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Class 12

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Sexual Reproduction in Flowering Plants

Pollen grain

Viability of pollen grains depends on

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Easy

Solving time: 2 mins

Viability of pollen grains depends on

Name: Video solution 1: Viability of pollen grains depends on
Uploaded: 2023-01-30T19:15:42.000Z
Duration: 5 min
Description: Video solution 1: Viability of pollen grains depends on

temperature

humidity

Both (a) and (b)

pressure

Views: 6,183 students

Updated on: Jan 30, 2023

Book:

CBSE New Pattern Biology Class 12 for 2021-22 Exam MCQs based book for Term 1

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The period in which the pollen grains remain viable is highly variable. It depends on the temperature and humidity.

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Practice questions from CBSE New Pattern Biology Class 12 for 2021-22 Exam MCQs based book for Term 1

Question 1

Easy

Continued self-pollination results in

inbreeding depression
outbreeding depression
hybrid vigour
better results in offsprings

Topic:

Sexual Reproduction in Flowering Plants

Book:

CBSE New Pattern Biology Class 12 for 2021-22 Exam MCQs based book for Term 1

View solution

Question 2

Easy

Find out the correct statement(s). I. Endosperm formation is the prior event than embryo formation. II. Angiospermic and gymnospermic endosperm are

3 n

and

n

, respectively. III. Endospermic seeds are found in castor, barley and coconut. IV. In albuminous seed, food is stored in endosperm and in non-albuminous seeds, it is stored in cotyledons.

I and IV
II and III
I and III
I, II, III and IV

Topic:

Sexual Reproduction in Flowering Plants

Book:

CBSE New Pattern Biology Class 12 for 2021-22 Exam MCQs based book for Term 1

View solution

Question 3

Easy

The functions of germ pore is/are

emergence of radicle
absorption of water for seed germination
initiation of pollen tube
All of the above

Topic:

Sexual Reproduction in Flowering Plants

Book:

CBSE New Pattern Biology Class 12 for 2021-22 Exam MCQs based book for Term 1

View solution

Question 4

Easy

Assertion (A) Insects visit flowers to gather nectar. Reason (R) Attraction to flowers prevents the insects from damaging other parts of the plant.

Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
Both $A$ and $R$ are true, but $R$ is not the correct explanation of $A$
$A$ is true, but $R$ is false
$A$ is false, but $R$ is true

Topic:

Sexual Reproduction in Flowering Plants

Book:

CBSE New Pattern Biology Class 12 for 2021-22 Exam MCQs based book for Term 1

View solution

Practice more questions from Sexual Reproduction in Flowering Plants

Question 1

Easy

In the embryos of a typical dicot and a grass, true homologous structures are :
(A) Coleorhiza and coleoptile
(B) Coleoptile and scutellum
(C) Cotyledons and scutellum
(D) Hypocotyl and radicle.

Topic:

Sexual Reproduction in Flowering Plants

Book:

Oswaal CBSE MCQs Chapterwise For Term I II, Class 12, Biology (Oswaal)

View solution

Question 2

Easy

Assertion: The two sperms in a pollen tube often change their shape.
Reason: The sperms are released in the synergid as intact cells but only their nuclei migrate.

If both Assertion and Reason are true and Reason is the correct explanation of Assertion
If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
If Assertion is true but Reason is false.
If both Assertion and Reason are false.

Topic:

Sexual Reproduction in Flowering Plants

Book:

NEET Question Bank i Biology for AIIMS (Disha)

View solution

Question 3

Easy

Make a list of any three out breeding devices that flowering plants have developed and explain how they help to encourage cross pollination.

Topic:

Sexual Reproduction in Flowering Plants

Book:

Oswaal CBSE Question Bank Class 12 Biology Chapterwise and Topicwise (Oswaal)

View solution

Question 4

Easy

The formation of embryo without fusion of gametes is termed, as

Apospory
Isogamy
Apogamy
Syngamy

Topic:

Sexual Reproduction in Flowering Plants

Book:

Biology-for-NEET-Volume-2-Class-XII

View solution

Practice questions on similar concepts asked by Filo students

Question 1

Texts: #22 (a,b,c) with explanation On the chromosome, genes introduced from a donor cell. As seen in the solution to Solved Problem I, only an even number of crossovers can produce viable recombinant chromosomes. Gene mapping is simplified if you remember that progeny classes that result from four crossovers are found much less often than progeny classes that require two crossovers. 22. Suppose you have two Hfr strains of E. coli (HfrA and HfrB), derived from a fully prototrophic streptomycin-sensitive (wild-type) F strain. In separate experiments, you allow these two Hfr strains to conjugate with an F recipient strain (Rcp) that is streptomycin resistant and auxotrophic for glycine (Gly), lysine (Lys), nicotinic acid (Nic), phenylalanine (Phe), tyrosine (Tyr), and uracil (Ura). By using an interrupted-mating protocol, you determine the earliest time after mating at which each of the markers can be detected in the streptomycin-resistant recipient strain, as shown here. 20. In Problem 19, do you think that most of the Pyr Arg* exconjugants are also Xyl* and Mal, or not? Explain your answer by considering the recombination events that would be required to generate colonies that are Pyr* Arg* Xyl* Mal* and those required to make PyrArgXylMal colonies. 21. One issue with interrupted-mating experiments such as that in Problem 19 is that gene order may be ambiguous if the genes are close together. Another shortcoming is that such experiments do not provide Cross Gly+ Lys* Phe+ Tyr* Ura* HfrA Rcp 3 * 8 3 3 3 HfrB Rcp 8 3 13 8 8 8 (The * indicates that no Lys* cells were recovered in the 60 minutes of the experiment.) 508 Chapter 16 Bacterial Genetics a. Draw the best map you can from these data, showing the relative locations of the markers and the origins of transfer in strains HfrA and HfrB. Show distances where possible. b. To resolve ambiguities in the preceding map, you studied cotransduction of the markers by the generalized transducing phage P1. You grew phage P1 on strain HfrB and then used the lysate to infect strain Rcp. You selected 1000 Phe* clones and tested them for the presence of unselected markers, with the following results: and cys were then scored in the Trp* recombinants yielding the following results: Number of recombinants Trp + + + + Pyr + + - - Cys + - + - 16L 145 60 5 Draw a map of the trp, pyr, and cys markers relative to each other. (Note that you cannot determine the order relative to the nic or his genes using these data.) Express map distances between adjacent genes as the frequency of crossing-over between them. 24. You can carry out matings between an Hfr and F strain by mixing the two cell types in a small patch on a plate and then replica plating to selective medium. This methodology was used to screen hundreds of different cells for a recombination-deficient recA mutant. Why is this an assay for RecA function? Would you be screening for a recA mutation in the Hfr strain using this protocol? Explain. 25. Genome sequences show that some pathogenic bacteria contain virulence genes that promote disease next to genes that originally came from bacterio- Number of transductants Phenotype Nic Phe - + - + - + Gly - - Lys - - - Tyr - 600 300 100 - + + + Draw the order of the genes as best you can based on the preceding cotransduction data. c. Suppose you wanted to use generalized transduction to map the gly gene relative to at least some of the other markers. How would you modify the cotransduction experiment just described to increase your chances of success? Describe the composition of the medium you would use.

Topic:

Biology

View solution

Question 2

8.5.8 Cilia (sing.: cilium) and flagella (sing.: flagellum) are hair-like outgrowths of the cell membrane. Cilia are small structures which work like(бars) causing the movement of either the cell or the surrounding fluid. Flagella are comparatively longer and responsible for cell movement. The prokaryotic bacteria also possess flagella but these are structurally different from that of the eukaryotic flagella. The electron microscopic study of a cilium or the flagellum show that they are covered with plasma membrane. Their core called the axoneme, possesses a number of microtubules running parallel to the long axis. The axoneme usually has nine doublets of radially arranged peripheral microtubules, and a pair of centrally located microtubules. Such an arrangement of axonemal microtubules is referred to as the

9 + 2

array (Figure

8.10

). The central tubules are connected by bridges and is also enclosed by@central sheath which is connected to one of the tubules of each peripheral doublets by a radial spoke. Thus, there are nine radial spokes. The peripheral doublets are also interconnected by linkers. Both the cilium and flagellum emerge from centriole-like structure called the basal bodies. 1

8.5

a Controname and Centrioles

Topic:

Cell: Structure and Functions

View 2 solutions

Question 3

Sue does not have the autosomal recessive disease cystic fibrosis but has a sister, Jamie with cystic fibrosis. Neither of Sue and Jamie's parents have cystic fibrosis. Sue has a baby with someone who does not have a family history of cystic fibrosis. However, you know that the disease cystic fibrosis occurs in 0.05% of people in the USA. Taking this into account, what is the probability their child will have cystic fibrosis?

0 %
0.043%
0.022%
0.978%
0.717%

Topic:

Biology

View solution

Question 4

During a dihybrid cross with contrasting characters in the

F_{2}

generation parental genotypes will appear in _ ratio.

$1/16$
$2/16$
$3/16$
$9/16$

Topic:

Genetics and Evolution

View 2 solutions

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Question Text	Viability of pollen grains depends on
Updated On	Jan 30, 2023
Topic	Sexual Reproduction in Flowering Plants
Subject	Biology
Class	Class 12
Answer Type	Text solution:1 Video solution: 1
Upvotes	177
Avg. Video Duration	5 min

Viability of pollen grains depends on

Text solutionVerified

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Practice questions from CBSE New Pattern Biology Class 12 for 2021-22 Exam MCQs based book for Term 1

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