Somatic embryogenesis is a process where a plant or embryo is derived from a single somatic cell or group of somatic cells. Somatic embryos are formed from plant cells that are not normally involved in the development of embryos, i.e. ordinary plant tissue. In plants, somatic embryogenesis comprises of embryo initiation followed by embryo production. ​​
Embryo initiation occurs on a medium rich in auxin, which induces differentiation of localized meristematic cells. The auxin typically used is 2,4-Dichlorophenoxyacetic acid (2,4-D). thus embryo initiation is dependent on 2,4-D. While these are transferred to a medium containing low or no auxin, then embryo production does not require high concentration of 2, 4-D. ​

Correct Option: C

Somatic embryogenesis is a process where a plant or embryo is derived from a single somatic cell or group of somatic cells. Somatic embryos are formed from plant cells that are not normally involved in the development of embryos, i.e. ordinary plant tissue. In plants, somatic embryogenesis comprises of embryo initiation followed by embryo production. ​​
Embryo initiation occurs on a medium rich in auxin, which induces differentiation of localized meristematic cells. The auxin typically used is 2,4-Dichlorophenoxyacetic acid (2,4-D). thus embryo initiation is dependent on 2,4-D. While these are transferred to a medium containing low or no auxin, then embryo production does not require high concentration of 2, 4-D. ​

Rapid amplification of cDNA ends (RACE) polymerase chain reactions (PCR) are used for independent high-throughput verification of transcriptional starting sites (TSSs) determined by genome-wide assays. 52 -RACE PCR is a well-established and widely used method to specifically amplify the 52 end of a transcript, facilitating mapping of the TSS and the approximate location of promoter elements. Conventionally, this mapping is done by cloning the 52 -RACE PCR product into a bacterial vector and sequencing a few clones by classic electrophoresis-based Sanger sequencing. Now-a-days, even 3’ end transcripts are being mapped using 3’-RACE PCR. Baculoviruses are a very diverse group of viruses with double-stranded, circular, supercoiled genomes, with sizes varying from about 80 to over 180 kb which encode between 90 and 180 genes. Baculoviruses have evolved to initiate infection in the insect midgut. Baculovirus-insect cell expression systems have the capacity to produce many recombinant proteins at high levels and they also provide significant eukaryotic protein processing capabilities.

Correct Option: C

Rapid amplification of cDNA ends (RACE) polymerase chain reactions (PCR) are used for independent high-throughput verification of transcriptional starting sites (TSSs) determined by genome-wide assays. 52 -RACE PCR is a well-established and widely used method to specifically amplify the 52 end of a transcript, facilitating mapping of the TSS and the approximate location of promoter elements. Conventionally, this mapping is done by cloning the 52 -RACE PCR product into a bacterial vector and sequencing a few clones by classic electrophoresis-based Sanger sequencing. Now-a-days, even 3’ end transcripts are being mapped using 3’-RACE PCR. Baculoviruses are a very diverse group of viruses with double-stranded, circular, supercoiled genomes, with sizes varying from about 80 to over 180 kb which encode between 90 and 180 genes. Baculoviruses have evolved to initiate infection in the insect midgut. Baculovirus-insect cell expression systems have the capacity to produce many recombinant proteins at high levels and they also provide significant eukaryotic protein processing capabilities.

Let N₀ be the initial number of molecules present in the aliquot and let N be the end product of PCR amplification. Let n be the number of cycles performed by the PCR.
We, know N = N₀ × 2ⁿ for n number of cycles since in each cycle the number of amplicons double the number of parent strands in an exponential manner. ​​
Here, N₀ = 100 for a 200 μL solution (given), ​​​​n ​= 10 ​​
Therefore, ​N​ = 100 × 2¹⁰
= 100 × 1024 ​​​​​= 1.024 × 10⁵ ​​
Hence, the required amplicons generated after 10 cycles of PCR run = 1.024 × 10⁵

Correct Option: B

Let N₀ be the initial number of molecules present in the aliquot and let N be the end product of PCR amplification. Let n be the number of cycles performed by the PCR.
We, know N = N₀ × 2ⁿ for n number of cycles since in each cycle the number of amplicons double the number of parent strands in an exponential manner. ​​
Here, N₀ = 100 for a 200 μL solution (given), ​​​​n ​= 10 ​​
Therefore, ​N​ = 100 × 2¹⁰
= 100 × 1024 ​​​​​= 1.024 × 10⁵ ​​
Hence, the required amplicons generated after 10 cycles of PCR run = 1.024 × 10⁵

​Hybridoma technology is a technology of forming hybrid cell lines (called hybridomas) by fusing a specific antibody-producing B cell with a myeloma (B cell cancer) cell that is selected for its ability to grow in tissue culture and for an absence of antibody chain synthesis. The antibodies produced by the hybridoma are all of a single specificity and are therefore monoclonal antibodies Fused cells are incubated in HAT medium (hypoxanthine-aminopterin-thymidine medium) for roughly 10 to 14 days. Aminopterin blocks the pathway that allows for nucleotide synthesis. Hence, unfused myeloma cells die, as they cannot produce nucleotides by the de novo or salvage pathways because they lack HGPRT. Removal of the unfused myeloma cells is necessary because they have the potential to outgrow other cells, especially weakly established hybridomas. Unfused B cells die as they have a short life span. In this way, only the B cell-myeloma hybrids survive, since the HGPRT gene coming from the B cells is functional. ​

Correct Option: C

​Hybridoma technology is a technology of forming hybrid cell lines (called hybridomas) by fusing a specific antibody-producing B cell with a myeloma (B cell cancer) cell that is selected for its ability to grow in tissue culture and for an absence of antibody chain synthesis. The antibodies produced by the hybridoma are all of a single specificity and are therefore monoclonal antibodies Fused cells are incubated in HAT medium (hypoxanthine-aminopterin-thymidine medium) for roughly 10 to 14 days. Aminopterin blocks the pathway that allows for nucleotide synthesis. Hence, unfused myeloma cells die, as they cannot produce nucleotides by the de novo or salvage pathways because they lack HGPRT. Removal of the unfused myeloma cells is necessary because they have the potential to outgrow other cells, especially weakly established hybridomas. Unfused B cells die as they have a short life span. In this way, only the B cell-myeloma hybrids survive, since the HGPRT gene coming from the B cells is functional. ​

From previous question, we have ​​
Number of amplicons in 200 μL of the solution after PCR run = 1.024 × 10⁵ ​​
Therefore, number of amplicons in 0.1 μL solution = 1.024 × 10⁵ × 0.1 / 200 = 51.2
​​Hence, the required number of amplicons present in 0.1 μL of the solution = 51.2

Correct Option: C

From previous question, we have ​​
Number of amplicons in 200 μL of the solution after PCR run = 1.024 × 10⁵ ​​
Therefore, number of amplicons in 0.1 μL solution = 1.024 × 10⁵ × 0.1 / 200 = 51.2
​​Hence, the required number of amplicons present in 0.1 μL of the solution = 51.2