void MousePos(int* x) { union REGS in,out; in.x.ax=3; int86(0x33,&in,&out); *x=(int)out.x.cx; } main() { int BallX, BallY, Base1, Base2, dx = 1, dy = -1, OldX, OldY ; int totallayer[5] = { 10, 20, 30, 40, 50 }, max = 50, layer = 4 ; int i, flag = 0, speed = 10, score = 0, chance = 4, areareq ; char *m1, *m2 ; //pointers to memory initgraph ( &graphdriver, &graphmode, "c:\\tc\\bgi" ) ; MaxX = getmaxx() ; MaxY = getmaxy() ; MidX = MaxX / 2 ; MidY = MaxY / 2 ; /* draw the four layer of bricks, the paddle and the ball */ rectangle ( 0, 0, MaxX, MaxY - 12 ) ; bricks() ; rectangle ( MidX - 25, MaxY - 19, MidX + 25, MaxY - 12 ) ; circle ( MidX, MaxY -25, 12 ) ; /* memory allocation for storing the image of the ball */ areareq = imagesize ( MidX - 12, MaxY - 18, MidX + 12, MaxY - 8 ) ; m1 =((char*) malloc ( areareq )) ; /* memory allocation for storing the image of the paddle */ areareq = imagesize ( MidX - 25, MaxY - 7, MidX + 25, MaxY - 1 ) ; m2 =((char *) malloc ( areareq ) ); /* image of the paddle and the ball is stored into allocated memory */ getimage ( MidX - 12, MaxY - 7 - 12 - 12 + 1, MidX + 12, MaxY - 8 -12,m1 ) ; getimage ( MidX - 25, MaxY - 7 - 12, MidX + 25, MaxY - 1 - 12, m2 ) ; /* store current position of the paddle and ball */ Base1 = MidX - 25 ; Base2 = MaxY - 19 ; BallX = MidX - 12 ; BallY = MaxY - 7 - 12 + 1 - 12 ; /* display balls remaining ( initially 3 ) */ gotoxy ( 45, 25 ) ; cout<< "Balls :" ; for ( i = 0 ; i < 3 ; i++ ) { circle ( 515 + i * 35, MaxY - 5, 12 ) ; } /* display starting score */ gotoxy ( 1, 25 ) ; cout<< "Score: "; gotoxy(16,25); cout<<score; while ( !kbhit() ) { flag = 0 ; /* saving current x and y coordinates of the ball */ OldX = BallX ; OldY = BallY ; /* update ballx and bally to move the ball in correct direction */ BallX = BallX + dx ; BallY = BallY + dy ; /* according to the position of ball the layer of bricks isdetermined*/ if ( BallY > 40 ) { max = 50 ; layer = 4 ; } else { if ( BallY > 30 ) { max = 40 ; layer = 3 ; } else { if ( BallY > 20 ) { max = 30 ; layer = 2 ; } else { if ( BallY > 10 ) { max = 20 ; layer = 1 ; } else { max = 10 ; layer = 0 ; } } } } /* if the ball hits the right boundary, move it to the left */ if ( BallX > ( MaxX - 24 - 1 ) ) { BallX = MaxX - 24 - 1 ; dx = -dx ; } /* if the ball hits the left boundary, move it to the right */ if ( BallX < 1 ) { BallX = 1 ; dx = -dx ; } /* if the ball hits the top boundary, move it down */ if ( BallY < 1 ) { BallY = 1 ; dy = -dy ; } /* if the ball is in the area of the bricks */ if ( BallY < max ) { /* if there is no brick at the top of the ball */ if ( bri[layer][ ( BallX + 10 ) / 32 ] 800 - ( ( 4 - chance ) * 10 ) ) { outtextxy ( MidX, MidY, "Winner !!" ) ; if ( score < 800 ) outtextxy ( MidX, MidY + 30, "Try to score 800" ) ; else outtextxy ( MidX, MidY + 30, " GREAT!" ) ; closegraph() ; restorecrtmode() ; exit ( 0 ) ; } /* introduce delay for few seconds */ delay ( speed ) ; /* put the image of the paddle at the old coordinates */ putimage ( Base1, Base2, m2, OR_PUT ) ; /* erase the image of the paddle at the old coordinates */ putimage ( Base1, Base2, m2, XOR_PUT ) ; MousePos(&Base1); /* if paddle goes beyond left boundary */ if ( Base1 < 1 ) Base1 = 1 ; /* if paddle goes beyond right boundary */ if ( Base1 > 588 ) Base1 = 588 ; /* put the image of the paddle at the proper position */ putimage ( Base1, Base2, m2, XOR_PUT ) ; } closegraph(); /* Return the system to text mode */ return(0); } bricks() { int i, j, lx = 0, ly = 0 ; for ( i = 0 ; i < 5 ; i++ ) /* 5 rows */ { for ( j = 0 ; j < 20 ; j++ ) /* 20 columns */ { rectangle ( lx, ly, lx + 20, ly + 7 ) ; floodfill ( lx + 1, ly + 1, 2 ) ; lx = lx + 32 ; } lx = 0 ; ly = ly + 10 ; } } delbrick ( int b, int l ) { /* b - brick number, l - layer */ setcolor ( BLACK ) ; rectangle ( b * 32, l * 10, ( b * 32 ) + 20 , ( l * 10 ) + 7 ) ; rectangle ( b * 32 + 1, l * 10, ( b * 32 ) + 20 - 1, ( l * 10 ) + 7 -1 ); rectangle ( b * 32 + 2, l * 10, ( b * 32 ) + 20 - 2, ( l * 10 ) + 7 -2 ); rectangle ( b * 32 + 3, l * 10, ( b * 32 ) + 20 - 3, ( l * 10 ) + 7 -3 ); rectangle ( b * 32 + 4, l * 10, ( b * 32 ) + 20 - 4, ( l * 10 ) + 7 -4 ); rectangle ( b * 32 + 5, l * 10, ( b * 32 ) + 20 - 5, ( l * 10 ) + 7 -5 ); rectangle ( b * 32 + 6, l * 10, ( b * 32 ) + 20 - 6, ( l * 10 ) + 7 -6 ); setcolor ( CGA_YELLOW ) ; }
Far Pointer is a pointer that is stored using four bytes (32 bits). The bytes are stored little endian or low to high order. A far pointer can access objects up to 16K in size in any memory area. Objects larger than 16K must be accessed using huge pointers This book is basic for c , download and Read this... must required !
TGC ATC CGA AGT CGA
The complementary DNA strand would be AGC CTG GTA GCT. In DNA, adenine pairs with thymine and cytosine pairs with guanine. Therefore, the complementary strand is formed by replacing each base with its complementary base.
The complementary strand of the DNA is TAA-GCT-ACG
the DNA strand GTT ACC would be transcribed to CAA UGG.
GCT AAG would produce the strand of mRNA of "CGA UUC" CGU AAU UGA CUG
The sequence of nucleotides of the complementary strand will be the nucleotides which bind to the nucleotides of the template. In DNA, adenine binds to thymine and cytosine binds to guanine. The complementary strand will therefore have an adenine where the template strand has a thymine, a guanine where the template has a cytosine, etc. For example: If the template strand is ATG-GGC-CTA-GCT Then the complementary strand would be TAC-CCG-GAT-CGA
The complementary strand of DNA for the sequence AGTT would be TCAA. In DNA, adenine pairs with thymine and guanine pairs with cytosine. So the complementary base for A is T, G is C, T is A, and T is A.
AGTCG (I'm assuming your strand was written in the normal 5' to 3' order, and I wrote mine in that order as well, which means the last residue in my strand pairs with the first residue in your strand, and vice versa).
If cga ct were used as a template strand for complementary DNA synthesis, the complementary DNA produced would be gct ga. This is because each nucleotide pairs with its complementary base: cytosine (c) pairs with guanine (g), guanine (g) pairs with cytosine (c), adenine (a) pairs with thymine (t), and thymine (t) pairs with adenine (a). Therefore, the complementary sequence would read from 5' to 3' as gct ga.
The DNA strand CAT-TAG would produce a complementary mRNA strand of GUA-AUC.
Aca tag gct aat gct aat cgt gca cga tct gaa cgatgt atc cga tta cga tta gca cgt gct aga ctt gct
it depends on the codon spcified. The tRNA will have the complementary strand along with an amino acid, for which is specified by the mRNA. if the mRNA codon was "CGA" the tRNA codon would have an amino acid and the complementary codon of "GCU"