QBManager

Principle of Inheritance and Variation

184283 When a tall pea plant crossed with dwarf pea plant the $F_{2}$ generation produced 390 tall and 130 dwarf plants, then how many plants will show Heterozygous condition.

1 130

2 260

3 390

4 520

Explanation:

$TT \Rightarrow$ Puretall $t$ Tt $tt$ ratio $\Rightarrow 1 : 2 : 1$
$Tt \Rightarrow$ Heterozygous $F_{2}$ generation
$tt \Rightarrow$ Purl dwarf
According to question $=$ pure fall $\Rightarrow 390$
pure dwarf $\Rightarrow 160$
Heterozygous $\Rightarrow 390 - 130 \Rightarrow 260$

Shift-II

Principle of Inheritance and Variation

184285 Heterozygous tall plants were crossed with dwarf plants, then the percentage of dwarf plants produced in the next progeny

1 25

2 75

3 50

4 65

Explanation:

Heterozygous tall plants were crossed with
Homozygous dwarf plants, then $50 %$ of dwarf plant and
$50 %$ of tall plant produced in the next progeny.
Tall
{{|c|c|c|}
| | | |
|---|---|---|
| | | |
| | | |
|
Tall $(Tt) - 50 %$ dominant
Dwraf $(tt) - 50 %$ recessive.

AP EAPCET-23.05.2023

Principle of Inheritance and Variation

184288 The ratio of homozygous tall and dwarf plant in $F_{2}$ generation in monohybrid cross is

1

2 : 1

2

1 : 2 : 1

3

3 : 1

4

1 : 1

Explanation:

The ratio of homozygous tall and dwraf plant in $F_{2}$ generation in monophbrid cross is $1 : 1$
{|c|c|c|}
|{c}{{c}
|
|---|
Tall |
|} & \multicolumn{2}{c}{ Dwarf } \
\hline & $T t$ \
\hline $t$ & $Tt$ & $t$ \
\hline $t$ & $Tt$ & $tt$ \
\hline
\end{tabular}
Tall $(Tt) - 50 %$ dominant
Dwraf (tt) $- 50 %$ recessive.

Shift-II

Principle of Inheritance and Variation

184289 The experimental material in Mendels experiment was

1 Pisum sativum

2 Oryza sativa

3 Mirabilis jalapa

4 None of these

AMU-2015 / J and K CET-2007

Principle of Inheritance and Variation

184297 In a population $N_{AA}$ and $N_{aa}$ are the numbers of homozygous individuals of allele ' $A$ ' and ' $a$ ' respectively, and $N_{Aa}$ is the number of heterozygous individuals. Which one of the following options is the allele frequency of ' $A$ ' and ' $a$ ' in population with $N_{AA} = 90, N_{Aa} = 40$ and $N_{aa} = 70$ ?

1

A = 0.55

and

a = 0.45

2

A = 0.40

and

a = 0.60

3

A = 0.35

and

a = 0.65

4

A = 0.25

and

a = 0.75

Explanation:

Allele a frequency $=$
$(2 \times$ numbers of AA homozygote $) +$ (Number of Aa heterozyote)
( $2 \times$ total number of individual)
Then, $F (A) = \frac{2 NAA + Aa}{2 N}$
Total number of individuals in population $= N$
$\begin{aligned} N = 90 + 40 + 70 = 200 \\ A = \frac{2 \times 90 + 40}{2 \times 200} \\ A = \frac{180 + 40}{400} = \frac{220}{400} = 0.55 \end{aligned}$
Since allelic frequency of ' $A$ ' $= 0.55$ .
Frequency of allele ' $a$ ' $= \frac{2 Naa + Aa}{2 N}$
$\begin{aligned} a & = \frac{2 \times 70 + 40}{2 \times 200} \\ a & = \frac{140 + 40}{400} = \frac{180}{400} = 0.45 \\ So, a^{'} & = 0.45 \end{aligned}$
Therefore $A = 0.55$ and $a = 0.45$ is the correct answer.

KVPY SB \& SX-2019

Principle of Inheritance and Variation

184283 When a tall pea plant crossed with dwarf pea plant the $F_{2}$ generation produced 390 tall and 130 dwarf plants, then how many plants will show Heterozygous condition.

1 130

2 260

3 390

4 520

Explanation:

$TT \Rightarrow$ Puretall $t$ Tt $tt$ ratio $\Rightarrow 1 : 2 : 1$
$Tt \Rightarrow$ Heterozygous $F_{2}$ generation
$tt \Rightarrow$ Purl dwarf
According to question $=$ pure fall $\Rightarrow 390$
pure dwarf $\Rightarrow 160$
Heterozygous $\Rightarrow 390 - 130 \Rightarrow 260$

Shift-II

Principle of Inheritance and Variation

184285 Heterozygous tall plants were crossed with dwarf plants, then the percentage of dwarf plants produced in the next progeny

1 25

2 75

3 50

4 65

Explanation:

Heterozygous tall plants were crossed with
Homozygous dwarf plants, then $50 %$ of dwarf plant and
$50 %$ of tall plant produced in the next progeny.
Tall
{{|c|c|c|}
| | | |
|---|---|---|
| | | |
| | | |
|
Tall $(Tt) - 50 %$ dominant
Dwraf $(tt) - 50 %$ recessive.

AP EAPCET-23.05.2023

Principle of Inheritance and Variation

184288 The ratio of homozygous tall and dwarf plant in $F_{2}$ generation in monohybrid cross is

1

2 : 1

2

1 : 2 : 1

3

3 : 1

4

1 : 1

Explanation:

The ratio of homozygous tall and dwraf plant in $F_{2}$ generation in monophbrid cross is $1 : 1$
{|c|c|c|}
|{c}{{c}
|
|---|
Tall |
|} & \multicolumn{2}{c}{ Dwarf } \
\hline & $T t$ \
\hline $t$ & $Tt$ & $t$ \
\hline $t$ & $Tt$ & $tt$ \
\hline
\end{tabular}
Tall $(Tt) - 50 %$ dominant
Dwraf (tt) $- 50 %$ recessive.

Shift-II

Principle of Inheritance and Variation

184289 The experimental material in Mendels experiment was

1 Pisum sativum

2 Oryza sativa

3 Mirabilis jalapa

4 None of these

AMU-2015 / J and K CET-2007

Principle of Inheritance and Variation

184297 In a population $N_{AA}$ and $N_{aa}$ are the numbers of homozygous individuals of allele ' $A$ ' and ' $a$ ' respectively, and $N_{Aa}$ is the number of heterozygous individuals. Which one of the following options is the allele frequency of ' $A$ ' and ' $a$ ' in population with $N_{AA} = 90, N_{Aa} = 40$ and $N_{aa} = 70$ ?

1

A = 0.55

and

a = 0.45

2

A = 0.40

and

a = 0.60

3

A = 0.35

and

a = 0.65

4

A = 0.25

and

a = 0.75

Explanation:

Allele a frequency $=$
$(2 \times$ numbers of AA homozygote $) +$ (Number of Aa heterozyote)
( $2 \times$ total number of individual)
Then, $F (A) = \frac{2 NAA + Aa}{2 N}$
Total number of individuals in population $= N$
$\begin{aligned} N = 90 + 40 + 70 = 200 \\ A = \frac{2 \times 90 + 40}{2 \times 200} \\ A = \frac{180 + 40}{400} = \frac{220}{400} = 0.55 \end{aligned}$
Since allelic frequency of ' $A$ ' $= 0.55$ .
Frequency of allele ' $a$ ' $= \frac{2 Naa + Aa}{2 N}$
$\begin{aligned} a & = \frac{2 \times 70 + 40}{2 \times 200} \\ a & = \frac{140 + 40}{400} = \frac{180}{400} = 0.45 \\ So, a^{'} & = 0.45 \end{aligned}$
Therefore $A = 0.55$ and $a = 0.45$ is the correct answer.

KVPY SB \& SX-2019

Principle of Inheritance and Variation

184283 When a tall pea plant crossed with dwarf pea plant the $F_{2}$ generation produced 390 tall and 130 dwarf plants, then how many plants will show Heterozygous condition.

1 130

2 260

3 390

4 520

Explanation:

$TT \Rightarrow$ Puretall $t$ Tt $tt$ ratio $\Rightarrow 1 : 2 : 1$
$Tt \Rightarrow$ Heterozygous $F_{2}$ generation
$tt \Rightarrow$ Purl dwarf
According to question $=$ pure fall $\Rightarrow 390$
pure dwarf $\Rightarrow 160$
Heterozygous $\Rightarrow 390 - 130 \Rightarrow 260$

Shift-II

Principle of Inheritance and Variation

184285 Heterozygous tall plants were crossed with dwarf plants, then the percentage of dwarf plants produced in the next progeny

1 25

2 75

3 50

4 65

Explanation:

Heterozygous tall plants were crossed with
Homozygous dwarf plants, then $50 %$ of dwarf plant and
$50 %$ of tall plant produced in the next progeny.
Tall
{{|c|c|c|}
| | | |
|---|---|---|
| | | |
| | | |
|
Tall $(Tt) - 50 %$ dominant
Dwraf $(tt) - 50 %$ recessive.

AP EAPCET-23.05.2023

Principle of Inheritance and Variation

184288 The ratio of homozygous tall and dwarf plant in $F_{2}$ generation in monohybrid cross is

1

2 : 1

2

1 : 2 : 1

3

3 : 1

4

1 : 1

Explanation:

The ratio of homozygous tall and dwraf plant in $F_{2}$ generation in monophbrid cross is $1 : 1$
{|c|c|c|}
|{c}{{c}
|
|---|
Tall |
|} & \multicolumn{2}{c}{ Dwarf } \
\hline & $T t$ \
\hline $t$ & $Tt$ & $t$ \
\hline $t$ & $Tt$ & $tt$ \
\hline
\end{tabular}
Tall $(Tt) - 50 %$ dominant
Dwraf (tt) $- 50 %$ recessive.

Shift-II

Principle of Inheritance and Variation

184289 The experimental material in Mendels experiment was

1 Pisum sativum

2 Oryza sativa

3 Mirabilis jalapa

4 None of these

AMU-2015 / J and K CET-2007

Principle of Inheritance and Variation

184297 In a population $N_{AA}$ and $N_{aa}$ are the numbers of homozygous individuals of allele ' $A$ ' and ' $a$ ' respectively, and $N_{Aa}$ is the number of heterozygous individuals. Which one of the following options is the allele frequency of ' $A$ ' and ' $a$ ' in population with $N_{AA} = 90, N_{Aa} = 40$ and $N_{aa} = 70$ ?

1

A = 0.55

and

a = 0.45

2

A = 0.40

and

a = 0.60

3

A = 0.35

and

a = 0.65

4

A = 0.25

and

a = 0.75

Explanation:

Allele a frequency $=$
$(2 \times$ numbers of AA homozygote $) +$ (Number of Aa heterozyote)
( $2 \times$ total number of individual)
Then, $F (A) = \frac{2 NAA + Aa}{2 N}$
Total number of individuals in population $= N$
$\begin{aligned} N = 90 + 40 + 70 = 200 \\ A = \frac{2 \times 90 + 40}{2 \times 200} \\ A = \frac{180 + 40}{400} = \frac{220}{400} = 0.55 \end{aligned}$
Since allelic frequency of ' $A$ ' $= 0.55$ .
Frequency of allele ' $a$ ' $= \frac{2 Naa + Aa}{2 N}$
$\begin{aligned} a & = \frac{2 \times 70 + 40}{2 \times 200} \\ a & = \frac{140 + 40}{400} = \frac{180}{400} = 0.45 \\ So, a^{'} & = 0.45 \end{aligned}$
Therefore $A = 0.55$ and $a = 0.45$ is the correct answer.

KVPY SB \& SX-2019

Principle of Inheritance and Variation

184283 When a tall pea plant crossed with dwarf pea plant the $F_{2}$ generation produced 390 tall and 130 dwarf plants, then how many plants will show Heterozygous condition.

1 130

2 260

3 390

4 520

Explanation:

$TT \Rightarrow$ Puretall $t$ Tt $tt$ ratio $\Rightarrow 1 : 2 : 1$
$Tt \Rightarrow$ Heterozygous $F_{2}$ generation
$tt \Rightarrow$ Purl dwarf
According to question $=$ pure fall $\Rightarrow 390$
pure dwarf $\Rightarrow 160$
Heterozygous $\Rightarrow 390 - 130 \Rightarrow 260$

Shift-II

Principle of Inheritance and Variation

184285 Heterozygous tall plants were crossed with dwarf plants, then the percentage of dwarf plants produced in the next progeny

1 25

2 75

3 50

4 65

Explanation:

Heterozygous tall plants were crossed with
Homozygous dwarf plants, then $50 %$ of dwarf plant and
$50 %$ of tall plant produced in the next progeny.
Tall
{{|c|c|c|}
| | | |
|---|---|---|
| | | |
| | | |
|
Tall $(Tt) - 50 %$ dominant
Dwraf $(tt) - 50 %$ recessive.

AP EAPCET-23.05.2023

Principle of Inheritance and Variation

184288 The ratio of homozygous tall and dwarf plant in $F_{2}$ generation in monohybrid cross is

1

2 : 1

2

1 : 2 : 1

3

3 : 1

4

1 : 1

Explanation:

The ratio of homozygous tall and dwraf plant in $F_{2}$ generation in monophbrid cross is $1 : 1$
{|c|c|c|}
|{c}{{c}
|
|---|
Tall |
|} & \multicolumn{2}{c}{ Dwarf } \
\hline & $T t$ \
\hline $t$ & $Tt$ & $t$ \
\hline $t$ & $Tt$ & $tt$ \
\hline
\end{tabular}
Tall $(Tt) - 50 %$ dominant
Dwraf (tt) $- 50 %$ recessive.

Shift-II

Principle of Inheritance and Variation

184289 The experimental material in Mendels experiment was

1 Pisum sativum

2 Oryza sativa

3 Mirabilis jalapa

4 None of these

AMU-2015 / J and K CET-2007

Principle of Inheritance and Variation

184297 In a population $N_{AA}$ and $N_{aa}$ are the numbers of homozygous individuals of allele ' $A$ ' and ' $a$ ' respectively, and $N_{Aa}$ is the number of heterozygous individuals. Which one of the following options is the allele frequency of ' $A$ ' and ' $a$ ' in population with $N_{AA} = 90, N_{Aa} = 40$ and $N_{aa} = 70$ ?

1

A = 0.55

and

a = 0.45

2

A = 0.40

and

a = 0.60

3

A = 0.35

and

a = 0.65

4

A = 0.25

and

a = 0.75

Explanation:

Allele a frequency $=$
$(2 \times$ numbers of AA homozygote $) +$ (Number of Aa heterozyote)
( $2 \times$ total number of individual)
Then, $F (A) = \frac{2 NAA + Aa}{2 N}$
Total number of individuals in population $= N$
$\begin{aligned} N = 90 + 40 + 70 = 200 \\ A = \frac{2 \times 90 + 40}{2 \times 200} \\ A = \frac{180 + 40}{400} = \frac{220}{400} = 0.55 \end{aligned}$
Since allelic frequency of ' $A$ ' $= 0.55$ .
Frequency of allele ' $a$ ' $= \frac{2 Naa + Aa}{2 N}$
$\begin{aligned} a & = \frac{2 \times 70 + 40}{2 \times 200} \\ a & = \frac{140 + 40}{400} = \frac{180}{400} = 0.45 \\ So, a^{'} & = 0.45 \end{aligned}$
Therefore $A = 0.55$ and $a = 0.45$ is the correct answer.

KVPY SB \& SX-2019

Principle of Inheritance and Variation

184283 When a tall pea plant crossed with dwarf pea plant the $F_{2}$ generation produced 390 tall and 130 dwarf plants, then how many plants will show Heterozygous condition.

1 130

2 260

3 390

4 520

Explanation:

$TT \Rightarrow$ Puretall $t$ Tt $tt$ ratio $\Rightarrow 1 : 2 : 1$
$Tt \Rightarrow$ Heterozygous $F_{2}$ generation
$tt \Rightarrow$ Purl dwarf
According to question $=$ pure fall $\Rightarrow 390$
pure dwarf $\Rightarrow 160$
Heterozygous $\Rightarrow 390 - 130 \Rightarrow 260$

Shift-II

Principle of Inheritance and Variation

184285 Heterozygous tall plants were crossed with dwarf plants, then the percentage of dwarf plants produced in the next progeny

1 25

2 75

3 50

4 65

Explanation:

Heterozygous tall plants were crossed with
Homozygous dwarf plants, then $50 %$ of dwarf plant and
$50 %$ of tall plant produced in the next progeny.
Tall
{{|c|c|c|}
| | | |
|---|---|---|
| | | |
| | | |
|
Tall $(Tt) - 50 %$ dominant
Dwraf $(tt) - 50 %$ recessive.

AP EAPCET-23.05.2023

Principle of Inheritance and Variation

184288 The ratio of homozygous tall and dwarf plant in $F_{2}$ generation in monohybrid cross is

1

2 : 1

2

1 : 2 : 1

3

3 : 1

4

1 : 1

Explanation:

The ratio of homozygous tall and dwraf plant in $F_{2}$ generation in monophbrid cross is $1 : 1$
{|c|c|c|}
|{c}{{c}
|
|---|
Tall |
|} & \multicolumn{2}{c}{ Dwarf } \
\hline & $T t$ \
\hline $t$ & $Tt$ & $t$ \
\hline $t$ & $Tt$ & $tt$ \
\hline
\end{tabular}
Tall $(Tt) - 50 %$ dominant
Dwraf (tt) $- 50 %$ recessive.

Shift-II

Principle of Inheritance and Variation

184289 The experimental material in Mendels experiment was

1 Pisum sativum

2 Oryza sativa

3 Mirabilis jalapa

4 None of these

AMU-2015 / J and K CET-2007

Principle of Inheritance and Variation

184297 In a population $N_{AA}$ and $N_{aa}$ are the numbers of homozygous individuals of allele ' $A$ ' and ' $a$ ' respectively, and $N_{Aa}$ is the number of heterozygous individuals. Which one of the following options is the allele frequency of ' $A$ ' and ' $a$ ' in population with $N_{AA} = 90, N_{Aa} = 40$ and $N_{aa} = 70$ ?

1

A = 0.55

and

a = 0.45

2

A = 0.40

and

a = 0.60

3

A = 0.35

and

a = 0.65

4

A = 0.25

and

a = 0.75

Explanation:

Allele a frequency $=$
$(2 \times$ numbers of AA homozygote $) +$ (Number of Aa heterozyote)
( $2 \times$ total number of individual)
Then, $F (A) = \frac{2 NAA + Aa}{2 N}$
Total number of individuals in population $= N$
$\begin{aligned} N = 90 + 40 + 70 = 200 \\ A = \frac{2 \times 90 + 40}{2 \times 200} \\ A = \frac{180 + 40}{400} = \frac{220}{400} = 0.55 \end{aligned}$
Since allelic frequency of ' $A$ ' $= 0.55$ .
Frequency of allele ' $a$ ' $= \frac{2 Naa + Aa}{2 N}$
$\begin{aligned} a & = \frac{2 \times 70 + 40}{2 \times 200} \\ a & = \frac{140 + 40}{400} = \frac{180}{400} = 0.45 \\ So, a^{'} & = 0.45 \end{aligned}$
Therefore $A = 0.55$ and $a = 0.45$ is the correct answer.

KVPY SB \& SX-2019