Chapter 21, Problem 21.105UTC

(a)

Interpretation Introduction

Interpretation:

The bases in the parent and new strands should be completed.

Parent strand

New stand

Concept introduction:

Adenine and thymine are complementary base, cytosine and guanine are the complementary base.

In the synthesis of mRNA, cytosine(C) pairs with guanine (G), thymine (T), pairs with adenine (A), adenine pairs with uracil (U) and guanine (G) with cytosine(C).

C $\cdots \cdots$ G, T $\cdots \cdots$ A, A $\cdots \cdots$ U, G $\cdots \cdots$ C

First letter	Second letter	Third letter
	u	C	A	G
U	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}U& U& U\end{array}\\ \begin{array}{ccc}U& U& C\end{array}\end{array}\}\text{Phe}\left(F\right)\\ \begin{array}{l}\begin{array}{ccc}U& U& A\end{array}\\ \begin{array}{ccc}U& U& G\end{array}\end{array}\}\text{Leu}\left(1\right)\end{array}$	$\begin{array}{l}\begin{array}{ccc}U& C& U\\ U& C& C\\ U& C& A\end{array}\\ \begin{array}{ccc}U& C& G\end{array}\end{array}\}\text{Ser}\left(S\right)$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}U& A& U\end{array}\\ \begin{array}{ccc}U& A& C\end{array}\end{array}\}Tyr\\ \begin{array}{l}\begin{array}{ccc}U& A& A\end{array}\\ \begin{array}{ccc}U& A& G\end{array}\end{array}\}Stop\end{array}$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}U& G& U\end{array}\\ \begin{array}{ccc}U& G& C\end{array}\end{array}\}Cys\\ \begin{array}{ccc}U& G& A\end{array}\text{ Stop}\\ \begin{array}{ccc}U& G& G\end{array}\text{ Trp}\end{array}$	U C A G
C	$\begin{array}{l}\begin{array}{ccc}C& U& U\\ C& U& C\\ C& U& A\end{array}\\ \begin{array}{ccc}C& U& G\end{array}\end{array}\}Leu\left(L\right)$	$\begin{array}{l}\begin{array}{ccc}C& C& U\\ C& C& C\\ C& C& A\end{array}\\ \begin{array}{ccc}C& C& G\end{array}\end{array}\}\mathrm{Pr}o\left(P\right)$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}C& A& U\end{array}\\ \begin{array}{ccc}C& A& C\end{array}\end{array}\}His\left(H\right)\\ \begin{array}{l}\begin{array}{ccc}C& G& A\end{array}\\ \begin{array}{ccc}C& A& G\end{array}\end{array}\}G\ln \left(Q\right)\end{array}$	$\begin{array}{l}\begin{array}{ccc}C& G& U\\ C& G& C\\ C& G& A\end{array}\\ \begin{array}{ccc}C& G& G\end{array}\end{array}\}Arg$	U C A G
A	$\begin{array}{l}\begin{array}{ccc}A& U& U\\ A& U& C\\ A& U& A\end{array}\}Lle\left(l\right)\\ \begin{array}{ccc}A& U& G\end{array}\end{array}$	$\begin{array}{l}\begin{array}{ccc}A& C& U\\ A& C& C\\ A& C& A\end{array}\\ \begin{array}{ccc}A& C& G\end{array}\end{array}\}Thr\left(T\right)$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}A& A& U\end{array}\\ \begin{array}{ccc}A& A& C\end{array}\end{array}\}Asn\left(N\right)\\ \begin{array}{l}\begin{array}{ccc}A& A& A\end{array}\\ \begin{array}{ccc}A& A& G\end{array}\end{array}\}Lys\left(k\right)\end{array}$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}A& G& U\end{array}\\ \begin{array}{ccc}A& G& C\end{array}\end{array}\}Ser\\ \begin{array}{l}\begin{array}{ccc}A& G& A\end{array}\\ \begin{array}{ccc}A& G& G\end{array}\end{array}\}Arg\end{array}$	U C A G
G	$\begin{array}{l}\begin{array}{ccc}G& U& U\\ G& U& C\\ G& U& A\end{array}\\ \begin{array}{ccc}G& U& G\end{array}\end{array}\}Val\left(V\right)$	$\begin{array}{l}\begin{array}{ccc}G& C& U\\ G& C& C\\ G& C& A\end{array}\\ \begin{array}{ccc}G& C& G\end{array}\end{array}\}Ala\left(A\right)$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}G& A& U\end{array}\\ \begin{array}{ccc}G& A& C\end{array}\end{array}\}Asp\left(D\right)\\ \begin{array}{l}\begin{array}{ccc}G& A& A\end{array}\\ \begin{array}{ccc}G& A& G\end{array}\end{array}\}Glu\left(E\right)\end{array}$	$\begin{array}{l}\begin{array}{ccc}G& G& U\end{array}\\ \begin{array}{ccc}G& G& C\end{array}\\ \begin{array}{ccc}G& G& A\end{array}\\ \begin{array}{ccc}G& G& G\end{array}\end{array}\}Glu\left(G\right)$	U C A G

Answer to Problem 21.105UTC

Solution: The base in the parent and new strands

Explanation of Solution

In DNA replication, the strands in the parent DNA and new strands are complementary to each other.

Adenine and thymine are complementary base, cytosine and guanine are the complementary base. During formation of the complementary base, thymine (T) pairs only with adenine (A) cytosine(C) pairs only with guanine (G).

Thus, the bases for the parent and new strands will be

(b)

Interpretation Introduction

Interpretation:

Using the new strand as a template, write the mRNA sequence

Concept introduction:

Adenine and thymine are complementary base, cytosine and guanine are the complementary base.

In the synthesis of mRNA, cytosine(C) pairs with guanine (G), thymine (T), pairs with adenine (A), adenine pairs with uracil (U) and guanine (G) with cytosine(C).

C $\cdots \cdots$ G, T $\cdots \cdots$ A, A $\cdots \cdots$ U, G $\cdots \cdots$ C

First letter	Second letter	Third letter
	u	C	A	G
U	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}U& U& U\end{array}\\ \begin{array}{ccc}U& U& C\end{array}\end{array}\}\text{Phe}\left(F\right)\\ \begin{array}{l}\begin{array}{ccc}U& U& A\end{array}\\ \begin{array}{ccc}U& U& G\end{array}\end{array}\}\text{Leu}\left(1\right)\end{array}$	$\begin{array}{l}\begin{array}{ccc}U& C& U\\ U& C& C\\ U& C& A\end{array}\\ \begin{array}{ccc}U& C& G\end{array}\end{array}\}\text{Ser}\left(S\right)$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}U& A& U\end{array}\\ \begin{array}{ccc}U& A& C\end{array}\end{array}\}Tyr\\ \begin{array}{l}\begin{array}{ccc}U& A& A\end{array}\\ \begin{array}{ccc}U& A& G\end{array}\end{array}\}Stop\end{array}$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}U& G& U\end{array}\\ \begin{array}{ccc}U& G& C\end{array}\end{array}\}Cys\\ \begin{array}{ccc}U& G& A\end{array}\text{ Stop}\\ \begin{array}{ccc}U& G& G\end{array}\text{ Trp}\end{array}$	U C A G
C	$\begin{array}{l}\begin{array}{ccc}C& U& U\\ C& U& C\\ C& U& A\end{array}\\ \begin{array}{ccc}C& U& G\end{array}\end{array}\}Leu\left(L\right)$	$\begin{array}{l}\begin{array}{ccc}C& C& U\\ C& C& C\\ C& C& A\end{array}\\ \begin{array}{ccc}C& C& G\end{array}\end{array}\}\mathrm{Pr}o\left(P\right)$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}C& A& U\end{array}\\ \begin{array}{ccc}C& A& C\end{array}\end{array}\}His\left(H\right)\\ \begin{array}{l}\begin{array}{ccc}C& G& A\end{array}\\ \begin{array}{ccc}C& A& G\end{array}\end{array}\}G\ln \left(Q\right)\end{array}$	$\begin{array}{l}\begin{array}{ccc}C& G& U\\ C& G& C\\ C& G& A\end{array}\\ \begin{array}{ccc}C& G& G\end{array}\end{array}\}Arg$	U C A G
A	$\begin{array}{l}\begin{array}{ccc}A& U& U\\ A& U& C\\ A& U& A\end{array}\}Lle\left(l\right)\\ \begin{array}{ccc}A& U& G\end{array}\end{array}$	$\begin{array}{l}\begin{array}{ccc}A& C& U\\ A& C& C\\ A& C& A\end{array}\\ \begin{array}{ccc}A& C& G\end{array}\end{array}\}Thr\left(T\right)$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}A& A& U\end{array}\\ \begin{array}{ccc}A& A& C\end{array}\end{array}\}Asn\left(N\right)\\ \begin{array}{l}\begin{array}{ccc}A& A& A\end{array}\\ \begin{array}{ccc}A& A& G\end{array}\end{array}\}Lys\left(k\right)\end{array}$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}A& G& U\end{array}\\ \begin{array}{ccc}A& G& C\end{array}\end{array}\}Ser\\ \begin{array}{l}\begin{array}{ccc}A& G& A\end{array}\\ \begin{array}{ccc}A& G& G\end{array}\end{array}\}Arg\end{array}$	U C A G
G	$\begin{array}{l}\begin{array}{ccc}G& U& U\\ G& U& C\\ G& U& A\end{array}\\ \begin{array}{ccc}G& U& G\end{array}\end{array}\}Val\left(V\right)$	$\begin{array}{l}\begin{array}{ccc}G& C& U\\ G& C& C\\ G& C& A\end{array}\\ \begin{array}{ccc}G& C& G\end{array}\end{array}\}Ala\left(A\right)$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}G& A& U\end{array}\\ \begin{array}{ccc}G& A& C\end{array}\end{array}\}Asp\left(D\right)\\ \begin{array}{l}\begin{array}{ccc}G& A& A\end{array}\\ \begin{array}{ccc}G& A& G\end{array}\end{array}\}Glu\left(E\right)\end{array}$	$\begin{array}{l}\begin{array}{ccc}G& G& U\end{array}\\ \begin{array}{ccc}G& G& C\end{array}\\ \begin{array}{ccc}G& G& A\end{array}\\ \begin{array}{ccc}G& G& G\end{array}\end{array}\}Glu\left(G\right)$	U C A G

Answer to Problem 21.105UTC

Solution: mRNA sequence

Explanation of Solution

In the synthesis of mRNA, cytosine(C) pairs with guanine (G), thymine (T) pairs with adenine (A) and adenine pairs up with uracil.

Thus, the complementary base sequence in mRNA is

(c)

Interpretation Introduction

Interpretation:

Write the three-letter symbols for the amino acids that would go into the peptide from the mRNA you wrote in part (b)

Concept introduction:

Adenine and thymine are complementary base, cytosine and guanine are the complementary base.

In the synthesis of mRNA, cytosine(C) pairs with guanine (G), thymine (T), pairs with adenine (A), adenine pairs with uracil (U) and guanine (G) with cytosine(C).

C $\cdots \cdots$ G, T $\cdots \cdots$ A, A $\cdots \cdots$ U, G $\cdots \cdots$ C

First letter	Second letter	Third letter
	u	C	A	G
U	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}U& U& U\end{array}\\ \begin{array}{ccc}U& U& C\end{array}\end{array}\}\text{Phe}\left(F\right)\\ \begin{array}{l}\begin{array}{ccc}U& U& A\end{array}\\ \begin{array}{ccc}U& U& G\end{array}\end{array}\}\text{Leu}\left(1\right)\end{array}$	$\begin{array}{l}\begin{array}{ccc}U& C& U\\ U& C& C\\ U& C& A\end{array}\\ \begin{array}{ccc}U& C& G\end{array}\end{array}\}\text{Ser}\left(S\right)$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}U& A& U\end{array}\\ \begin{array}{ccc}U& A& C\end{array}\end{array}\}Tyr\\ \begin{array}{l}\begin{array}{ccc}U& A& A\end{array}\\ \begin{array}{ccc}U& A& G\end{array}\end{array}\}Stop\end{array}$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}U& G& U\end{array}\\ \begin{array}{ccc}U& G& C\end{array}\end{array}\}Cys\\ \begin{array}{ccc}U& G& A\end{array}\text{ Stop}\\ \begin{array}{ccc}U& G& G\end{array}\text{ Trp}\end{array}$	U C A G
C	$\begin{array}{l}\begin{array}{ccc}C& U& U\\ C& U& C\\ C& U& A\end{array}\\ \begin{array}{ccc}C& U& G\end{array}\end{array}\}Leu\left(L\right)$	$\begin{array}{l}\begin{array}{ccc}C& C& U\\ C& C& C\\ C& C& A\end{array}\\ \begin{array}{ccc}C& C& G\end{array}\end{array}\}\mathrm{Pr}o\left(P\right)$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}C& A& U\end{array}\\ \begin{array}{ccc}C& A& C\end{array}\end{array}\}His\left(H\right)\\ \begin{array}{l}\begin{array}{ccc}C& G& A\end{array}\\ \begin{array}{ccc}C& A& G\end{array}\end{array}\}G\ln \left(Q\right)\end{array}$	$\begin{array}{l}\begin{array}{ccc}C& G& U\\ C& G& C\\ C& G& A\end{array}\\ \begin{array}{ccc}C& G& G\end{array}\end{array}\}Arg$	U C A G
A	$\begin{array}{l}\begin{array}{ccc}A& U& U\\ A& U& C\\ A& U& A\end{array}\}Lle\left(l\right)\\ \begin{array}{ccc}A& U& G\end{array}\end{array}$	$\begin{array}{l}\begin{array}{ccc}A& C& U\\ A& C& C\\ A& C& A\end{array}\\ \begin{array}{ccc}A& C& G\end{array}\end{array}\}Thr\left(T\right)$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}A& A& U\end{array}\\ \begin{array}{ccc}A& A& C\end{array}\end{array}\}Asn\left(N\right)\\ \begin{array}{l}\begin{array}{ccc}A& A& A\end{array}\\ \begin{array}{ccc}A& A& G\end{array}\end{array}\}Lys\left(k\right)\end{array}$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}A& G& U\end{array}\\ \begin{array}{ccc}A& G& C\end{array}\end{array}\}Ser\\ \begin{array}{l}\begin{array}{ccc}A& G& A\end{array}\\ \begin{array}{ccc}A& G& G\end{array}\end{array}\}Arg\end{array}$	U C A G
G	$\begin{array}{l}\begin{array}{ccc}G& U& U\\ G& U& C\\ G& U& A\end{array}\\ \begin{array}{ccc}G& U& G\end{array}\end{array}\}Val\left(V\right)$	$\begin{array}{l}\begin{array}{ccc}G& C& U\\ G& C& C\\ G& C& A\end{array}\\ \begin{array}{ccc}G& C& G\end{array}\end{array}\}Ala\left(A\right)$	$\begin{array}{l}\begin{array}{l}\begin{array}{ccc}G& A& U\end{array}\\ \begin{array}{ccc}G& A& C\end{array}\end{array}\}Asp\left(D\right)\\ \begin{array}{l}\begin{array}{ccc}G& A& A\end{array}\\ \begin{array}{ccc}G& A& G\end{array}\end{array}\}Glu\left(E\right)\end{array}$	$\begin{array}{l}\begin{array}{ccc}G& G& U\end{array}\\ \begin{array}{ccc}G& G& C\end{array}\\ \begin{array}{ccc}G& G& A\end{array}\\ \begin{array}{ccc}G& G& G\end{array}\end{array}\}Glu\left(G\right)$	U C A G

Answer to Problem 21.105UTC

Solution: Three letter symbols for the amino acids that would go into peptide from the mRNA would be

Explanation of Solution

Genetic code consists of a series of three nucleotides called codons. Thus, a triplet of mRNA contains the codons AGG, UCG, and CCU.

AGG is decoded for arginine amine amino acid (Arg). UCG is decoded for serine (Ser)

CCU is decoded for proline (Pro).

Therefore, three letter symbols for the amino acids that would go into the peptide is