Genetics

Question 1

DNA

Answer 1

Deoxyribose nucleic acid, stores genetic information. Provides instructions for the development and function of cells

Question 2

Nucleotide

Answer 2

A nitrogenous base attached to a sugar which is attached to a phosphate group. The sugar and phosphate form a phosphate backbone, sugar is deoxyribose. DNA is negative because of the phosphate.

Question 3

Nitrogenous bases

Answer 3

adenine, thymine, cytosine guanine. Undergoes complimentary base pairing, C always binds with G, A and T bind together

Question 4

How DNA strands effect structure

Answer 4

One of the strands is 3 prime, the other is 5 prime. The DNA therefore has directionality. DNA is double stranded and ant-parallel.

Question 5

DNA structure

Answer 5

Forms a double helix, directionality causes a major and minor groove. The bases are easier to access at the major grooves which is the site for transcription factor binding.

Question 6

RNA (ribonucleic acid)

Answer 6

Translates genetic information to proteins. Single stranded with a sugar-phosphate backbone connected to the bases. The four bases are Adenine, Uracil, Cytosine and Guanine. The A pairs with U and C pairs with G. The sugar is ribose. RNA has directionality with a 5 prime end and a 3 prime end. Because phosphate its negatively charged.

Question 7

Difference between RNA and DNA

Answer 7

RNA is small DNA is large. DNA is highly stable and fixed. RNA is mass-produced and disposable.

Question 8

DNA replication

Answer 8

1) Double helix is unwound by DNA helicase
2) New bases are added to each strand through complimentary base pairing using the enzyme DNA polymerase
3) On the leading strand which is 5’-3’ synthesis is continuous
4) For the 3’-5’ strand it is discontinuous and is synthesised in fragments. DNA primase adds RNA primers to the lagging strand and the DNA polymerase uses them as starting points for synthesis.
5) On the lagging strand you get multiple short segments of DNA known as Okazaki fragments. DNA ligase then fills the gaps between these fragments and RNA primers are removed.
6) Double helix is reformed and the DNA is checked for errors, they are fixed by repair mechanisms

Question 9

What is a duplicated chromosome

Answer 9

Two sister chromatids, they are identical to each other and form one chromosome

Question 10

Structure of histone proteins in chromosomes

Answer 10

DNA is packed around nucleosomes which consist of 8 histone proteins. You get a gap between histone where there is linker DNA, this provides traction so you can wind the nucleosomes around each other. The coils are round around each other again repeatedly to form a supercoil.

Question 11

Telomeres

Answer 11

At the end of each chromosome. Because the end of the chromosome can not be copied junk DNA (telomeres) are added so that they can be lost instead of useful DNA. The more the cell divides the shorter the telomere gets

Question 12

Centromere

Answer 12

Where the duplicated chromosomes attach during DNA replication, not always in the centre. The centromere creates a shorter arm of the chromosome (P arm) and a longer arm (Q arm)

Question 13

Function of chromosome

Answer 13

Provide storage for DNA, allows DNA to be accurately distributed between cells during cell division

Question 14

Karyotype

Answer 14

Normal chromosome set, 23 pairs of chromosomes (46 in total). Half are inherited from each parent

Question 15

Haploid cells

Answer 15

egg and sperm, only have 23 chromosomes so half the normal amount

Question 16

Is all of the DNA genes

Answer 16

no, some of the DNA encodes genes but most of the DNA is made of non-coding DNA which have roles in regulation.

Question 17

Enhancers

Answer 17

Where the proteins bind to enhance the expression of genes

Question 18

Silencers

Answer 18

Where repressor proteins bind to silence the expression of genes

Question 19

Promoters

Answer 19

where gene expression (transcription) is initiated from, though the promoter and enhancer have overlapping functions

Question 20

Start codon

Answer 20

Where the gene sequencing begins, tells the cell where the start of the gene is

Question 21

open reading frame

Answer 21

The gene itself, composed of introns and exons

Question 22

Stop codon

Answer 22

Tells the cell where the end of the gene is and where gene sequencing stops.

Question 23

Coding strand

Answer 23

5’-3’ mRNA is identical to coding strand except T is replaced with U

Question 24

Template strand

Answer 24

3’-5;

Question 25

Transcription-initiation

Answer 25

RNA polymerase binds to the promoter region, unwinds the DNA so it can bind to the template strand

Question 26

Transcription-elongation

Answer 26

The RNA polymerase uses the template strand to create an RNA copy by adding nucleotides using complimentary base pairing. Reading the strand in the 3’-5’ direction and copying in the 5’-3’ direction. The RNA polymerase forms a DNA-RNA hybrid region within the transcription bubble, where the DNA binds to the RNA as its produced. Produces a pre-mRNA strand

Question 27

Transcription-termination

Answer 27

The new mRNA strand comes of the DNA and the DNA rewinds to form a double helix. The pre-RNA is sent for nuclear post-transcriptional processing to form mature mRNA. The new RNA strand is orientated in 5’-3’ direction

Question 28

Transcription

Answer 28

The process of copying DNA into RNA

Question 29

Translation

Answer 29

The process of copying mRNA into a proteins

Question 30

Translation initiation

Answer 30

mRNA leaves the nucleus and enters the cytoplasm where there are ribosomes. The small and large ribosomal subunit join on the mRNA. The 5’UTR contains a ribosome binding site. The tRNA which detects the start codon is the initiator tRNA, allowing the ribosomal subunit to form over the mRNA creating the translational initiator complex

Question 31

Translation elongation

Answer 31

The tRNA brings amino acids to the A site. The anticodon of the tRNA pairs up with the codon on the mRNA, through complimentary base pairing. The amino acids are joined by peptide bonds, to form a polypeptide at the P site. The empty tRNA is released at the E site. The ribosome moves along the chain bringing in more tRNA molecules till a stop codon is reached.

Question 32

Translation termination

Answer 32

Once the stop codon is reached release factors bind. No tRNA’s recognise the stop codon

Question 33

Regulatory sequence of the gene

Answer 33

silencer/enhancer, promoter and 5’UTR. There is further regulatory sequence after the gene with the 3’UTR

Question 34

Role of 3’UTR and 5’UTR

Answer 34

control of mRNA translation

Question 35

Post transcriptional processing= RNA splicing

Answer 35

Introns are removed leaving the exons which do the coding. Some of the exons can be removed to form a different version of the protein

Question 36

Post transcriptional processing= 5’ capping

Answer 36

A 5’ cap is added to the 5’UTR it is a modified G base. It regulates nuclear export, prevents degradation of the mRNA by exonuclease by providing protection. It promotes translation

Question 37

Post transcriptional translation= Polyadenylation

Answer 37

The addition of a poly A tail at the 3’UTR end of the mRNA, this is a string of A bases. Makes the mRNA more stable and prevents degradation. Allows export from the nucleus as it shows that the mRNA is mature.

Question 38

Codon

Answer 38

The three bases of mRNA which code for a specific amino acid. May be more then one codon for an amino acid but each codon codes for a specific amino acid

Question 39

Missense mutation

Answer 39

When one nucleotide is replaced with another changing the amino acid. This would affect protein structure.

Question 40

Silent mutation

Answer 40

When one nucleotide is replaced with another but the codon produced codes for the same amino acid, meaning the protein would still be functional.

Question 41

Nonsense mutation

Answer 41

When one nucleotide is replaced with another and a stop codon is produced. The amino acids after this stop codon will not be produced causing a shorter protein which would affect its function and lead to disease.

Question 42

Frameshift mutation

Answer 42

When one base is removed from the DNA sequence, when the DNA is read one letter is missed out so you read from the next codon. Every amino acid from that point will be different till you get to a stop codon, leading to a drastically different protein.

Question 43

How do cells control their function

Answer 43

By formation of different sets of mRNA and proteins which is done through regulating gene expression

Question 44

How is gene expression controlled

Answer 44

1) Transcription factors
2) RNA processing (splicing)
3) Nuclear export of mRNA into the cytosol can be regulated and stopped.
4) Translational repressors can bind to the 5’ or 3’ UTR of the mRNA and stop translation
5) mRNA degradation and half life
6) Post translational modification

Question 45

Gene expression- post translational modification

Answer 45

Can either activate or inactivate a protein. For example, in phosphorylation, by adding or taking away phosphate groups you can activate or deactivate the protein

Question 46

Amino acids

Answer 46

Have an amine, carboxylic and R group. Act as zwitterions and have a negative and positive charge (NH3+, COO-) but they are neutral overall. The R groups effect the characteristic of the protein. The N-terminus is the amine group the C-terminus is the carboxyl group.