Multiple Choice Questions on DNA Packaging

Q1. Which of the following statement is true?

a)    Histone proteins are of low molecular weight

b)    Non-histone proteins are of high molecular weight

c)    Both histone and non-histone proteins are of low molecular weight

d)    Both histone and non-histone proteins are of high molecular weight

e)    Both a and b

Q2. Which of the following statement is true?

a)    Histone proteins are basic in nature

b)    Histone proteins are acidic in nature

c)    Non-histone proteins are acidic in nature

d)    Both histone and non-histone proteins are acidic in nature

e)    Both a and c

Q3. Which of the following statements are true?

a)    Histone proteins are rich in lysine and arginine

b)    Lysine and arginine are basic amino acids

c)    Non-histone proteins are rich in tyrosine and tryptophan

d)    Tyrosine and tryptophan are aromatic  amino acids

e)    All of the above

Q4. Out of the five histone proteins, which are rich in lysine and arginine?

a)    H1, H2A, and H2B are rich in lysine and H3 and H4 are rich in arginine

b)    H1, H2A, and H2B are rich in arginine and H3 and H4 are rich in lysine

c)    All five amino acids are rich in lysine

d)    All five amino acids are rich in arginine

Q5. Out of the three types of non-histone proteins, what is the function of structural / scaffold protein?

a)    Scaffold proteins are involved in the packaging of chromatids at a higher level

b)    Help in the replication of DNA

c)    Involved in gene expression

d)    None of the above

Q6. What are the three different types of non-histone proteins involved in DNA packaging?

A)   Structural proteins, scaffold proteins, enzymatic proteins

B)   Scaffold proteins/structural proteins, enzymatic proteins, regulatory proteins

C)   Scaffold proteins, enzymatic proteins, inhibitory proteins

D)   H1, H2A, H2B

E)   H2B, H3, H4

Q7. DNAP and RNAP are examples of:

a)    Structural proteins

b)    Scaffold proteins

c)    Enzymatic proteins

d)    Regulatory proteins

e)    Inhibitory proteins

Q8. Until the time DNA is attached to histone proteins, DNA does not express itself. What happens when regulatory proteins interact with histone proteins?

a)    The regulatory proteins make the histone protein unstable so that they are removed from the DNA

b)    Regulatory proteins help in the expression of DNA

c)    Histone proteins are considered as repressor proteins as they do not allow the DNA to express

d)    Regulatory proteins are also known as derepressor proteins

e)    All of the above

Q9. What is the diameter of a double-stranded DNA?

a)    2 nm

b)    3 nm

c)    2 angstrom

d)    4 nm

e)    None of the above

Q10. Out of the five histone proteins, two molecules of each H2A, H2B, H3 and H4 form a structure called:

a)    Octamer

b)    Nu-body

c)    Core molecule

d)    Hexamer

Q11. How many turns of DNA are present in each octamer?

a)    2 complete turns

b)    1 complete turns and 3/4^th turn

c)    1 complete turn and a half turn

d)    3 complete turns

e)    3 and a half turn

Q12. Which of the following protein is involved in the clipping of DNA ends:

a)    H1

b)    H2A

c)    H3

d)    H4

e)    H2B

Q13. Nucleosome structure is formed by combining:

a)    Octamer and DNA only

b)    Octamer and H1 only

c)    Octamer, DNA and H1

d)    DNA and H1 only

e)    DNA, RNA, Octamer and H1

Q14. What is the diameter and length of the octamer?

a)    10 nm diameter and 6 nm length

b)    6 nm diameter and 10 nm length

c)    10 nm diameter and 7 nm length

d)    7 nm diameter and 10 nm length

e)    None of the above

Q15. How many base pairs of DNA are present in a nucleosome:

a)    300 bp

b)    200 bp

c)    150 bp

d)    400 bp

e)    900 bp

Q16. After the first level of packaging of DNA, beads like strings are seen under the microscope. The diameter of this string is about:

a)    10 nm

b)    15 nm

c)    10.5 nm

d)    15.5 nm

e)    30 nm

Q17. The second coiling of beads like strings forms solenoid structure or chromatin fiber with a diameter of:

a)    30 nm

b)    300 nm

c)    150 nm

d)    15 nm

e)    100 nm

Q18. In chromatin fibre or solenoid, each coil consists of:

a)    2-4 nucleosomes

b)    4 -6 nucleosomes

c)    5-6 nucleosomes

d)    6-8 nucleosomes

e)    10-12 nucleosomes

A19. The third coiling results in the formation of the super solenoid with a diameter of:

a)    200 nm

b)    300 nm

c)    400 nm

d)    500 nm

e)    600 nm

Q20. The fourth level of coiling results in the formation of:

a)    Chromatid with a diameter of 700 nm

b)    Chromatid with a diameter of 1400 nm

c)    Chromosome with a diameter of 700 nm

d)    Chromosome with a diameter of 1400 nm

e)    None of the above

Q21. Two regions are present in DNA i.e., darkly stained and lightly stained. Which of the following statement is true?

a)    The darkly stained region is called heterochromatin and it is transcriptionally active region

b)    The darkly stained region is called euchromatin and it is transcriptionally active region

c)    The light stained region is called euchromatin and it is transcriptionally inactive region

d)    The light region is euchromatin and dark region is heterochromatin where euchromtin is active and hetrochromtin is inactive

e)    All statements are correct

     Answer key: 

1. e

2. e

3. e

4. a

5. a

6. b

7. c

8. e

9. a

10. a

11. b

12. a

13. c

14. a

15. b

16. a

17. a

18. c

19. b

20. a

21. d

Explanation:
Q1. Protein size is measured in Daltons, a measure of molecular weight. One Dalton is defined as the mass of a hydrogen atom, which is 1.66 x 10^-24 grams. Most proteins have masses on the order of thousands of Daltons, so the term KD. The average weight of an amino acid is 110 Daltons – the no, of amino acids can be approximated.  Appropriate molecular weight of protein = no. of amino acids x 110 Daltons

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Concept of Hardy-Weinberg Equilibrium Explained

Before we start the Hardy-Weinberg equilibrium, let’s talk about the meaning of evolution. Does evolution mean the development of new species? Well, the answer is absolutely no. A slight change within a species from one generation to another over a long period of time can result in the gradual transition of new species.

Biological sciences define evolution as the sum total of all the genetically inherited changes in individuals who are the members of a population’s gene pool (the sum total of all the alleles in a population). It is clear that the effects of evolution are experienced by individuals, but it is the overall population that actually evolves. So, we can define evolution as the change in frequencies of alleles in the gene pool of a population.

In the early 20^th century, English mathematician Godfrey Hardy and the German physician Wilhelm Weinberg independently provided the explanation of concept called Hardy-Weinberg Equilibrium. This theory states that a population’s allele and genotype frequencies are inherently stable unless some kind of evolutionary force is acting on the population, the population would carry the same alleles in the same populations generation after generation. Individuals in a population would look essentially the same and this would be unrelated to whether the alleles are dominant or recessive. However, this concept is true only when the five evolutionary forces are absent as these forces can disrupt the equilibrium. These five forces include:

• Natural Selection
• Mutation
• Genetic Drift
• Gene migration or gene flow
• Genetic recombination

It means, if the above five factors are absent in a population, no evolution will occur in that particular population. However, in a practical sense, it is not possible that any of these factors would not be present in nature. In other words, we can say that evolution is the inevitable result.

Hardy and Weinberg developed a simple equation that can be used to find the probable genotype frequencies in a population and changes if any can be tracked from one generation to another. This is known as the Hardy-Weinberg Equilibrium equation. 

P² + 2pq + q² = 1          or, (p + q)² = 1

Where p represents dominant allele A; q represents recessive allele a

In other words, p = AA + ½ Aa;     q = aa + ½ Aa    or p + q = 1

In the above equation, p² is the predicted frequency of homozygous dominant (AA) people in a population; 2pq is the predicted frequency of heterozygous (Aa) people and q² is the predicted frequency of homozygous recessive (aa) people.

It is interesting to note that only the frequency of homozygous recessive people can be calculated. As far as a dominant trait is considered, it is represented by either homozygous dominant (p²) or heterozygous (2pq). However, using the Hardy-Weinberg equation, we can also calculate the frequency of p² and 2pq without any hassle. Since, p + q = 1 and p = 1-q where q is known, so we can calculate the frequency of p² and 2pq easily. 

Q1. In a given population, the percentage of homozygous recessive genotype (aa) is 36%. What would be the frequency of:

a)      a allele

b)      A allele

c)       AA and Aa genotype

Solution: Since the percentage of recessive genotype is given (36%), it means q² = 36% = 0.36

So, q = 0.6, and since q= a; the frequency of a = 60%

The frequency of A can be calculated from the equation p + q = 1

P = 1 – q

P = 1 – 0.6 = 0.4; so the frequency of A is 40%

The frequency of genotype AA and Aa

The frequency of AA equals to p²  (0.4x0.4 = 0.16) = 16%

The frequency of Aa equals 2pq = (2 x 0.4 x 0.6) = 0.48 = 48%

Q2. Sickle-cell anemia is a genetic disease. Normal homozygous individuals (SS) having normal red blood cells are susceptible to the malaria parasite. On the other hand, individuals with sickle-cell trait (ss) have red blood cells that easily collapse under deoxygenated condition. These individuals are resistant to malaria parasite as such parasites cannot grow in the sickle-celled red blood cells. Individuals with sickle-cell red blood cells often die because of the genetic defect. However, individuals with heterozygous condition (Ss) have some sickle-celled red blood cells, but not enough to cause mortality. Besides, malaria parasite can also survive within these partially defective red blood cells. In other words, a heterozygous condition is better than either of the homozygous conditions. If 9% of an African population is born with sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene?

Ans: ss = q² = 9% = 0.09; q = 0.3

P = 1 – q = 1 – 0.3 = 0.7

2pq = 2 x 0.7 x 0.3 = 0.42 = 42 % (heterozygous carrier)

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Examples of Homologous and Analogous Structures

What do you mean by homologous structures? Well, homology represents common ancestry. In other words, when we say homologous structure, it means we are talking about two structures which have a similar origin. However, it is important to mention that though two structures have a similar origin, that does not mean they will have a similar and function as well. For instance, thorn and tendrils of Bougainvillea and Cucurbita. Both are modifications of stem buds, but they perform different functions. For example, a thorn is a structure for protection while tendrils are structures for climbing and support to the substratum.

If we take an example in animals, humans, cheetah, whale, and bat show similar anatomical structure -- all of them possess humerus, radius, ulna, carpals, metacarpals, and phalanges, but they perform different functions.

Besides, homology represents divergent evolution because though their origins are same, but their structure and functions are different.

Now, let's discuss about analogous structures. Well, an analogy is just the opposite of homology meaning, two structures show analogy when they have a different origin, but similar structure and function. For instance, the eye of octopus and mammals; flipper of penguins and dolphins. sweet potato (root modification) and potato (stem modification). 

Octopus belongs to the phylum Mollusca and cat belongs to the phylum Mammalia. However, both possess eyes which have a similar structure and function. Besides, analogous structure leads to convergent evolution because though they have a different origin, but structurally and functionally, they are the same. 

Q. What do you think about vertebrate heart and brains? Do they show homology or analogy?

Ans: Both vertebrate heart and brain show homology. For example, chicken heart and fish heart show homology because they evolved under common ancestry as they show structures like auricle and ventricle. On the other hand, vertebrate brains are divided into forebrain, midbrain, and hindbrain - common subdivisions of the brain in all vertebrate animals. 

Now, I let you think about other homologous and analogous structure.

Heart of Fish and Heart of Crocodile

Wings of butterfly and wings of birds

Thorns of Bougainvillea and spines of Opuntia

Please write your answer in the comment box.

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Introduction to the Complementary Gene Interaction

Complementary gene interaction is an intergenic or interallelic gene interaction in which two dominant genes express themselves to the same character; independently;  however, when they are combined together, they express to a different character. In other words, two dominant genes when combined, complement to each other.

Complementary gene interaction is present in sweet pea (Lathyrus odoratus). There are two varieties of sweet pea plants -- both produce white flowers. However, when two varieties are crossed together, purple flowers are formed. I have described everything about complementary gene interaction in the video below:

Please like and subscribe to the channel for my upcoming videos on genetics and other important topics in Biology. 

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Concept of Intergenic and Intragenic Gene Interaction

There are mainly two types of gene interactions namely intragenic and intergenic gene interaction. First, we will discuss the intergenic gene interaction. It is an interaction between two different genes located on either the same chromosome or different chromosome. To understand this concept, I have discussed it in detail on my video below: 

In the case of intragenic gene interaction, it is an interaction between two alleles of a gene present on the homologous chromosome. Please watch the video until the end to understand the concept in a very easy way.

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Why are night blooming flowers generally white?

White is considered as the lightest color and achromatic without any hue as it fully reflects and scatters all the visible spectrum of light. White stands out dramatically with limited light. Besides, color, smell and overall form of any flower are produced for the sole purpose of pollination. During the daytime, flowers with colored petals generally attract butterflies, bees, and other insects. On the other hand, night-blooming flowers usually attract moths and other nocturnal pollinators. Night blooming flowers are generally white and fragrant as well because there is no purpose for them to be colorful at night. White flowers also reflect moonlight. Why they flower at night is because they avoid competition from day blooming pollinators. It’s nature’s win-win situation for plants as well as pollinating animals. 

Q. How does the bee know which flower has nectar?

is a sugar-rich liquid produced by plants in glands called nectaries. Nectar is produced by plants to attract pollinators including bees. Besides, nectar-bearing flowers are attractive and produce fragrance as well. These features attract bees to such flower-bearing plants. 

Q. Why does cactus have so many thorns?

The cactus thorns are actually modified leaves. This modification is meant to reduce the transpiration rate. Cactus grows in deserts where water is a limiting factor. If cactus has normal leaves, the loss of water through transpiration would be more. That’s why the leaves modified into spines and the stem modified to a flattened green structure responsible for photosynthesis. Besides, the thorns protect plants from animals.

Q. How does the chick recognize her own mother?

The chick recognizes her own mother through the process called imprinting that involves vision, hearing, and olfaction along with touch. Once the egg hatches and the chicks come out – they see the mother, her smell and sound get imprinted within 24 to 48 hours. This type of learning is limited to discrete and sensitive periods in an animal’s life, often when the animal is very young and it has long-lasting effects. Imprinting usually means that the animals learn to identify; approach and follow something or someone, usually a parent. And that helps the animal to find food, shelter, and warmth, etc. Chicks show this type of learning within the first 48 hours after hatching. 

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