Principles of Inheritance & Variation Test

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Principles of Inheritance & Variation Test - 3

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Weekly Quiz Competition

Question 1
4 / -1

"When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters is independent of the other pair of characters." The statement is independent of the other pair of characters. The statement explains which of the following laws/principles of Mendel?

A
Principle of paired factors

B
Principle of dominance

C
Law of segregation

D
Law of independent assortment

Solution

Cross involving two contrasting character is called dihybrid cross or a two factor cross. The two factors of each trait assort at random and independent of the factors of other traits at the time of meiosis (gametogenesis) and get randomly as well as independently rearranged in the offspring producing both parental and new combinations of traits. This forms the basis of the law of independent assortment given by Mendel.
Question 2
4 / -1

Match column -I with column-ii and select the correct option from the codes given below

A
A - (iii), B - (iv), C - (ii), D - (v)

B
A - (i), B - (iv), C - (ii), D - (v)

C
A - (iii), B - (ii), C - (iv), D - (v)

D
A - (ii), B - (v), C - (iii), D - (i)
Question 3
4 / -1

In Mendelian dihybrid cross, when heterozygous Round Yellow are self crossed, Round Green offsprings are represented by the genotype

A
RrYy, RrYY, RRYy

B
Rryy, Rryy, rryy

C
rrYy, rrYY

D
Rryy, RRyy

Solution

A typical Mendelian dihybrid cross between yellow round (dominant) and green wrinkled (recessive plants) can be represented as follows:
Thus, round green (green round) offsprings are represented by genotypes yyRR and yyRr (or RRyy and Rryy).
Question 4
4 / -1

A man having the genotype EEFfGgHH can produce P number of genetically different sperms, and a woman of genotype liLLMmNn can generate Q number of genetically different eggs. Determine the values P and Q.

A
P = 4, Q = 4

B
P = 4,Q = 8

C
P = 8, Q = 4

D
P = 8,Q = 8

Solution

Types of gametes = 2ⁿ, where n is number of heterozygous loci. Thus, man will produe 2²= 4 types of gametes and woman will produce 2³= 8 types of gametes.
Question 5
4 / -1

When a cross is made between a tall plant with yellow seeds (Tt Yy) and a tall plant with green seeds (Tt yy), what is true regarding the proportions of phenotypes of the offsprings in F₁ generation?

A
a

B
b

C
c

D
d

Solution

A cross between a tall plant with yellow seeds (TtYy) and a tall plant with green seeds (Ttyy) will be:
Pnenotypic ratio is
Tall and yellow : Tall and green : Dwarf and yellow : Dwarf and green :: 3 : 3: 1: 1 or 3/8, 3/8,1/8,1/8.
Question 6
4 / -1

Refer the given statements and select the correct option,
(i) Percentage of homozygous dominant individuals obtained by selfing Aa individuals is 25%.
(ii) Types of genetically different gametes produced by genotype AABbcc are 2.
(iii) Phenotypic ratio of monohybrid F₂ progeny in case of Mirabilis jalapa is 3:1.

A
All the statements are true

B
Statements (i) and (ii) are true, but statement (iii) is false

C
Statements (i) and (iii) are true, but statement (ii) is false

D
Statements (ii) and (iii) are true, but statement (i) is false

Solution

The phenotypic ratio of monohybrid F₂ generation in case of Mirabilis jalapa is 1 Red : 2 Pink : 1 White, due to incomplete dominance.
Question 7
4 / -1

Law of independent assortment can be explained with the help of

A
dihybrid cross

B
test cross

C
back cross

D
monohybrid cross

Solution

The law of independent assortment applies only to factor or genes present on different pairs or distantly on the same chromosome or pairs of homologous chromosomes. The principle or law of independent assortment can be studied by means of dihybrid cross, e.g., between pure breeding pea plants having yellow round seeds (YYRR) and pure breeding pea plants having green wrinkled seeds (yyrr).
Question 8
4 / -1

The percentage of ab gamete produced by AaBb parent will be

A
25%

B
50%

C
75%

D
12.5%

Solution

Gametes produced by AaBb parent would be 25 % AB, 25 % aB, 25 % Ab and 25 % ab.
Question 9
4 / -1

The given Punnett's square represents the pattern of inheritance in a dihybrid cross where yellow (Y) and round (R) seed condition is dominant over white (y) and wrinkled (r) seed condition.
A plant of type 'H' will produce seeds with the genotype identical to seeds produced by the plants of

A
Type M

B
Type J

C
Type P

D
Type N

Solution

Plant H is formed by fusion of gametes yR and YR and hence has the genotype YyRR. Plant N is formed by fusion of gametes YR and yR and hence will have the same genotype as plant N i.e., YyRR.
Question 10
4 / -1

In maize, coloured endosperm (C) is dominant over colourless (c); and full endosperm (R) is dominant over shrunken (r). When a dihybrid of F₁generation were test crossed, it produced four phenotypes in the following percentage:
Coloured full 48%, Coloured shrunken 5%,
Colourless full 7%, and Colourless shrunken 40%.
From this data, what will be the distance between two non-allelic genes?

A
48 units

B
5 units

C
7 units

D
12 units

Solution

Given that recombination percentage is 7% and 5%, therefore, total recombinants would be 7 + 5 = 12%. It is known that one map unit is the distance that yields 1% recombinant chromosomes. Hence the distance between two non-allelic genes is 12 map units.
Question 11
4 / -1

What proportion of the offsprings obtained from cross AABBCC x AaBbCc will be completely heterozygous for all the genes segregated independently?

A
1/8

B
1/4

C
1/2

D
1/16

Solution

Since one parent is homozygous AABBCC the genotype formed will be only 1 type i.e. ABC. The other parent is heterozygous for all genes. About 1/8th or 12.5% of total offspring will be heterozygous at all three-locus.
Question 12
4 / -1

True-breeding red-eyed Drosophila flies with plain thoraxes were crossed with pink-eyed flies with striped thoraxes
Red eye plain thorax × Pink eye striped thorax
The F₁ flies were then test crossed against the double recessive.
The following F₂ generation resulted from the cross:
What percentage number of recombinants resulted from the test cross ?

A
12

B
14

C
16

D
28

Solution

Recombinants obtained from the cross are 16 red eye striped thorax flies and 12 pink eye plain thorax flies.
Total number of flies =200
Total number of recombinants =28
Therefore, percentage of recombinants
Question 13
4 / -1

Mendel law of independent assortment does not hold true for the genes that are located closely on

A
same chromosome

B
non-homologous chromosomes

C
X - chromosome

D
autosomes

Solution

As per linkage experiments carried out by Morgan, the two linked genes do not always segregate independently of each other and F₂ ratio deviated very significantly from 9:3:3:1 ratio (expected when two genes are independent). Hence, if linkage was known at the time of Mendel, he would riot have been able to explain law of independent assortment.
Question 14
4 / -1

Read the given paragraph to answer
In a certain plant, yellow fruit colour (Y) is dominant to green fruit colour (y) and round shape (R) is dominant to oval shape (r). The two genes involved are located on different chromosomes.
Which of the following will result when plant YyRr is self-pollinated?

A
9:3:3:1 ratio of phenotypes only

B
9:3:3:1 ratio of genotypes only

C
1:1:1:1 ratio of phenotypes only

D
1:1:1:1 ratio of phenotypes and genotypes

Solution

When plant YyRr is self pollinated, 9:3:3:1 ratio of phenotypes will be observed.
Question 15
4 / -1

Read the given paragraph to answer
In a certain plant, yellow fruit colour (Y) is dominant to green fruit colour (y) and round shape (R) is dominant to oval shape (r). The two genes involved are located on different chromosomes.
Which of the following is correct for the condition when plant YyRr is back crossed with the double recessive parent?

A
9:3:3:1 ratio of phenotypes only

B
9:3:3:1 ratio of genotypes only

C
1:1:1:1 ratio of phenotypes only

D
1:1:1:1 ratio of phenotypes and genotypes

Solution

When YyRr is crossed with double recessive parent then genotypic and phenotypic ratio will be 1:1:1:1
Phenotypic ratio = 1 yellow round : 1 yellow oval : 1 green round : 1 green and oval. This represent a test cross.