Content Curator updated
GATE Biotechnology (BT) Question Paper PDFs are available for download here. Candidates will be able to download GATE BT 2023 Question Paper after the completion of the exam on Feburary 11, 2023 (Afternoon Session). GATE Question Paper for Biotechnology consists of three sections: General Aptitude (15 marks), Engineering Mathematics, and the core subject (85 marks). Questions cover six core subjects of Biotechnology- Genetics, cellular and Molecular Biology; Fundamental of biological Engineering; Bioprocess Engineering and Process Biotechnology; Plant, Microbial, and Animal Biotechnology; Recombinant DNA and other tools. Check GATE Biotechnology Syllabus
Also Check:
GATE 2023 Biotechnology Question Paper with Answer Key
Candidates will be able to download GATE 2023 BT Question Paper PDF from the direct link provided below.
| Exam Date | Session | Question Paper PDF |
|---|---|---|
| February 11 | Afternoon Session | Check Here |
GATE BT Previous Years’ Question Papers with Answer Key
GATE BT Question Paper comprises a total of 65 questions are asked in the exam carrying a total weightage of 100 marks. Candidates appearing for GATE Exam can download and practice using the previous year GATE Biotechnology Question Paper PDF available in the article below.
GATE 2022 Biotechnology Question Paper with Answer Key
| Exam Date | Session | Question Paper PDF |
|---|---|---|
| February 12 | Forenoon Session | Check Here |
GATE 2021 Biotechnology Question Paper with Answer Key
| Exam Date | Session | Question Paper PDF |
|---|---|---|
| February 13 | Forenoon Session | Check Here |
GATE 2020 Biotechnology Question Paper with Answer Key
| Exam Date | Session | Question Paper PDF |
|---|---|---|
| February 2 | Forenoon Session | Check Here |
GATE 2019 Biotechnology Question Paper with Answer Key
| Exam Date | Session | Question Paper PDF |
|---|---|---|
| February 3 | Afternoon Session | Check Here |
GATE 2018 Biotechnology Question Paper with Answer Key
| Exam Date | Session | Question Paper PDF |
|---|---|---|
| February 4 | Afternoon Session | Check Here |
GATE 2017 Biotechnology Question Paper with Answer Key
| Exam Date | Session | Question Paper PDF |
|---|---|---|
| February 5 | Forenoon Session | Check Here |
GATE 2016 Biotechnology Question Paper with Answer Key
| Exam Date | Session | Question Paper PDF |
|---|---|---|
| January 30 | Afternoon Session | Check Here |
GATE 2015 Biotechnology Question Paper with Answer Key
| Exam Date | Session | Question Paper PDF |
|---|---|---|
| January 31 | Forenoon Session | Check Here |
GATE 2014 Biotechnology Question Paper with Answer Key
| Exam Date | Session | Question Paper PDF |
|---|---|---|
| February 2 | Afternoon Session | Check Here |
GATE 2013 Biotechnology Question Paper with Answer Key
| Exam Date | Session | Question Paper PDF |
|---|---|---|
| January 20 | Afternoon Session | Check Here |
GATE 2012 Biotechnology Question Paper with Answer Key
| Exam Date | Session | Question Paper PDF |
|---|---|---|
| February 12 | Forenoon Session | Check Here |
GATE 2011 Biotechnology Question Paper with Answer Key
| Exam Date | Session | Question Paper PDF |
|---|---|---|
| February 13 | Forenoon Session | Check Here |
GATE 2010 Biotechnology Question Paper with Answer Key
| Exam Date | Session | Question Paper PDF |
|---|---|---|
| February 14 | Forenoon Session | Check Here |
GATE Question Papers: Download Paper-wise Previous Year Papers
GATE Questions
2. The area of a rhombus whose vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order, is:
The area of a rhombus whose vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order, is:
12 sq.units
24 sq.units
3. Write a balanced equation for the reaction of molecular nitrogen (N2) and oxygen (O2) to form dinitrogen pentoxide. \((f(x)=x^2)\)
Write a balanced equation for the reaction of molecular nitrogen (N2) and oxygen (O2) to form dinitrogen pentoxide. \((f(x)=x^2)\)
- N2 + O2 → N2O5 (unbalanced equation)
- 2N2 + 5O2 → 2N2O5
- option 3
4. For the reaction, $H_{2} + I_{2} {\rightleftharpoons} 2HI, K= 47.6.$ If the initial number of moles of each reactant and product is 1 mole then at equilibrium
- $\left[I_{2}\right]=\left[H_{2}\right], \left[I_{2}\right] > \left[HI\right]$
- $({\frac{x^3}{9}})$
\(\left[I_{2}\right]>\left[H_{2}\right], \left[I_{2}\right] = \left[HI\right]\)- $\omega\propto\,n^{\frac{1}{3}}$
Comments