Biology:2015:CBSE:[Delhi]:Set-II
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Q1
What is Biopiracy?
Marks:1Answer:
The illegal exploitation of biological material of a country by organisations or multinational companies without proper authorisation from the concerned country is known as biopiracy.
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Q2
State a reason for the increased population of dark coloured moths coinciding with the loss of lichens (on tree barks) during industrialization period in England.
Marks:1Answer:
Industrial melanism was the reason for the increase in the population of dark-coloured moths along with the loss of lichens during industrialisation in England.
After industrialization, the tree trunks became darker with the deposits of soot and smoke. The number of dark moths increased as they were not easily visible to their predators however, the white-winged moths were easily picked up by the predators. Thus, dark ones were naturally selected and light ones fail to survive. -
Q3
Indiscrimate diagnostic practices using X-rays etc., should be avoided. Give one reason.
Marks:1Answer:
X-rays are potent mutagens and can cause adverse effects in children in the form of genital changes; i.e., they alter the genetic make-up of an organism.
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Q4
Name the transcriptionally active region of chromatin in a nucleus.
Marks:1Answer:
Euchromatin is the transcriptionally active region of chromatin in a nucleus.
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Q5
A geneticist interested in studying variations and patterns of inheritance in living beings prefers to choose organisms for experiments with shorter life cycle. Provide a reason.
Marks:1Answer:
A geneticist interested in studying variations and patterns of inheritance in living beings prefers to choose organisms with short life cycles for experiments because it enables the geneticist to study many generations of those organisms in their short lifespans.
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Q6
Many fresh water animals cannot survive in marine environment. Explain.
Marks:2Answer:
Many freshwater fish cannot survive in the marine environment as their bodies are not adapted to the marine environment. Freshwater fish lose body water because of the hypertonic environment that has a higher concentration of salt in the surrounding water, making their survival difficult.
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Q7
How are productivity, gross productivity, net primary productivity and secondary productivity interrelated?
Marks:2Answer:
Productivity: It is the rate of production of biomass at any trophic level at any given interval of time.
Gross productivity: It is the rate of production of organic matter by green plants per unit time per unit area. In other words, it is the total amount of productivity.
Net primary productivity: It is the difference between gross primary productivity and the loss due to respiration.Net primary productivity is the difference of gross primary productivity and respiratory loss. -
Q8
Name any two common Indian millet crops. State one characteristic of millets that has been improved as a result of hybrid breeding so as to produce high yielding millet crops.
Marks:2Answer:
Two common Indian millet crops are maize and jowar. Hybrid breeding has resulted in the production of high yielding millet varieties that are resistant to water stress.
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Q9
Mention a product of human welfare obtained with the help of each one of the following microbes:
(a) LAB
(b) Saccharomyces cerevisiae
(c) Propionibacterium shermanii
(d) Aspergillus nigerMarks:2Answer:
a) Lactic acid bacteria (LAB); Lactobacillus is used in the production of curd from milk.
b) Saccharomyces cerevisiae is used in baking industry.
c) Propionibacterium shermanii is used in the production of cheese.
d) Aspergillus niger is used in the production of citric acid.
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Q10
Explain the mechanism of sex-determination in birds.
Marks:2Answer:
In case of birds, Females are responsible for sex determination as they are heterozygous having the sex chromosome pair as Z and W and hence, produce two types of ova, one with autosomes and Z-chromosome and the other with autosomes and W-chromosome.
Males are homozygous producing only one type of sperms, having autosomes and one Z-chromosome.
Therefore, a bird with sex chromosome genotype ‘ZW’ becomes female and the bird with sex chromosome genotype ‘ZZ’ becomes male.