Question

1) T/F In order to amplify a gene by PCR, one needs to know the exact DNA sequence of the regions flanking the gene.

2) T/F The initiation complex for translation includes transfer RNA-MET.

3) T/F A mutation that changes the amino acid sequence in a protein sometimes increases the fitness of the individual.

4) T/F A mutation in the TATA box will change the level of gene expression.

5) T/F Pyrimidine dimer formation is a common cause of DNA mutations.

6) Homologous recombination

a. produces new linkage groups

b. results in independent assortment

c. both are correct

d. neither are correct

7) Two parents both have the following genotype: XxYyZz

What is the probability that they will have a child with this genotype: XxYyZz

a. 1/4

b.1/2

c. 1/8

d. 0

8) An internal deletion removes a segment of one copy of chromosome 15. If FISH analysis is performed using a probe that hybridizes only to the deleted segment, what will be observed?

a.one copy of chromosome 15 will be detected by FISH

b. no copy of chromosome 15 will be detected

c. FISH analysis will be inconclusive

d. two copies of chromosome 15 will be detected

9)If a centromere probe is used in FISH analysis how many spots will light up in the cells of an individual with trisomy 21?

a. 46

b.47

c. 23

d. 24

10) T/F The repressor is constitutively synthesized [all the time] in both the lac and trp operons.

11) T/F In order to amplify a gene by PCR, one needs to know the exact DNA sequence of the regions flanking the gene.

Answer 1

1. False

To amplify a gene, a need to have primers which are complementary to a certain segment of gene or certain part of the flanking region if gene and not whole gene sequence. Primers will bind to that certain segment and DNA polymerase 3 will elongate the whole DNA segment.

2. True

Formation of 70S initiation complex in prokaryotes

1) IF1 binds to A site and IF3 binds to the site that Will become the E site on smaller ribosomal subunit (30S subunit).

2) Now, IF2-GTP binds to this on P site. This whole assembly of 30S ribosomal subunit and the three initiation factors forms the 30S initiation complex.

3) This binds to mRNA so that the AUG on mRNA faces the P site of ribosome.

4) IF2-GTP facilitates the binding of initiator tRNA, fmet-tRNAi fmet

5) The start codon of mRNA and anticodon of tRNA base pairs with each other, causing a conformational change in the smaller subunit. This leads to release of IF3.

6) The larger subunit (50S) with its cargo IF1, IF2, mRNA and fmet-tRNAi fmet now binds to the smaller subunit.

7) The binding of larger subunit stimulates the GTPase activity of IF2 causing the hydrolysis of GTP.

8) This IF2-GDP has less affinity for ribosome and initiator tRNA leading to the release of IF2-GDP and IF1.

9) Thus, an intact ribosome is assembled at the start site of mRNA with the initiator tRNA. This complex is referred to as the 70S initiation complex.

3. True

Mutations are changes in DNA which may alter the phenotype of an organism and if they occur in germline cell of an organism then they are passed on to the next generation thereby affecting the progeny of that individual. Mutations are not always harmful sometimes they are beneficial and are selected in a positive way and help the individual to adapt better in its environment.

4. True

Tata box is the promoter present in eukaryotes as well as in prokaryotes. It is the sequence where RNA polymerase binds and start the transcription of the DNA. If Tata box is mutated and the sequence is altered then there may be an increase or decrease in the level of expression of the gene. This is because in case of mutation RNA polymerase may or may not bind to the the promoter properly and result in an alteration in the expression of gene.

5. True

Pyrimidine dimerization is the formation of covalent bonds between two adjacent pyrimidines located at the same strand of DNA. It occurs due to to UV rays. It is corrected by the enzymes called photolyases. A common example of pyrimidine dimerization is thymine dimers.

6. Both are correct

Homologous recombination is the exchange of genetic material between homologous chromosomes. It results in formation of chromosomes having both paternal and maternal genes. Also it does not violate the law of independent assortment according to which the chromosomes in a homologous pair assort independently of each other during meiosis.

7. 1/8

8. Option A

There are two chromosome number 15. One of them has the segment and the other shows deletion of the segment. So, complementary probe will bind to the chromosome which has the segment and will not bind to the chromosome which does not have the segment. So we will be able to see only one chromosome 15.

9. 47

Trisomy of 21 means there are 3 copies of chromosome number 21. It means normally, humans have 46 chromosomes and 1 chromosome have 1 centromere and we see 46 centromeres by fish. So, in trisomy, total number of chromosomes will be 47. So we will see 47 centromeres.

10. False

Lac Operon is inducible type of Operon because normally it is switched off because of presence of lac repressor and it gets switched on whenever lactose is there in the medium. Show this statement is correct for Lac Operon.

Tryptophan Operon is repressible type of Operon because normally it is switched on and gets switched off when ever tryptophan is present in the medium. When trytophan is present in the medium tryptophan repressor binds to the promoter of tryptophan Operon. This statement is incorrect for tryptophan Operon.

11. False

Same as question 1

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