DNA, DNA Technology, Genetic Diease Quiz

Consider these DNA features

1. linear

2. circular

3. associated with histone

4. inside nucleus

Which feature(s) is/are associated with chromosomal DNA of Escherichia coli?

• 1

• 2

• 2 & 3

• 1, 3 & 4

Histone proteins in chromosomes are composed of amino acids, which are mostly negatively charged. Makes it able to bind to the DNA strand well.

• TRUE

• FALSE

Answer explanation

Histone proteins in chromosomes are composed of amino acids, which are mostly negatively charged. Makes it able to bind to the DNA strand well.

Answer: positively

DNA is a polynucleotide strand that has two ends, one called the 5' end and the other called the 3' end. The 3' end has a phosphate group attached to a deoxyribose at carbon position 3'.

• TRUE

• FALSE

Answer explanation

NA is a polynucleotide strand that has two ends, one called the 5' end and the other called the 3' end. The 3' end has a phosphate group attached to a deoxyribose at carbon position 3'.

Answer: hydroxyl

In protein synthesis, the code on the mRNA strand is translated from the 5′ end to the 3′ end. The amino acid methionine (MET) is translated from the code, which has a base sequence on the mRNA strand, to AUG, which binds to the tRNA that has the base sequence 5′ UAC 3′.

• TRUE

• FALSE

Answer explanation

In protein synthesis, the code on the mRNA strand is translated from the 5′ end to the 3′ end. The amino acid methionine (MET) is translated from the code, which has a base sequence on the mRNA strand, to AUG, which binds to the tRNA that has the base sequence 5′ UAC 3′.

Answer: 5′ CAU 3′ or 3′ UAC 5′

​ ​ A DNA simulation started with 1 DNA molecule. After 3 rounds of simulation, we will get ​ (a)   new DNA molecules and ​ (b)   molecule of starting DNA.

• 8
• 2
• 3
• 4
• 1
• 5
• 6
• 7
• 9
• 10

The picture shows the process of protein synthesis in bacteria. If bacteria are exposed to various types of antibiotics, protein synthesis will be inhibited. Match names of antibiotics to relate to the steps that inhibit bacterial protein synthesis in different ways as follows.

Chloramphenicol - binds to the large ribosome subunit and inhibits peptide bond formation.

Streptomycin - changes the shape of the small subunit of the ribosome, resulting in the misreading of the codon of the mRNA.

Tetracycline - interferes with the binding of mRNA and tRNA.

Erythromycin - binds to the large ribosome subunit, immobilizing the ribosome.

• Streptomycin

• 3

• Erythromycin

• 1

• Tetracycline

• 2

• Chloramphenicol

• 4

Information of 5 types of double-strand DNA (molecule M, N, O, P, and Q) with 250 base pairs each:

M has 150 A and 100 G

N has 75 C

O has 125 T

P has 200 A

Q has 225 T and 25 C

• M uses less energy to separate the complementarity of DNA strands.

• N has base pairs with 2 hydrogen bonds more than that of Q.

• O has base pairs with 3 hydrogen bonds more than that of other molecules.

• P has the most phosphate diester bonds.

• Q has the most numbers of pyrimidine.

Answer explanation

There are 2 hydrogen bonds between A and T.

There are 3 hydrogen bonds between G and C.

A researcher cultured cells to increase their number and used X-rays to force mutation. They found that some cells had their nucleotides altered compared to those before the X-ray treatment. The mutated cells are classified as type A – E, as shown in the table.

• Cell A – mutation occurred at chromosomal level

• Cell B – mutation occurred due to the addition or removal of 3 nucleotides

• Cell C – mutation occurred due to disjunction

• Cell D – mutation occurred due to the addition or removal of 1 – 2 nucleotides

• Cell E – frameshift mutation had occurred

Which information is correct?

• If there was NO mutation, the yielded polypeptide after translation would have 8 amino acids.

• If the base the position III was replaced by C, the corresponding phenotype would be altered.

• If the base the position III was replaced by A, the translated polypeptide would be shorter.

• If the base the position I is replaced by G, there would be no effect on the translated polypeptide.

• If 1 nucleotide was added to the position II, frameshift mutation would happen and the translated polypeptide with 5 amino acids would be obtained.

A genetic disease is controlled by an allele on an X chromosome (X-linked recessive). The dominant allele is 700 bp. However, the recessive allele is only 200 bp due to some missing nucleotides. After the uses of PCR and gel-electrophoresis, 3 types of DNA are found as shown in the figure. If a carrier female has an offspring with an unaffected male, which is/are the possible DNA(s) for the child.

• Son – A & C

  Daughter – A & B

• Son – A & C

  Daughter – B & C

• Son – A

  Daughter – A B & C

• Son – C

  Daughter – A B & C

• Son and Daughter – A B & C