Chapter 8, Problem 36P

The human gene for ß2 lens crystallin has the components listed below. The numbers represent nucleotide pairs that make up the particular component. Assume for simplicity that no alternative splicing is involved.

5′ UTR	174
1st exon	119
1st intron	532
2nd exon	337
2nd intron	1431
3rd exon	208
3rd intron	380
4th exon	444
4th intron	99
5th exon	546
3′ UTR	715

Answer the following questions about the ß2 lens crystallin gene, primary transcript, and gene product. Questions asking where should be answered with one of the 11 components from the list or with None. Assume poly-A tails contain 150 As.

a	How large is the ß2 lens crystallin gene in bp (base pairs)?
b.	How large is the primary transcript for ß2 lens crystallin in bases?
c.	How large is the mature mRNA for ß2 lens crystallin in bases?
d.	Where would you find the base pairs encoding the initiation codon?
e.	Where would you find the base pairs encoding the stop codon?
f.	Where would you find the base pairs encoding the 5′ cap?
g.	Where would you find the base pairs that constitute the promoter?
h.	Which intron interrupts the 3′ UTR?
i.	Where would you find the sequences encoding the C terminus?
j.	Where would you find the sequence encoding the poly-A tail?
k.	How large is the coding region of the gene in bp (base pairs)?
l.	How many amino acids are in the ß2 lens crystallin protein?
m.	Which intron interrupts a codon?
n.	Which intron is located between codons?
o.	Where would you be likely to find the site specifying poly-A addition? You find in lens-forming cells from several different people small amounts of a polypeptide that has the same N terminus as the normal ß lens crystallin, but it has a different C terminus. The polypeptide is 114 amino acids long, of which 94 are shared with the normal protein and 20 are unrelated junk. No mutation is involved in the production of this 114-amino acid protein.
p.	Outline a hypothesis for the process that would produce this protein. Your hypothesis should explain why 94 amino acids are the same as in the normal ß2 lens crystallin.
q.	Explain why you would expect that a polypeptide such as the 114-mer described above would on average have 20 amino acids of junk

Summary Introduction

a.

To determine:

The length of the human gene for β₂ lens crystallin.

Introduction:

The length of a gene is determined by the total number of base sequences present in the given gene region. Genes consist of coding as well as non-coding regions. Splicing helps in joining the coding sequences for the expression of the gene.

Explanation of Solution

The length of the β₂ lens crystalline gene is calculated by adding the total number of nucleotides present in the exons and introns of the gene.

$\begin{array}{rll}\text{Total}\text{gene}\text{length}& =& 119+532+337+1431+208+380+444+99+546\\ & =& 4096\text{bp}\end{array}$

Summary Introduction

b.

To determine:

The length of the primary transcript of β₂ lens crystalline gene.

Introduction:

Exons are the coding regions present on a gene and introns are the non coding regions. Transcription of the entire gene results into pre-mRNA which is also called primary transcript.

Explanation of Solution

The length of primary transcript of β₂ lens crystalline gene can be calculated by adding the total number of nucleotides present in the exons and introns of the gene.

$\begin{array}{rll}\text{Total}\text{gene}\text{length}& =& 119+532+337+1431+208+380+444+99+546\\ & =& 4096\text{bp}\end{array}$

Summary Introduction

c.

To determine:

The length of mature mRNA of β₂ lens crystalline gene.

Introduction:

Mature mRNA is formed after the splicing of the introns and joining the exons of the gene. In addition to splicing, mature mRNA is also associated with 5' cap and poly-A tail.

Explanation of Solution

The length of the mature mRNA of β₂ lens crystallin gene is calculated by adding all the exons along with the 5' cap and poly-A tail.

$\begin{array}{rll}\text{Length}\text{of}\text{mature}\text{mRNA}& =& 119+337+208+444+546+150\\ & =& 1804\text{bp}\end{array}$

Summary Introduction

d.

To determine:

The position of the base pair which encodes for initiation codon.

Introduction:

Initiation codon is called start codon which comprises AUG base pair in eukaryotes. The start codon codes for the amino acid methionine

Explanation of Solution

The base pairs which encode for the initiating codon can be found on the second exon with 337 base pairs.

Summary Introduction

e.

To determine:

The position of the base pair which encodes for stop codon.

Introduction:

Stop codon present within a gene encodes for the termination of the translation process. There are three stop codons present in eukaryotic cell. These are UAG, UAA, and UGA.

Explanation of Solution

The base pairs which encode for the stop codon can be found on the fourth exon with 444 base pairs.

Summary Introduction

f.

To determine:

The position of the base pair which encodes for 5' cap.

Introduction:

5' cap comprises altered nucleotides which are present on the 5' end of the mRNA. 5' cap protects the nascent mRNA from degradation.

Explanation of Solution

None of the base pairs encodes for 5' cap because the ends of the pre-mRNA are modified into a group of nucleotides through some chemicals. There are no base pairs which encodes for these nucleotides.

Summary Introduction

g.

To determine:

The position of the base pair which comprises the promoter.

Introduction:

The promoter region is present on the DNA, which specifies the initiation of transcription. These are located upstream on the DNA towards the 5' end of the sense strand.

Explanation of Solution

None of the base pairs encodes for promoter site because the promoter site is present in the DNA to initiate the process of transcription. Promoter site is not present in case of primary transcript or mature mRNA.

Summary Introduction

h.

To determine:

The intron which interrupts the 3' UTR.

Introduction:

The three prime UTR stands for the untranslated region, which is a part of mRNA present after the termination codon. These regions contain some regulatory regions which influence gene expression.

Explanation of Solution

The 3' UTR is found immediately after the stop codon is translated. In the given gene, the stop codon is the 4^th exon. Therefore, the 3' UTR must be present after that, but the fourth intron present after the stop codon interrupts the 3' UTR.

Summary Introduction

i.

To determine:

The position of the sequence which encodes for C terminus.

Introduction:

Protein structure consists of two terminals, the N terminus, which consists of a free amino group and C terminus, which consists of a carboxyl group (COOH).

Explanation of Solution

The C terminus of the polypeptide chain is present at the 3' end of the amino acid chain. The codon which is present near the 3' end of the mRNA will code for C terminus of the polypeptide chain. In the given gene, the 4^th exon is nearest to the 3' end. Hence it will code for C terminus.

Summary Introduction

j.

To determine:

The position of the sequence which encodes for poly-A tail.

Introduction:

Poly-A tail of a messenger RNA is made of a stretch of adenine base which is attached at the 3' end of the mRNA. The poly-A tail helps in protection of mRNA from degradation and also helps in the termination of translation.

Explanation of Solution

None of the base pairs encodes for poly-A tail of the mRNA because poly-A tail is synthesized by a series of reactions called polyadenylation. The adenine bases are synthesized by polyadenylate polymerase at the 3' end of the mRNA; they are not encoded by base pairs.

Summary Introduction

k.

To determine:

The length of the coding region of the gene.

Introduction:

Coding region consisting of nucleotide bases as a part of a gene that codes for protein. The coding region on DNA or RNA begins from start codon and ends at the stop codon.

Explanation of Solution

The coding region in the given gene can be calculated by adding the base pairs in the exons starting from start codon to stop codon.

$\begin{array}{rll}\text{Length}\text{of}\text{coding}\text{region}& =& 337+208+444\\ & =& 989\text{bp}\end{array}$

Summary Introduction

l.

To determine:

The number of amino acids present in the β₂ lens crystalline protein.

Introduction:

Three DNA or RNA nucleotides in a sequence form a codon. Each codon corresponds with an amino acid during the translation process. There are 64 total codons present in the genome of the eukaryotes. Out of these, 61 codons codes for the amino acids.

Explanation of Solution

Three base-pair combines to form a codon which corresponds to a particular amino acid. In the case of β₂ lens crystalline gene, the length of the coding sequence on the gene is 989 bp. Therefore the number of amino acids can be calculated by dividing the total number by 3.

$\begin{array}{rll}\text{Number}\text{of}\text{amino}\text{acid}& =& \frac{989}{3}\\ & =& 329\end{array}$

Summary Introduction

m.

To determine:

The intron which interrupts a codon.

Introduction:

Introns are considered as intervening sequences that are present on a gene. These sequences are removed from the mature mRNA because they are considered as non-coding regions of a gene.

Explanation of Solution

The coding region of a gene is considered from the start codon to stop codon. In the β₂ lens crystalline gene, the 4^th exon is the stop codon, so the intron before the 4^th exon interrupts the codon. Therefore, the 3^rd intron interrupts the codon of the β₂ lens crystalline gene.

Summary Introduction

n.

To determine:

The intron which is located between a codon.

Introduction:

Introns and exons are the nucleotide sequences present on the DNA, whereas codon is the feature of RNA. Reading of codon is based on the genetic code or the RNA code.

Explanation of Solution

Codons on mRNA are equivalent to the exons on a DNA. In the β₂ lens crystalline gene, the codons are the 2^nd exon which is start codon, and 4^th exon is the stop codon. The protein-coding codons are 2^nd and 3^rd. Therefore, the intron which is present between these codons is 2^nd intron.

Summary Introduction

o.

To determine:

The region where site specifying poly-A-addition is present.

Introduction:

Poly-A tail of a messenger RNA is made of a stretch of adenine base which is attached at the 3' end of the mRNA. The poly-A tail helps in protection of mRNA from degradation and also helps in the termination of translation.

Explanation of Solution

Site specifying poly-A-addition can be found in the region of 5^th exon and 3' UTR because poly-A tail is synthesized at the 3' end of the mRNA. The region nearest to the 3' end is 3' UTR and 5^th exon.

Summary Introduction

p.

To determine:

A hypothesis for the process of production of protein.

Introduction:

There are 64 total codons present in the genome of the eukaryotes. Out of these, 61 codons codes for the amino acids. A codon is formed from three DNA or RNA nucleotides in a sequence. Each codon corresponds with an amino acid during the translation process.

Explanation of Solution

Alternate splicing is a process during gene expression in which a single gene code for multiple proteins. This condition arises when a particular exon is either included or excluded from the final mature mRNA. It is given that the polypeptide obtained from β₂ lens crystalline cell collected from different people had the same N terminus but different C terminus. 94 amino acids were shared with the normal proteins, and 20 were junk. This is because of alternate splicing which could occur if the 2^nd exon is connected either to the 4^th or 5^th exon.

Summary Introduction

q.

To determine:

The explanation for the presence of 20 junk amino acids in a polypeptide of 114 amino acids.

Introduction:

The non-functional or the unfolded proteins are considered as junk proteins, and the amino acids which code for these proteins are called junk amino acids.

Explanation of Solution

In the genetic code, approx 1/20 codons are the stop codons, which cause termination of the translation process. Therefore, approximately 1/20 triplets will be stop codons in the case of a random DNA sequence. In the given case, out of the total coding sequence, the 4^th exon is stopped codon so; it is out of the reading frame with respect to the 2^nd exon. The 5^th exon of the β₂ lens crystalline gene does not code for any protein. This results in improper processing of the mRNA and 20 junk amino acids are expected to be produced.