Inheritance of two gene Questions in English

(D) In a Mendelian dihybrid cross involving two traits,the $F_{2}$ phenotypic ratio is $9:3:3:1$.
Total number of combinations = $16$.
Parental phenotypes (resembling parents $TTRR$ and $ttrr$) are $9$ (dominant-dominant) and $1$ (recessive-recessive),totaling $10$.
Non-parental (recombinant) phenotypes are $3$ and $3$,totaling $6$.
The ratio of parental to non-parental plants is $10:6$,which simplifies to $5:3$.
Option $D$ correctly identifies the standard Mendelian phenotypic ratio for a dihybrid cross.

(C) In a Mendelian dihybrid cross involving two traits (e.g.,seed shape and seed color),the $F_2$ generation is produced by selfing the $F_1$ dihybrid $(RrYy \times RrYy)$.
According to the Punnett square for a dihybrid cross,the total number of combinations is $16$.
The genotype $RrYy$ appears $4$ times in the $16$ possible combinations.
Therefore,the percentage is calculated as: $(4 / 16) \times 100 = 25 \,\%$.

(A) Mendel performed his dihybrid cross experiment using the garden pea plant $(Pisum \text{ } sativum)$.
He chose two pairs of contrasting traits to study the inheritance of two genes simultaneously.
The two traits selected were the seed color (yellow vs. green) and the seed shape (round vs. wrinkled).
Therefore, the correct option is $A$.

(B) In Mendel's dihybrid cross involving seed shape (Round/Wrinkled) and seed color (Yellow/Green),the parental phenotypes are Round-Yellow $(RRYY)$ and Wrinkled-Green $(rryy)$.
In the $F_2$ generation,the phenotypic ratio is $9:3:3:1$.
The total number of phenotypes is $16$.
The parental types are Round-Yellow $(9)$ and Wrinkled-Green $(1)$.
The recombinant (non-parental) types are Round-Green $(3)$ and Wrinkled-Yellow $(3)$.
Thus,there are $2$ types of offspring that are different from the parental types.

(A) Mendel performed dihybrid crosses involving two traits (e.g.,seed shape and seed color) to understand how different genes are inherited together.
Based on the results of these experiments,he proposed the $Law$ $of$ $Independent$ $Assortment$.
This law states that when two pairs of traits are combined in a hybrid,segregation of one pair of characters is independent of the other pair of characters at the time of gamete formation.

(D) In Mendel's dihybrid cross,the phenotypic ratio is $9: 3: 3: 1$.
However,the genotypic ratio for the $F_2$ generation is determined by the combination of two monohybrid crosses $(1: 2: 1) \times (1: 2: 1)$.
This results in the ratio $1: 2: 1: 2: 4: 2: 1: 2: 1$.
Therefore,the correct option is $D$.

(C) In a dihybrid cross,Mendel studied the inheritance of two pairs of contrasting traits.
For $F_2$ generation:
$1$. The phenotypic ratio is $9:3:3:1$,which results in $4$ distinct phenotypes.
$2$. The genotypic ratio is $(1:2:1:2:4:2:1:2:1)$,which results in $9$ distinct genotypes.
Therefore,the number of genotypes $(P)$ is $9$ and the number of phenotypes $(Q)$ is $4$.

(D) In Mendel's dihybrid cross,the traits for seed shape are round $(R)$ and wrinkled $(r)$,and for seed color are yellow $(Y)$ and green $(y)$.
Green and wrinkled traits are recessive,represented by the genotype $rryy$.
Looking at the Punnett square provided:
- $P$ corresponds to $RrYy$ (Round,Yellow).
- $Q$ corresponds to $Rryy$ (Round,Green).
- $R$ corresponds to $rrYy$ (Wrinkled,Yellow).
- $S$ corresponds to $rryy$ (Wrinkled,Green).
Therefore,the green and wrinkled phenotype is represented by $S$.

(C) In pea plants,green pod colour $(G)$ is dominant over yellow $(g)$,and inflated pod shape $(I)$ is dominant over constricted $(i)$.
The parental cross is between green-inflated $(GGII)$ and yellow-constricted $(ggii)$.
The $F_1$ generation is $GgIi$ (green-inflated).
In a dihybrid cross,the $F_2$ phenotypic ratio is $9:3:3:1$.
The phenotypes are:
$9$ (Green,Inflated) : $3$ (Green,Constricted) : $3$ (Yellow,Inflated) : $1$ (Yellow,Constricted).
The total number of parts is $9+3+3+1 = 16$.
The proportion of yellow and inflated pods is $3/16$.
Percentage = $(3/16) \times 100 = 18.75 \%$.

(C) According to Mendel's Law of Independent Assortment,when two pairs of contrasting traits are considered simultaneously (dihybrid cross),the inheritance of one pair of traits is independent of the other.
In this cross,the parent generation $(P)$ consists of $RRYY$ (round yellow) and $rryy$ (wrinkled green).
The $F1$ generation results in $RrYy$ (round yellow) individuals.
When $F1$ individuals $(RrYy)$ are self-crossed,the gametes produced are $RY, Ry, rY,$ and $ry$.
Using a Punnett square to cross these gametes,the resulting $F2$ phenotypic distribution is $9$ round yellow,$3$ round green,$3$ wrinkled yellow,and $1$ wrinkled green.
Therefore,the phenotypic ratio is $9: 3: 3: 1$.

(C) In a dihybrid cross,two pairs of contrasting traits are considered.
For a dihybrid cross,the number of phenotypes is calculated by the formula $2^n$,where $n$ is the number of gene pairs. Here,$n = 2$,so phenotypes = $2^2 = 4$.
The number of genotypes is calculated by the formula $3^n$. Here,$n = 2$,so genotypes = $3^2 = 9$.
Therefore,in a dihybrid Mendelian cross,there are $9$ types of genotypes and $4$ types of phenotypes.

(C) To determine the genotype,we look at the gametes produced by the individual. The gametes are $RY$ and $Ry$.
Since both gametes contain the dominant allele $R$ for the first trait,the individual must be homozygous dominant for this trait,which is $RR$.
For the second trait,the gametes contain both $Y$ and $y$ alleles. Therefore,the individual must be heterozygous for the second trait,which is $Yy$.
Combining these,the genotype of the individual is $RRYy$.

(C) In a Mendelian dihybrid cross,the $F_2$ generation consists of $16$ possible combinations of genotypes.
Among these,only $4$ plants are homozygous for both traits.
These genotypes are $YYRR$,$YYrr$,$yyRR$,and $yyrr$.
Therefore,the correct answer is $4$.

(B) This cross is a classic example of a dihybrid test cross.
In a dihybrid test cross,a dihybrid individual $(YyRr)$ is crossed with a homozygous recessive individual $(yyrr)$.
The dihybrid parent $(YyRr)$ produces four types of gametes: $YR$,$Yr$,$yR$,and $yr$ in equal proportions.
The homozygous recessive parent $(yyrr)$ produces only one type of gamete: $yr$.
When these gametes fuse,the resulting offspring genotypes are $YyRr$,$Yyrr$,$yyRr$,and $yyrr$ in a $1:1:1:1$ ratio.
These genotypes correspond to the phenotypes: Yellow-round,Yellow-wrinkled,Green-round,and Green-wrinkled,respectively.
Therefore,all four phenotypes are produced in equal proportions.

(A) The cross between two dihybrid pea plants $(RrTt \times RrTt)$ follows Mendel's Law of Independent Assortment.

	$RT$	$Rt$	$rT$	$rt$
$RT$	$RRTT$	$RRTt$	$RrTT$	$RrTt$
$Rt$	$RRTt$	$RRtt$	$RrTt$	$Rrtt$
$rT$	$RrTT$	$RrTt$	$rrTT$	$rrTt$
$rt$	$RrTt$	$Rrtt$	$rrTt$	$rrtt$

The resulting phenotypes are:
$1$. Tall plant with round seeds $(R-T-)$
$2$. Tall plant with wrinkled seeds $(R-tt)$
$3$. Dwarf plant with round seeds $(rrT-)$
$4$. Dwarf plant with wrinkled seeds $(rrtt)$
Thus,there are $4$ distinct phenotypes.

(C) In a dihybrid cross,two pairs of contrasting traits are considered. For $n$ pairs of contrasting traits,the number of genotypes is $3^n$ and the number of phenotypes is $2^n$.
For a dihybrid cross,$n = 2$.
Number of genotypes = $3^2 = 9$.
Number of phenotypes = $2^2 = 4$.
Therefore,in the $F_2$ generation,there are $9$ genotypes and $4$ phenotypes.

(C) In a dihybrid cross,when two pairs of contrasting traits are considered,the $F_2$ generation exhibits not only the parental combinations but also new combinations of traits (recombinants).
This occurs because the alleles of different genes segregate independently of each other during gamete formation.
This phenomenon is described by Mendel's Law of Independent Assortment,which states that when two pairs of traits are combined in a hybrid,segregation of one pair of characters is independent of the other pair of characters.

(B) In a dihybrid cross between $YYRR$ (yellow round) and $yyrr$ (green wrinkled),the $F_1$ generation produces $YyRr$ (yellow round) seeds.
When $F_1$ generation plants are self-pollinated,the $F_2$ phenotypic ratio is $9:3:3:1$.
- Yellow round = $9$
- Yellow wrinkled = $3$
- Green round = $3$
- Green wrinkled = $1$
To find the ratio of yellow to green seeds:
- Total yellow seeds = $9 + 3 = 12$
- Total green seeds = $3 + 1 = 4$
Thus,the ratio of yellow to green seeds in the $F_2$ generation is $12:4$ (which simplifies to $3:1$).

(B) test cross is a cross between an individual with a dominant phenotype (but unknown genotype) and a homozygous recessive individual.
In a dihybrid test cross,an individual heterozygous for two traits $(YyRr)$ is crossed with a homozygous recessive individual $(yyrr)$.
The gametes produced by $YyRr$ are $YR, Yr, yR,$ and $yr$.
The gametes produced by $yyrr$ are only $yr$.
The resulting progeny are $YyRr$ (Yellow Round),$Yyrr$ (Yellow wrinkled),$yyRr$ (Green round),and $yyrr$ (Green wrinkled) in a $1:1:1:1$ ratio.
Therefore,the cross $YyRr \times yyrr$ results in a $1:1:1:1$ phenotypic ratio.

(D) The number of different types of gametes produced by an organism can be calculated using the formula $2^n$,where $n$ is the number of heterozygous gene pairs.
For the genotype $TtYy$,there are $2$ heterozygous pairs ($Tt$ and $Yy$),so $n = 2$.
The number of gamete types = $2^2 = 4$.
The specific types of gametes formed are $TY$,$Ty$,$tY$,and $ty$.

(C) In a dihybrid cross between $YYRR$ (yellow round) and $yyrr$ (green wrinkled), the phenotypic ratio of the $F_2$ generation is $9:3:3:1$.

$9/16$ Yellow round (Parental)	$3/16$ Yellow wrinkled (Recombinant)
$3/16$ Green round (Recombinant)	$1/16$ Green wrinkled (Parental)

The parental phenotypes are yellow round and green wrinkled, which account for $9/16 + 1/16 = 10/16$ of the total offspring.
The recombinant phenotypes are yellow wrinkled and green round, which account for $3/16 + 3/16 = 6/16$ of the total offspring.
Therefore, the proportion of recombinant offspring is $6/16$.

(A) In pea plants,tall $(T)$ is dominant over dwarf $(t)$,and axial flowers $(A)$ are dominant over terminal flowers $(a)$.
The dwarf plant with terminal flowers must have the genotype $ttaa$.
The offspring are tall with axial flowers $(TtAa)$ and tall with terminal flowers $(Ttaa)$ in a $1:1$ ratio.
Since all offspring are tall $(Tt)$,the tall parent must be homozygous dominant $(TT)$.
Since the offspring show a $1:1$ ratio for flower position (axial:terminal),the tall parent must be heterozygous for flower position $(Aa)$.
Therefore,the cross is $TTAa \times ttaa$.

(B) In a dihybrid cross involving two traits (e.g.,$AaBb \times AaBb$),the $F_2$ generation produces $16$ combinations of genotypes.
Among these,the parental genotypes are the homozygous dominant $(AABB)$ and the homozygous recessive $(aabb)$.
There is $1$ individual with the $AABB$ genotype and $1$ individual with the $aabb$ genotype.
Therefore,the total number of individuals with parental genotypes is $1 + 1 = 2$.
The proportion of these individuals in the $F_2$ generation is $2/16$.

(B) In a dihybrid cross, the $F_2$ phenotypic ratio is $9:3:3:1$.
Individuals showing one dominant and one recessive character are represented by the two groups of $3$ in the ratio (i.e., $3 + 3 = 6$).
These phenotypes correspond to the genotypes $AAbb$ and $aaBB$ (or $Yyrr$ and $yyRr$ in the case of seed color and shape).
Therefore, the total proportion of individuals showing one dominant and one recessive trait is $6/16$.

(A) The genotype of the individual is $AaBbcc$.
To calculate the number of different types of gametes produced,we use the formula $2^n$,where $n$ represents the number of heterozygous gene pairs.
In the genotype $AaBbcc$,the heterozygous pairs are $Aa$ and $Bb$,while $cc$ is homozygous.
Therefore,$n = 2$.
The number of gametes produced is $2^2 = 4$.
The four types of gametes are $Abc, Abc, aBc, abc$.

(C) In a dihybrid cross involving two traits (flower color and flower position),the $F_1$ generation produces dihybrid plants with the genotype $VvAa$.
When these $F_1$ individuals are self-crossed to produce the $F_2$ generation,the phenotypic ratio follows the Mendelian dihybrid ratio of $9:3:3:1$.
The traits are:
$9$ (Violet,Axial)
$3$ (Violet,Terminal)
$3$ (White,Axial)
$1$ (White,Terminal)
White flowers are recessive $(vv)$. In the $F_2$ generation,the total proportion of white-flowered plants is the sum of those that are white-axial and white-terminal,which is $3/16 + 1/16 = 4/16$ (or $1/4$).
Therefore,the phenotypic ratio of white flowers is $4/16$.

(C) In a dihybrid cross,Mendel studied the inheritance of two pairs of contrasting traits simultaneously. For example,when crossing a plant with round and yellow seeds $(RRYY)$ with a plant having wrinkled and green seeds $(rryy)$,the $F_1$ generation produces all round and yellow seeds $(RrYy)$. Upon self-pollination of the $F_1$ generation,the $F_2$ generation exhibits four different phenotypes in the ratio of $9 : 3 : 3 : 1$. These represent:
$1$. Round-Yellow $(9)$
$2$. Round-Green $(3)$
$3$. Wrinkled-Yellow $(3)$
$4$. Wrinkled-Green $(1)$

(A) In a dihybrid cross,the phenotypic ratio in the $F_2$ generation is $9:3:3:1$.
The parental combinations are Round-Yellow $(9/16)$ and Wrinkled-Green $(1/16)$.
The non-parental (recombinant) combinations are Round-Green $(3/16)$ and Wrinkled-Yellow $(3/16)$.
Total proportion of non-parental characters = $3/16 + 3/16 = 6/16 = 3/8$.
Given that the total number of $F_2$ progenies is $64$.
Number of non-parental progenies = $(3/8) \times 64 = 3 \times 8 = 24$.

(B) In a dihybrid cross,the inheritance of two traits is considered independently. According to Mendel's Law of Independent Assortment,the alleles for seed color (Yellow $Y$ and Green $y$) segregate independently of the alleles for seed shape (Round $R$ and Wrinkled $r$).
For the seed color trait,the $F_1$ generation is heterozygous $(Yy)$. When $F_1$ plants are self-hybridized $(Yy \times Yy)$,the resulting $F_2$ generation follows a monohybrid phenotypic ratio.
The Punnett square for seed color yields: $1 YY$ (Yellow) : $2 Yy$ (Yellow) : $1 yy$ (Green).
Thus,the total number of Yellow seeds is $3$ $(1 YY + 2 Yy)$ and the total number of Green seeds is $1$ $(1 yy)$.
Therefore,the phenotypic segregation ratio of Yellow to Green seeds in the $F_2$ generation is $3:1$.

(A) In a typical Mendelian dihybrid cross,the $F_2$ phenotypic ratio is $9:3:3:1$.
Here,the parental combinations are the phenotypes that resemble the original parents (homozygous dominant and homozygous recessive),which are represented by the $9$ (both dominant) and $1$ (both recessive) classes.
Total parental combinations = $9 + 1 = 10$.
The recombinations are the new phenotypes that differ from the parents,represented by the two $3$ classes (one dominant and one recessive trait each).
Total recombinations = $3 + 3 = 6$.
Therefore,the ratio of parental combinations to recombinations is $10:6$.