Food Molecule Quiz

1. What type of food molecule is this?

\setcrambond{2pt}{}{} \chemfig[,1.4]{ C?[a](-[2,0.4])(-[6,0.5,,2]HO) <[7,1]C(-[6,0.4])(-[2,0.5]OH) -[,,,,line width=2pt]C(-[2,.4])(-[6,.5]OH) >[1,1]C(-[6,.4]) (-[3,1]O-[4]C?[a](-[2,0.8]C(-[,.5]OH)(-[4,0.4])-[2,0.4])(-[6,0.4])) -[,0.7]O -[,.7]C?[z](-[2,0.4]) <[7,1]C(-[6,0.4])(-[2,0.5]OH) -[,,,,line width=2pt]C(-[2,.4])(-[6,.5]OH) >[1,1]C(-[6,.4]) (-[3,1]O-[4]C?[z](-[2,0.8]C(-[,.5]OH)(-[4,0.4])-[2,0.4])(-[6,0.4])) -[2,0.5]OH }
 
 
 
 
 

Question 1 of 12

2. What type of food molecule is this?

\setcrambond{2pt}{}{} \chemfig{HO-[2,0.5,2]?<[7,0.7](-[2,0.5]OH)-[,,,, line width=2pt](-[6,0.5]OH)>[1,0.7](-[6,0.5]OH)-[3,0.7] O-[4]?(-[2,0.3]-[3,0.5]OH)}
 
 
 
 
 

Question 2 of 12

3. What type of food molecule is this?

\definesubmol{a}{-C(-[2,.4])(-[6,.4])} \definesubmol{d}{-C(-[6,.4])=C(-[6,.4])} \definesubmol{e}{-C(-[2,.4])=C(-[2,.4])} \definesubmol{s}{-O-C(=[6]O)} \definesubmol{FA}{!s!a!a!e!a!a!d!a!a!e!a!a-[,.4]} \chemfig{[,.5]C(-[4,.4])(-[2,.4])(!{FA}) -[6,1.2]C(-[4,.4])(!{FA}) -[6,1.2]C(-[4,.4])(-[6,.4])(!{FA})}
 
 
 
 
 

Question 3 of 12

4. What type of food molecule is this?

\definesubmol{a}{-[:30]-} \definesubmol{b}{-[:30]=[0]-} \chemfig{[:-30,0.6]HO-(=[6]O)!a!a!b!a!a!a!b!a}
 
 
 
 
 

Question 4 of 12

5. What type of food molecule is this?

\definesubmol{a}{-C(-[2,.4])(-[6,.4])} \definesubmol{d}{-C(-[6,.4])=C(-[6,.4])} \definesubmol{u}{-C(-[2,.4])=C(-[2,.4])} \chemfig{[4,.6]C(=[7]O)(-[1,.6]OH)!a!a!a!d!a!u!a!a!d!a!a-[,.4]}
 
 
 
 
 

Question 5 of 12

6. What type of food molecule is this?

\definesubmol{a}{-C(-[2,.4])(-[6,.4])} \definesubmol{e}{-C(-[2,.4])=C(-[2,.4])} \definesubmol{s}{-O-C(=[2]O)} \definesubmol{FA}{!s!a!a!a!a!e!a!a!a!a!a-[,.4]} \chemfig{[,.5]C(-[4,.4])(-[2,.4])(!{FA}) -[6,1.2]C(-[4,.4])(!{FA}) -[6,1.2]C(-[4,.4])(-[6,.4])(!{FA})}
 
 
 
 
 

Question 6 of 12

7. What type of food molecule is this?

\definesubmol{n}{-[,.7]N(-[2]H)-[,.7]C(-[2]H)} \definesubmol{a}{-[,.7]C(=[2]O)} \definesubmol{gly}{!n([6]-H)!a} \chemfig{[,0.6]H!{gly}-OH}
 
 
 
 
 

Question 7 of 12

8. What type of food molecule is this?

\setcrambond{2pt}{}{} \chemfig[,1.4]{ C?[a](-[2,0.4])(-[6,0.5,,2]HO) <[7,1]C(-[6,0.4])(-[2,0.5]OH) -[,,,,line width=2pt]C(-[2,.4])(-[6,.5]OH) >[1,1]C(-[6,.4]) (-[3,1]O-[4]C?[a](-[2,0.8]C(-[,.5]OH)(-[4,0.4])-[2,0.4])(-[6,0.4])) -[,0.7]O -[,.7]C?[b](-[2,0.4]) <[7,1]C(-[6,0.4])(-[2,0.5]OH) -[,,,,line width=2pt]C(-[2,.4])(-[6,.5]OH) >[1,1]C(-[6,.4]) (-[3,1]O-[4]C?[b](-[2,0.8]C(-[,.5]OH)(-[4,0.4])-[2,0.4])(-[6,0.4])) -[,0.7]O -[,.7]C?[z](-[2,0.4]) <[7,1]C(-[6,0.4])(-[2,0.5]OH) -[,,,,line width=2pt]C(-[2,.4])(-[6,.5]OH) >[1,1]C(-[6,.4]) (-[3,1]O-[4]C?[z](-[2,0.8]C(-[,.5]OH)(-[4,0.4])-[2,0.4])(-[6,0.4])) -[2,0.5]OH }
 
 
 
 
 

Question 8 of 12

9. What type of food molecule is this?

\definesubmol{n}{-[,.7]N(-[2]H)-[,.7]C(-[2]H)} \definesubmol{a}{-[,.7]C(=[2]O)} \definesubmol{phe}{!n([6]-CH_2-*6(=-=-=-))!a} \chemfig{[,0.6]H!{phe}-OH}
 
 
 
 
 

Question 9 of 12

10. What type of food molecule is this?

\definesubmol{a}{-[:30]-} \definesubmol{b}{-[:30]=[0]-} \chemfig{[:-30,0.6]HO-C(=[6]O)!a!a!a!a!a!a!a!a!a}
 
 
 
 
 

Question 10 of 12

11. What type of food molecule is this?

\definesubmol{a}{-C(-[2,.4])(-[6,.4])} \definesubmol{e}{-C(-[2,.4])=C(-[2,.4])} \definesubmol{s}{-O-C(=[6]O)} \definesubmol{FA}{!s!a!a!a!a!a!e!a!a!e!a!a-[,.4]} \chemfig{[4,.5]C(-[0,.4])(-[2,.4])(!{FA}) -[6,1.2]C(-[0,.4])(!{FA}) -[6,1.2]C(-[0,.4])(-[6,.4])(!{FA})}
 
 
 
 
 

Question 11 of 12

12. What type of food molecule is this?

\definesubmol{n}{-[,.7]N(-[2]H)-[,.7]C(-[2]H)} \definesubmol{a}{-[,.7]C(=[2]O)} \definesubmol{gly}{!n([6]-H)!a} \definesubmol{ala}{!n(-[6]CH_3)!a} \definesubmol{val}{!n([6]-CH(-[5,.7,,1]CH_3)-[7,.7,1]CH_3)!a} \definesubmol{leu}{!n([6,.8]-CH_2-CH(-[5,,,1]CH_3)-[7,,1]CH_3)!a} \definesubmol{ile}{!n([6]-CH(-[7,0.7,1,1]CH_3)-[5,.7,1,1]CH_2-CH_3)!a} \definesubmol{ser}{!n([6,.6]-CH_2-OH)!a} \definesubmol{cys}{!n([6,.6]-CH_2-SH)!a} \definesubmol{thr}{!n([6]-CH(-[5,.7,,1]CH_3)-[7,.7,1]OH)!a} \definesubmol{asp}{!n([6]-CH_2-C(-[7]OH)=[5]O)!a} \chemfig{[,0.6]H!{ala}!{gly}!{asp}!{ser}-OH}
 
 
 
 
 

Question 12 of 12