Principles of Inheritance and Variation Test

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Principles of Inheritance and Variation Test - 83

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Weekly Quiz Competition

Question 1
1 / -0

Mendel performed his experiment on garden pea species and gave rules of heridity. Give the scientific name of the species.

A
Zea mays

B
Pisum sativum

C
Phaseolus mungo

D
Rana tigrina

Solution

The scientific name of the garden pea plant on which Mendel performed his experiments is Pisum sativum.
So, the correct answer is 'Pisum sativum'.
Question 2
1 / -0

How many true breeding pea plant varieties did Mendel select as pairs, which were similar except in onecharacter with contrasting traits?

A
2

B
14

C
8

D
4
Solution
- Mendel selected 14 True breeding plant varieties.
So, the correct answer is (B).
Question 3
1 / -0

In a human male, a sperm will contain :

A
Both X and Y chromosomes

B
Only Y Chromosomes

C
Only X chromosome

D
Either X or Y Chromosomes

Solution

The human sperm is the male gamete formed after meiotic division and hence it contains half the number of autosomes and allosomes. The human males are heterogametic, with X and Y chromosomes as the allosome pair. Since, gamete consists of half the number of chromosomes. The sperm wil have either X or Y chromosome.
So, the correct answer is 'Either X or Y Chromosomes'
Question 4
1 / -0

How many different types of gametes can be formed by $${F_1}$$ progeny, resulting from the following cross:
AA BB CC $${\times}$$ aa bb cc.

A
3

B
8

C
27

D
64
Solution
- The F$$_{1}$$ generation formed by the cross between
- AA BB CC x aa bb cc is Aa Bb Cc.
- Here, there are three different types of genes involved. The maximum number of gametes possible is given by 2$$^n$$.
- In the given case, n = 3. Therefore the maximum number of combinations = 2$$^n$$ = 8.
- The possible combinations: ABC, ABc, AbC, Abc, aBC, aBc, abC and abc.
  
  Hence, option b is correct.
Question 5
1 / -0

Phenotypic and genotypic ratios according to Mendel's pattern of inheritance is

A
Different

B
Same

C
Average

D
None of the above

Solution

The genotype ratio is the set of genes in our DNA which is responsible for a particular trait. The phenotype ratio is the physical expression, or characteristics, of that trait. For example, two organisms that have even the minutest difference in their genes are said to have different genotypes. Tt and TT are two different genotypes. So, the ratio is 1:1 but the phenotypes for them are same i.e., tall. Hence, the genotypic and phenotypic ratios can be different.
So, the correct answer is option A.
Question 6
1 / -0

Mrs. Verma has an autosomal gene pair 'Bb' and she contain x-linked gene 'd'. What is the percentage of gamete, which contain 'bd' genes?

A
1/2 or 50%

B
1/4 or 25%

C
3/4 or 75%

D
1 or 100%

Solution

It has been given, that the gametes produced by a female of genotype Bb is to be investigated. In addition to it, it is also given that she contains an X-linked gene 'd'. Thus, when we analyze the sex chromosome combination of human females - XX. Thus, the female will have 50% chance of bearing ovum with Bd genes and another 50% with bd genes.
Question 7
1 / -0

Which of the following discoveries resulted in a Nobel Prize?

A
Genetic engineering

B
X-rays induce sex-linked recessive lethal mutations

C
Cytoplasmic inheritance

D
Recombination of linked genes

Solution

H. J. Muller in 1946 was awarded Nobel Prize for his discovery of the production of mutation by X-ray irradiation.
He was an American geneticist, educator, and Nobel laureate best known for his work on the physiological and genetic effects of radiation, which in the following years led to increased scrutiny to the use of radioactive elements and safety procedures.
Therefore, the correct answer is option B.

Question 8

1 / -0

Match the following and select the correct match.

	List I		List II
A	Mendel performed .................. between plants having alternate traits of a character, e.g., tall and dwarf, coloured flower and white flower.	1	Second filial
B	For this he removed .................. (emasculation) from 50% of the flowers of each type and used them as female parents.	2	Reciprocal crosses
C	.................. from the other flower were collected and dusted over the stigma of these flowers. The pollinated flowers were covered with paper bags and allowed to form pods with seeds.	3	Monohybrid cross
D	The cross involving two traits of a single character is called .....................	4	First filial generation
E	Cross involving traits of two characters is known as ..................	5	Anthers
F	The seeds of the cross were collected and sown next year. The plants were hybrid in nature. They constituted the ................ or $$F_1$$ generation.	6	Pollen
G	Self Breeding for $$F_2$$ and $$F_3$$ generations. Plants of $$F_1$$ generation were allowed to perform ................... or self-fertilization.	7	Third filial generation
H	Mendel collected seeds of the new generations. Plants raised from them were called ...................	8	Dihybrid cross
I	or $$F_2$$ ..............	9	Self-pollination
J	Further self-pollination of $$F_2$$ plants produced $$F_3$$	10	Generation

A- 2, B- 5, C- 6, D- 3, E- 8, F- 4, G- 9, H- 1, I- 10, J- 7

A- 2, B- 5, C- 6, D- 3, E- 8, F- 9, G- 4, H- 1, I- 10, J- 7

A- 2, B- 5, C- 6, D- 8, E- 3, F- 4, G- 9, H- 1, I- 10, J- 7

A- 2, B- 6, C- 5, D- 3, E- 8, F- 4, G- 9, H- 1, I- 10, J- 7

Solution

Mendel performed Reciprocal crosses between plants having alternate traits of a character, e.g., tall and dwarf, coloured flower and white flower.

For this he removed Anthers (emasculation) from 50% of the flowers of each type and used them as female parents.

Pollen from the other trait were collected and dusted over the stigma of these flowers. The pollinated flowers were covered with paper bags and allowed to form pods with seeds.

The cross involving two traits of a single character is called Monohybrid cross

Cross involving traits of two characters is known as Dihybrid cross

The seeds of the cross were collected and sown next year. The plants were hybrid in nature. They constituted the First filial generation or F1 generation.

Self Breeding for F2 and F3 generations. Plants of F1 generation were allowed to perform Self pollination or self fertilization.

Mendel collected seeds of the new generations. Plants raised from them were called Second filial

or F2 Generation

Further self pollination of F2 plants produced F3 Third filial generation

Question 9
1 / -0

The genome of Caenorhabditis elgans consists of how many base pairs and genes?

A
3 billion base pairs and 30,000 genes

B
180 million base pairs and 13,000 genes

C
4.7 million base pairs and 4,000 genes

D
97 million base pairs and 18,000 genes

E
12 million base pairs and 6,000 genes

Solution

The Caenorhabditis elegans genome sequence is now complete, fully contiguous telomere to telomere and totaling 97 million base pairs. The sequence has catalyzed the collection of systematic data sets and analyses, including protein-coding genes, giving a total number of 18000 genes.
Question 10
1 / -0

Human skin colour is controlled by several gene pairs. Let us assume here, that there are just three gene pairs on different chromosomes and that for each pair there are two alleles - an incompletely dominant one that codes for melanin deposition and an incompletely recessive one that codes for no melanin deposition. If a very dark skinned person marries a very light skinned woman, what will be the chance that their offspring will have very dark skin?

A
0

B
1/4

C
5/8

D
9/64

Solution

Human skin color is a polygenic trait, which is controlled by a number of genes. According to the question, there are just three gene pairs on different chromosomes and that for each pair there are two alleles - an incompletely dominant one that codes for melanin deposition and an incompletely recessive one that codes for no melanin deposition.
Let the dominant allele for the three genes controlling the skin color be represented as M, N and O and their respective recessive forms can be represented as m, n and o.
Since, these alleles exhibit incomplete dominance, a very dark skinned man should have all the alleles in dominant form and must have the genotype MMNNOO. Similarly, a very light skinned woman should have all the alleles in recessive form mmnnoo.
P generation: MMNNOO X mmnnoo
Gametes: MNO mno
F$$_{1}$$ generation: MmNnOo
None of the offsprings have very dark skin because the allele for melanin deposition is incompletely dominant.

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Re-Attempt Weekly Quiz Competition

Principles of Inheritance and Variation Test - 83

Result

Principles of Inheritance and Variation Test - 83

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SHARING IS CARING

Question 1

Zea mays

Pisum sativum

Phaseolus mungo

Rana tigrina

Solution

Question 2

2

14

8

4

Solution

Question 3

Both X and Y chromosomes

Only Y Chromosomes

Only X chromosome

Either X or Y Chromosomes

Solution

Question 4

3

8

27

64

Solution

Question 5

Different

Same

Average

None of the above

Solution

Question 6

1/2 or 50%

1/4 or 25%

3/4 or 75%

1 or 100%

Solution

Question 7

Genetic engineering

X-rays induce sex-linked recessive lethal mutations

Cytoplasmic inheritance

Recombination of linked genes

Solution

Question 8

A- 2, B- 5, C- 6, D- 3, E- 8, F- 4, G- 9, H- 1, I- 10, J- 7

A- 2, B- 5, C- 6, D- 3, E- 8, F- 9, G- 4, H- 1, I- 10, J- 7

A- 2, B- 5, C- 6, D- 8, E- 3, F- 4, G- 9, H- 1, I- 10, J- 7

A- 2, B- 6, C- 5, D- 3, E- 8, F- 4, G- 9, H- 1, I- 10, J- 7

Solution

Question 9

3 billion base pairs and 30,000 genes

180 million base pairs and 13,000 genes

4.7 million base pairs and 4,000 genes

97 million base pairs and 18,000 genes

12 million base pairs and 6,000 genes

Solution

Question 10

0

1/4

5/8

9/64

Solution

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