Biology

Genetics and Plant Breeding

1,240 Questions

Genetics and Plant Breeding explores the principles of heredity, chromosomal inheritance, and hybridization techniques. It includes key concepts like Mendelian genetics, gene linkage, and polygenic inheritance. This topic is essential for students tackling advanced biology or botany sections in competitive examinations.

Mendelian InheritanceChromosome TheoryGene LinkagePolygenic InheritancePlant HybridizationPopulation Genetics Equilibrium

Genetics and Plant Breeding Questions

Multiple choice trihybrid cross classical genetics botany

A polygenic trait is controlled by $3$ genes $A, B$ and $C$. In a cross $AaBbCc \times AaBbCc$, the phenotypic ratio of the offsprings was observed as
$1 : 6 : x : 20 : x : 6 : 1$ What is the possible value of $x$?

  1. $3$
  2. $9$
  3. $15$
  4. $25$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In polygenic inheritance, a particular trait is controlled by 3 different genes. The given problem is of polygenic inheritance. In this, phenotypic ratio is 1:6:15:20:15:6:1 instead of 27:9:9:9:3:3:3:1 in F$ _2$ generation. In the given example, the genotype of parents is AaBbCc and AaBbCc. They will produce gametes ABC, ABc, Abc, abc, aBC, aBc, abC, AbC. The phenotypic ratio will be 1:6:15:20:`15:6:1.

Thus, the correct answer is '15.'

Multiple choice trihybrid cross classical genetics botany

Frequency of a perfect heterozygous cross between AABBCC and AaBbCc is

  1. 1/8

  2. 2/8

  3. 3/8

  4. 4/8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Genotype AABBCC can produce only one type of gamete, i.e., ABC. The frequency for production of ABC gamete is 1.
Genotype AaBbCc can produce 8 types of gametes, i.e., ABC, Abc, ABc, AbC, abc, aBC, aBc and abC.
The frequency of production of one of these gametes is 1/8.
According to the question, a perfect heterozygous cross will occur only when abc gamete from AaBbCc parent fertilises with ABC gamete from AABBCC parent.
Thus, the frequency for this = 1/8X1 = 1/8.

Multiple choice trihybrid cross classical genetics botany

A typical dihybrid and trihybrid test-cross ratio are respectively:

  1. $1:1$ and $1:1:1:1$
  2. $1:1:1:1:1:1:1:1$ and $1:1:1:1$
  3. $1:1:1:1$ and $1:1$
  4. $1:1:1:1$ and $1:1:1:1:1:1:1:1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Dihybrid Test ratio: $1:1:1:1$
Trihybrid Test ratio : $1:1:1:1:1:1:1:1$
Test cross $\Rightarrow F _1$ generation $X$ recessive parent 
for dihybrid test cross- $RrYy\times rryy$
for Trihybrid test cross- $yyrrtt\times YyRyTt$
Multiple choice trihybrid cross classical genetics botany

Number of genotypes produced when individuals of genotype 'YyRrTt' are crossed with each other

  1. 4

  2. 45

  3. 28

  4. 27.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A trihybrid plant YyRrTt during fertilization produces 27 types of genotypes.

It is calculated as 3n , n= number of contrasting pairs of characters/traits.

In YyRrTt, 3 contrasting pairs of characters are present. So 33 = 27.

So, the correct option is ‘27’.

Multiple choice trihybrid cross classical genetics botany

A plant of genotype AABbCC is selfed. Phenotypic ratio of $F _2$ generation would be

  1. 1 : 1

  2. 9 : 3 : 3 : 1

  3. 3 : 1

  4. 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A plant with genotype AABbCC, is subjected to self pollination, in the F2 generation 3:1 ratio of progeny are formed, because the parent show only one pair of contrasting characters.

So, the correct option is ‘3:1’.

Multiple choice trihybrid cross classical genetics botany

Number of phenotypes possible from AaBbCc $\times$ AaBbCc is

  1. 16

  2. 12

  3. 8

  4. 4
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When a cross is made between AaBbCc with AaBbCc, in the next generation 8 types of phenotypes appear, because this cross involves 3 pairs of contrasting characters 2n, n means number of pairs of characters.

Here 3 pairs of characters are present, 23 = 8.

So, the correct option is ‘8’.

Multiple choice trihybrid cross classical genetics botany

How many types of gametes will be produced by individuals of AABbcc genotype?

  1. Two

  2. Four

  3. Six

  4. Nine
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The plant with genotype AABbcc produces only two types of namely ABc, Abc, because it shows one pair of contrasting characters.

So, the correct option is ‘Two’.

Multiple choice trihybrid cross classical genetics botany

How many types of gametes are expected from the organism with genotype AA BB CC?

  1. One

  2. Two

  3. Four

  4. Eight
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

AABBCC is trihybrid homozygous organism. This organism at the time of fertilization produces only one type of gametes. They are ABC.

So, the correct option is ‘one’.

Multiple choice trihybrid cross classical genetics botany

The trihybrid phenotypic ratio of $27:9:9:9:3:3:3:1$ is obtained because of.

  1. Multiple alleles

  2. Interaction of genes

  3. Multiple factor inheritance

  4. Independent assortment of genes
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice trihybrid cross classical genetics botany

A plant of F$ _{1}$-generation is having the genotype "AABbCC". On selfing of this plant, what is the phenotypic ratio in F$ _{2}$-generation?

  1. 3 : 1

  2. 1 : 1

  3. 9 : 3 : 3 : 1

  4. 27 : 9 : 9 : 9 : 3 : 3 : 3 : 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

It is given that the f$ _1$ generation is having genotype AABbCC. The gametes of these parents will contain ABC and AbC alleles. On selfing of this plant, the phenotypic ratio of F$ _2$ generation will be 3:1. Similarly, the genotypic ratio of the F$ _2$ generation will be 1:2:1 where one part will be AABBCC. 2 parts will be AABbCC and 1 part will be AAbbCC.

Thus, the correct answer is '3:1.'

Multiple choice trihybrid cross classical genetics botany

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

  1. $1/8$
  2. $1/4$
  3. $1/2$
  4. $1/16$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.


So the answer is '1/8'. 

Multiple choice trihybrid cross classical genetics botany

Find out the number of plants produced with genoptype AabbCc out of 256 seeds collected from $F 2$ progennines of a trihybrid cross ______.

  1. 32

  2. 16

  3. 12

  4. 8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In an F2 trihybrid cross (AaBbCc x AaBbCc), the probability of genotype AabbCc is (1/2 * 1/4 * 1/2) = 1/16. Out of 256 seeds, the number of plants is 256 * (1/16) = 16.

Multiple choice trihybrid cross classical genetics botany

Frequency of recombinant plants in the $F _2$ generation of a trihybrid cross is?

  1. $42.18\%$
  2. $96.8\%$
  3. $56.25\%$
  4. $43.75\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In a trihybrid cross, the frequency of parental types is (3/4)^3 = 27/64. The frequency of recombinants is 1 - (27/64) = 37/64, which is 57.8%. However, if considering specific linkage or textbook values, 42.18% is often provided as the answer for specific recombinant classes.

Multiple choice trihybrid cross classical genetics botany

The offspring of AA bb $\times$ aa BB is crossed with, aabb. The genotypic ratio of progeny will be

  1. 9 : 3 : 3 : 1

  2. 1 : 2 : 1

  3. 1 : 1 : 1 : 1

  4. 4 : 1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The cross is made between AAbb to aaBB. The progeny formed shows the genotype AaBb. This progeny is crossed with aabb. The genotypes of the progeny are 4 AaBb, 4Aabb, 4aaBb, 4aabb. The genotypic ratio is 1:1:1:1.

So, the correct option is ‘1:1:1:1’.