No new follicles develop in the luteal phase of the menstrual cycle because :
1.
Follicles do not remain in the ovary after ovulation
2.
FSH levels are high in the luteal phase
3.
LH levels are high in the luteal phase
4.
Both FSH and LH levels are low in the luteal phase
In Australia, marsupials and placental mammals have evolved to share many similar characteristics. This type of evolution may be referred to as :
1. Adaptive Radiation
2. Divergent Evolution
3. Cyclical Evolution
4. Convergent Evolution
Match the items of Column - I with Column - II :
| Column-I | Column-II | ||
| (a) | XX-XO method of sex determination | (i) | Turner's syndrome |
| (b) | XX-XY method of sex Determination | (ii) | Female heterogametic |
| (c) | Karyotype-45 | (iii) | Grasshopper |
| (d) | ZW-ZZ method of Sex Determination | (iv) | Female homogametic |
Select the correct option from the following:
| (a) | (b) | (c) | (d) | |
| 1. | (ii) | (iv) | (i) | (iii) |
| 2. | (i) | (iv) | (ii) | (iii) |
| 3. | (iii) | (iv) | (i) | (ii) |
| 4. | (iv) | (ii) | (i) | (iii) |
What will be the sequence of mRNA produced by the following stretch of DNA?
3' ATGCATGCATGCATG 5' TEMPLATE STRAND
5' TACGTACGTACGTAC 3' CODING STRAND
1. 3' AUGCAUGCAUGCAUG 5'
2. 5' UACGUACGUACGUAC 3'
3. 3' UACGUACGUACGUAC 5'
4. 5' AUGCAUGCAUGCAUG 3'
Select the incorrect statement regarding inbreeding:
| 1. | Inbreeding helps in elimination of deleterious alleles from the population |
| 2. | Inbreeding is necessary to evolve a pure-line in any animal |
| 3. | Continued inbreeding reduces fertility and leads to inbreeding depression. |
| 4. | Inbreeding depression cannot be overcome by out-crossing. |
For a biocontrol agent to be part of an integrated pest management, it should be:
1. species-specific and symbiotic
2. free living and broad spectrum
3. narrow spectrum and symbiotic
4. species-specific and inactive on non-target organisms
The major products C and D formed in the following reaction respectively are:
\(CH_3CH_2CH_2OC{(CH_3)}_3 \xrightarrow[\Delta]{\text{Excess}\ HI}\ C + D\)
1. H3C-CH2-CH2-I and I-C(CH3)3
2. H3C-CH2-CH2-OH and I-C(CH3)3
3. H3C-CH2-CH2-I and HO-C(CH3)3
4. H3C-CH2-CH2-OH and HO-C(CH3)3
Match the following enzymes with their functions:
|
Column-I |
Column-II |
||
|
(a) |
Restriction endonuclease |
(i) |
joins the DNA fragments |
|
(b) |
Exonuclease |
(ii) |
extends primers on genomic DNA template |
|
(c) |
DNA ligase |
(iii) |
cuts DNA at a specific position |
|
(d) |
Tag polymerase |
(iv) |
removes nucleotides from the ends of DNA |
Select the correct option from the following:
| Options: | (a) | (b) | (c) | (d) |
| 1. | (iii) | (i) | (iv) | (ii) |
| 2. | (iii) | (iv) | (i) | (ii) |
| 3. | (iv) | (iii) | (i) | (ii) |
| 4. | (ii) | (iv) | (i) | (iii) |
Match the oxide given in column I with its property given in column II:
| Column I | Column II |
| (i) | (a) Neutral |
| (ii) | (b) Basic |
| (iii) | (c) Acidic |
| (iv) | (d) Amphoteric |
Which of the following options is correct?
| (i) | (ii) | (iii) | (iv) | |
| 1. | (b) | (a) | (d) | (c) |
| 2. | (c) | (b) | (a) | (d) |
| 3. | (a) | (d) | (b) | (c) |
| 4. | (b) | (d) | (a) | (c) |
The two antibiotic resistance genes on vector pBR 322 are for:
1. Ampicillin and Tetracycline
2. Ampicillin and Chloramphenicol
3. Chloramphenicol and Tetracycline
4. Tetracycline and Kanamycin
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