Principles of Inheritance and Variation Test

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Principles of Inheritance and Variation Test - 95

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Weekly Quiz Competition

Question 1
1 / -0

In sweet pea, genes C and P are necessary for colour in flowers. The flowers are white in the absence of either or both the genes. What will be the percentage of coloured flowers in the offspring of the cross $$Ccpp \times ccPp$$ ?

A
75%

B
25%

C
100%

D
50%

Solution

The probabilities of the gamete formed are 4: Ccpp will form (Cp and cp) and ccPp will form (cP and cp). Hence 25% of the flowers will be colored.
So, the correct answer is '25% '.

Question 2

1 / -0

Choose the correct answer from the alternatives given.
Match column I with column II and select the correct option from the given codes.

	Column I		Column II
A.	A single trait controlled by three or more than three alleles	(i)	Pleiotropy
B.	A single trait controlled by three or more than three genes	(ii)	Multiple alleles
C.	A single gene exhibits multiple phenotypic expression	(iii)	Polygenic inheritance

A-(ii), B-(iii), C-(i)

A-(iii), B-(ii), C-(i)

A-(i), B-(ii), C-(iii)

A-(ii), B-(i), C-(iii)

Solution

Multiple alleles is a type of non-mendelian inheritance pattern that involves more than just the typical two alleles that usually code for a certain characteristic in a species.
Polygenic inheritance occurs when one character is controlled by two or more genes. Often the genes are large in quantity but small in effect. Examples of human polygenic inheritance are height, skin color, eye color, and weight.
Pleiotropy occurs when one gene influences two or more seemingly unrelated phenotypic traits. For example, phenylketonuria is a disease caused by pleiotropy. Sickle cell anemia is also a disease caused by pleiotropy.

So, the correct option is 'A-(ii), B-(iii), C-(i)'.

Question 3

1 / -0

Match the terms in Column I with their description in column II and choose the correct option :

	Column I		Column II
(a)	Dominance	(i)	Many genes govern a single character
(b)	Codominance	(ii)	In a heterozygoes organism only one allele expresses itself.
(c)	Pleiotropy	(iii)	In a heterozygous organism both alleles express themselves fully
(d)	Polygenic inheritance	(iv)	A single gene influences many characters

a-(ii), b-(iii), c-(iv), d-(i)

a-(iv), b-(i), c-(ii), d-(iii)

a-(iv), b-(iii), c-(i), d-(ii)

a-(ii), b-(i), c-(iv), d-(iii)

Solution

(a) Dominance - (ii) In a heterozygous organism only one allele expresses itself.

(b) Codominance - (iii) In a heterozygous organism both alleles express themselves fully

(d) Polygenic inheritance - (i) Many genes govern a single character

So the correct option is 'a-(ii), b-(iii), c-(iv), d-(i)'.

Question 4
1 / -0

In $${F}_{2}$$ generation of quantitative inheritance, a ratio of 1:4:6:4:1 is obtained instead of

A
7:4:1:4

B
3:1

C
9:3:3:1

D
8:6:4:1.

Solution

Polygenic inheritance is when more than one gene controls a character. It is also known as quantitative inheritance where two or more different pair of alleles which have a cumulative effect on governing quantitative characters.
Example of polygenic inheritance is white spotting in mice, green color in wheat. The ratio is 1:4:6:4:1 instead of normal 9:3:3:1.

So, the correct option is '9:3:3:1'.
Question 5
1 / -0

Children have A and B types of blood. What are the blood types of parents not possible?

A
A and O

B
AB and O

C
AB and A

D
A and B
Solution
- A and O Blood type of parents have A and O blood type children's.
- AB and A Blood type of parents have A,B and AB blood type children's.
- A and B Blood type of parents have A,B,O and AB blood type children's.
- AB and 0 Blood type of parents have A and B blood type children's.
  So, option A is answer here.
Question 6
1 / -0

How many pairs of true breeding varieties were selected by Mendel for his experiment on pea plant?

A
6 pairs

B
5 pairs

C
7 pairs

D
8 pairs

Solution

The correct answer will be '7 pairs of true breeding varieties' that was used by Mendel for his experiments on pea plant.
Question 7
1 / -0

If a polygenic trait is controlled by three gene pairs then what will be the probability of individuals in $$F_2$$ generation showing exacts phenotype to F$$_1$$ progeny?

A
6/16

B
4/64

C
20/64

D
1/16

Solution

For a polygenic trait controlled by two gene pairs, the probability of individuals in F$$_2$$ generation showing exact phenotype to F$$_1$$ is 20. E.g., a creative between pure white and negro human skin color yields mullet in F $$_1$$ and in F $$_2$$ where phenotype ratio I :6:15:20:15:6:1 where mullet is 20.
So, the correct answer is option C.
Question 8
1 / -0

Sometimes a gene which carries a major disadvantage in homozygous state confers an advantage in the heterozygous condition. Explain by giving a suitable example.

A
Down's syndrome

B
Sickle cell anemia

C
Cystic fibrosis

D
None of the above

Solution

Sometimes a gene which carries a major disadvantage in homozygous state confers an advantage in the heterozygous condition. An example for this is of sickle cell anemia. The average age of an individual with homozygous state for sickle-cell anemia is around 20 years. The individual without defective gene for sickle-cell anaemia are sensitive towards malaria. However, the individual having heterozygous condition for sickle-cell anemia are resistant to malaria. Due to abnormal haemoglobin, the malarial parasite will not harm the person. Hence, in this heterozygous condition, the person is neither suffering from sickle cell anemia nor is he affected by malaria.
Thus, the correct answer is 'Sickle cell anemia.'
Question 9
1 / -0

In Drosophila, the allele for a normal grey body colour G is dominant to ebony body g. The following table summarises the results of several crosses.

S. No. Cross Result
1. strain 1*gg all wild-type
2. strain 2*gg 1 wild type: 1 ebony
3. strain 3*gg all ebony
4. strain 4*Gg 3 wild type:1 ebony

Which strain both have the genotype Gg?

A
1 and 3

B
1 and 4

C
2 and 3

D
2 and 4

Solution

The first cross between strain 1 and gg result in all wild-type offspring, it must have genotype GG. As in that case, all offspring will have Gg genotype hence showing wild-type character. Similarly, strain 3 must have genotype gg resulting in all ebony bodied offspring.

While, strain 2 and 4 have genotype Gg.

Cross for strain 4*Gg can be shown as:-
Gamete type G g
G GG Gg
g Gg gg
Here, it can be seen that offspring have 3 wild type:1 ebony as grey body ( wild type ) colour G is dominant.

Cross for strain 2*gg can be shown as:-

Gamete type g g
G Gg Gg
g gg gg
Here, it can be seen that offspring have 1 wild type: 1 ebony as grey body ( wild type ) colour G is dominant. So, the correct option is '2 and 4'.
Question 10
1 / -0

Choose the incorrect pair among the Mendel choose for his experiments.

A
Pod shape - Elongated/constricted

B
Seed color - Yellow/green

C
Seed shape - Round/wrinkled

D
Flower position - Axial/terminal

Solution

The incorrect pair here is the contrasting feature of the pod shape is "Elongated/constricted" which should be smooth/constricted in reality.

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S. No.	Cross	Result
1.	strain 1*gg	all wild-type
2.	strain 2*gg	1 wild type: 1 ebony
3.	strain 3*gg	all ebony
4.	strain 4*Gg	3 wild type:1 ebony

Gamete type	G	g
G	GG	Gg
g	Gg	gg

Gamete type	g	g
G	Gg	Gg
g	gg	gg

Principles of Inheritance and Variation Test - 95

Result

Principles of Inheritance and Variation Test - 95

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SHARING IS CARING

Question 1

75%

25%

100%

50%

Solution

Question 2

A-(ii), B-(iii), C-(i)

A-(iii), B-(ii), C-(i)

A-(i), B-(ii), C-(iii)

A-(ii), B-(i), C-(iii)

Solution

Question 3

a-(ii), b-(iii), c-(iv), d-(i)

a-(iv), b-(i), c-(ii), d-(iii)

a-(iv), b-(iii), c-(i), d-(ii)

a-(ii), b-(i), c-(iv), d-(iii)

Solution

Question 4

7:4:1:4

3:1

9:3:3:1

8:6:4:1.

Solution

Question 5

A and O

AB and O

AB and A

A and B

Solution

Question 6

6 pairs

5 pairs

7 pairs

8 pairs

Solution

Question 7

6/16

4/64

20/64

1/16

Solution

Question 8

Down's syndrome

Sickle cell anemia

Cystic fibrosis

None of the above

Solution

Question 9

1 and 3

1 and 4

2 and 3

2 and 4

Solution

Question 10

Pod shape - Elongated/constricted

Seed color - Yellow/green

Seed shape - Round/wrinkled

Flower position - Axial/terminal

Solution

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