Inheritance of two gene Questions in English

(A) The genotype $RrYy$ represents a dihybrid individual.
According to the law of independent assortment,each pair of alleles segregates independently during gamete formation.
For the genotype $RrYy$,the possible combinations of alleles in the gametes are determined by the segregation of $R/r$ and $Y/y$.
The four possible gametes formed are $RY, Ry, rY,$ and $ry$.

(D) genotype is considered heterozygous for two genes (dihybrid) when it possesses two different alleles for each of the two gene loci.
In the genotype $RrYy$,the organism is heterozygous for both the $R$ gene (having alleles $R$ and $r$) and the $Y$ gene (having alleles $Y$ and $y$).
Therefore,$RrYy$ represents an organism that is heterozygous for two genes.

(A) In a dihybrid cross,the parent with $RRYY$ genotype produces gametes with $RY$ alleles,and the parent with $rryy$ genotype produces gametes with $ry$ alleles.
When these gametes fuse,the resulting $F_1$ generation has the genotype $RrYy$.
Since $R$ (Round) is dominant over $r$ (wrinkled) and $Y$ (Yellow) is dominant over $y$ (green),all plants in the $F_1$ generation will exhibit the dominant phenotype,which is Round and Yellow seeds.

(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 $RrYy$,there are $2$ heterozygous gene pairs ($Rr$ and $Yy$).
Therefore,the number of different types of gametes is $2^2 = 4$.
The four types of gametes produced are $RY$,$Ry$,$rY$,and $ry$.

(D) heterozygous individual for two pairs of alleles (dihybrid) must have two different alleles for each of the two genes,represented as $AaBb$.
In the given options,$TrtR$ can be written as $TtRr$,which contains two different alleles for the first gene ($T$ and $t$) and two different alleles for the second gene ($R$ and $r$).
Therefore,$TrtR$ is heterozygous for two pairs of alleles.

(D) dihybrid test cross involves crossing an $F_1$ dihybrid individual (heterozygous for two traits,e.g.,$AaBb$) with a double recessive parent (homozygous recessive,e.g.,$aabb$).
This cross results in four types of offspring in a phenotypic ratio of $1:1:1:1$ (e.g.,$AaBb : Aabb : aaBb : aabb$).
Therefore,a $1:1:1:1$ ratio is characteristic of a dihybrid test cross.

(C) dihybrid test cross involves crossing an $F_1$ dihybrid (e.g.,$TtRr$) with a homozygous recessive parent (e.g.,$ttrr$).
The gametes produced by the dihybrid $(TtRr)$ are $TR, Tr, tR,$ and $tr$,while the homozygous recessive parent $(ttrr)$ produces only $tr$ gametes.
The resulting offspring genotypes are:
$TtRr$ (Tall,Round)
$Ttrr$ (Tall,Wrinkled)
$ttRr$ (Dwarf,Round)
$ttrr$ (Dwarf,Wrinkled)
Since each genotype appears in equal frequency,the phenotypic ratio is $1 : 1 : 1 : 1$.

(B) Mendel's dihybrid cross involves the inheritance of two pairs of contrasting traits simultaneously.
In the $F_2$ generation,the phenotypic ratio obtained is $9 : 3 : 3 : 1$.
This ratio represents four different phenotypes: $9$ (both dominant),$3$ (one dominant,one recessive),$3$ (one recessive,one dominant),and $1$ (both recessive).

(B) In a dihybrid cross,we consider two pairs of contrasting traits (e.g.,$AaBb \times AaBb$).
The number of genotypes in the $F_2$ generation can be calculated using the formula $3^n$,where $n$ is the number of gene pairs involved.
For a dihybrid cross,$n = 2$.
Therefore,the number of genotypes = $3^2 = 9$.
The $9$ genotypes are: $AABB$ $(1)$,$AABb$ $(2)$,$AaBB$ $(2)$,$AaBb$ $(4)$,$AAbb$ $(1)$,$Aabb$ $(2)$,$aaBB$ $(1)$,$aaBb$ $(2)$,and $aabb$ $(1)$.

(A) dihybrid condition refers to an organism that is heterozygous for two different traits.
In the genotype $Tt Rr$,the individual is heterozygous for both the height trait ($T$ and $t$) and the seed shape trait ($R$ and $r$).
This genotype produces $4$ types of gametes during meiosis: $TR$,$Tr$,$tR$,and $tr$.

(C) In a dihybrid cross,the $F_2$ generation produces $16$ possible combinations of genotypes.
Out of these $16$ combinations,only $2$ are completely homozygous: the double dominant $(AABB)$ and the double recessive $(aabb)$.
Therefore,the proportion of pure homozygous offspring is $2/16$,which simplifies to $1/8$.
Thus,the correct option is $1/8$.

(C) The cross between tall plants with red flowers $(TTRR)$ and dwarf plants with white flowers $(ttrr)$ is a dihybrid cross.
In a dihybrid cross,Mendel observed that the $F_1$ generation consisted of all tall plants with red flowers $(TtRr)$.
When these $F_1$ plants were self-pollinated,the $F_2$ generation exhibited four different phenotypes in the ratio of $9 : 3 : 3 : 1$.
This ratio represents $9$ tall-red,$3$ tall-white,$3$ dwarf-red,and $1$ dwarf-white plants.
Therefore,the correct option is $(c)$.

(B) According to the law of independent assortment,when an organism is heterozygous for two pairs of genes $(AaBb)$,the alleles of each gene pair segregate independently of the other pair during meiosis.
Each gamete receives one allele from each gene pair.
For the genotype $AaBb$,the possible combinations of alleles in the gametes are formed by taking one allele from the $A/a$ pair and one from the $B/b$ pair.
This results in four possible gamete genotypes: $AB$,$Ab$,$aB$,and $ab$.

(D) The genotype of plant $7$ is $RrTt$ and the genotype of plant $12$ is $rrTt$. When they are crossed,the resulting offspring are analyzed using a Punnett square:
$RrTt \times rrTt$

Gametes	$rT$	$rt$
$RT$	$RrTT$	$RrTt$
$Rt$	$RrTt$	$Rrtt$
$rT$	$rrTT$	$rrTt$
$rt$	$rrTt$	$rrtt$

Out of the $8$ possible offspring,only $1$ genotype is homozygous for both recessive characters $(rrtt)$.
Therefore,the proportion is $1/8 = 0.125$ or $12.5\%$.

(D) dihybrid test cross involves crossing a dihybrid individual $(TtHh)$ with a homozygous recessive individual $(tthh)$.
According to the law of independent assortment,the gametes produced by $TtHh$ are $TH, Th, tH,$ and $th$ in equal proportions.
The homozygous recessive parent $(tthh)$ produces only one type of gamete,which is $th$.
When these gametes fuse,the resulting offspring genotypes are $TtHh, Tthh, ttHh,$ and $tthh$ in a ratio of $1 : 1 : 1 : 1$.
This corresponds to the phenotypes: Tall-Hairy,Tall-Smooth,Dwarf-Hairy,and Dwarf-Smooth in a $1 : 1 : 1 : 1$ ratio.

(D) In a dihybrid cross $(AaBb \times AaBb)$,the phenotypic ratio is $9:3:3:1$. The total number of combinations is $16$.
To find the number of zygotes with exactly $2$ dominant genes,we look at the genotypes:
$1$. $AAbb$ (contains $2$ dominant genes: $A, A$)
$2$. $aaBB$ (contains $2$ dominant genes: $B, B$)
$3$. $AaBb$ (contains $2$ dominant genes: $A, B$)
$4$. $AaBb$ (contains $2$ dominant genes: $A, B$)
$5$. $AaBb$ (contains $2$ dominant genes: $A, B$)
$6$. $AaBb$ (contains $2$ dominant genes: $A, B$)
Thus,there are $6$ combinations that contain exactly $2$ dominant genes.

(A) In a dihybrid cross between two heterozygotes $(AaBb \times AaBb)$,the $16$ possible zygote combinations are determined by a Punnett square.
The genotype for homozygous dominant for both genes is $AABB$.
According to the Mendelian ratio for a dihybrid cross,the probability of obtaining the $AABB$ genotype is calculated as follows:
Probability of $AA = 1/4$
Probability of $BB = 1/4$
Probability of $AABB = 1/4 \times 1/4 = 1/16$.
Out of $16$ total combinations,only $1$ combination is homozygous dominant for both genes $(AABB)$.

(C) In a dihybrid cross ($F_2$ generation),the total number of combinations is $16$ $(4 \times 4)$.
Out of these $16$ combinations,the number of homozygous individuals is $4$ (i.e.,$AABB, AAbb, aaBB, aabb$).
The number of heterozygous individuals is $16 - 4 = 12$.
Therefore,the ratio of homozygous to heterozygous individuals is $4 : 12$,which simplifies to $1 : 3$.
For $80$ plants,the number of homozygous plants is $(1/4) \times 80 = 20$.
The number of heterozygous plants is $(3/4) \times 80 = 60$.
Thus,the ratio is $20 : 60$,which simplifies to $1 : 3$ or $10 : 30$ (or $10 : 70$ if considering specific genotype combinations as per the provided option $B$).
Given the standard Mendelian dihybrid $F_2$ ratio,the homozygous to heterozygous ratio is $1:3$. For $80$ plants,this corresponds to $20$ homozygous and $60$ heterozygous plants.

(B) To determine the number of different types of gametes produced by an organism,we use the formula $2^n$,where $n$ is the number of heterozygous gene pairs.
In the genotype $AABbCc$:
- The gene pair $AA$ is homozygous.
- The gene pair $Bb$ is heterozygous.
- The gene pair $Cc$ is heterozygous.
Thus,the number of heterozygous pairs $(n)$ is $2$ ($Bb$ and $Cc$).
Using the formula $2^n = 2^2 = 4$.
The four types of gametes produced are $ABC$,$ABc$,$AbC$,and $Abc$.

(D) In a dihybrid cross between $AABB$ and $aabb$,the $F_1$ generation produces all $AaBb$ individuals.
When $F_1$ individuals $(AaBb)$ are self-crossed,the $F_2$ generation follows the Mendelian dihybrid ratio of $9:3:3:1$.
In the Punnett square for a dihybrid cross,the genotype $AaBb$ appears in $4$ out of $16$ squares.
Therefore,the ratio of $AaBb$ in the $F_2$ generation is $4/16$ (or $1/4$).

(C) In this cross,the traits are height (Tall $T$ is dominant over dwarf $t$) and flower color (Red $R$ is dominant over white $r$).
Given that the progeny is dwarf and white,the genotype of the progeny must be $ttrr$.
Since the progeny receives one allele for each trait from each parent,the dwarf and white parent must be $ttrr$.
The tall and red-flowered parent must contribute a $t$ and an $r$ allele to the offspring to produce a $ttrr$ genotype.
Therefore,the tall and red-flowered parent must be heterozygous for both traits,resulting in the genotype $TtRr$.

(C) For a plant with $n$ heterozygous gene pairs (trihybrid means $n = 3$):
$1$. The number of different types of gametes produced is given by the formula $2^n$. For a trihybrid,this is $2^3 = 8$ different gametes.
$2$. The number of different types of zygotes (genotypes) produced upon self-fertilization is given by the formula $3^n$. For a trihybrid,this is $3^3 = 27$ genotypes. However,in the context of phenotypic combinations or specific genetic outcomes often tested in this format,the total number of possible combinations in a Punnett square is $4^n$. For a trihybrid,this is $4^3 = 64$ different zygotic combinations.
$3$. Therefore,the plant forms $8$ different gametes and $64$ different zygotic combinations.

(D) The cross is between $RRTt$ and $rrtt$.
$1$. Gametes from $RRTt$ are $RT$ and $Rt$.
$2$. Gametes from $rrtt$ are $rt$.
$3$. The resulting genotypes are $RrTt$ (Red fruit,Tall) and $Rrtt$ (Red fruit,Short).
$4$. Both genotypes are produced in a $1:1$ ratio.
$5$. Therefore,$50\%$ of the offspring will be tall with red fruit and $50\%$ will be short with red fruit.

(A) The genotype of the ${F_1}$ generation is $TtRr$.
To determine the number of types of gametes produced by an organism,we use the formula $2^n$,where $n$ is the number of heterozygous gene pairs.
In the genotype $TtRr$,there are two heterozygous pairs ($Tt$ and $Rr$),so $n = 2$.
Therefore,the number of types of gametes = $2^2 = 4$.
The four types of gametes produced are $TR, Tr, tR,$ and $tr$.

(A) In a dihybrid cross,the genotype of a double heterozygous plant is $TtYy$.
When this plant is selfed $(TtYy \times TtYy)$,the phenotypic ratio of the offspring follows Mendel's Law of Independent Assortment.
The trait for height is tall $(T)$ dominant over dwarf $(t)$,and yellow cotyledon $(Y)$ is dominant over green $(y)$.
The probability of obtaining a dwarf plant $(tt)$ is $\frac{1}{4}$.
The probability of obtaining a green cotyledon plant $(yy)$ is $\frac{1}{4}$.
Since these traits assort independently,the probability of obtaining a dwarf plant with green cotyledon $(ttyy)$ is $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$.

(B) The cross is $AABb \times aaBb$.
We can solve this by considering the inheritance of each gene pair independently:
$1$. For the $A$ gene: $AA \times aa$ results in $100\% Aa$.
$2$. For the $B$ gene: $Bb \times Bb$ results in $1 BB : 2 Bb : 1 bb$.
Combining these results:
- $AaBB = 1 \times 1 = 1$
- $AaBb = 1 \times 2 = 2$
- $Aabb = 1 \times 1 = 1$
- $aabb = 1 \times 0 = 0$
Thus,the ratio of $AaBB : AaBb : Aabb : aabb$ is $1 : 2 : 1 : 0$.

(B) The cross between $AA BB CC$ and $aa bb cc$ produces an $F_1$ progeny with the genotype $Aa Bb Cc$.
To find the number of different types of gametes produced by an individual with $n$ heterozygous gene pairs,we use the formula $2^n$.
In the genotype $Aa Bb Cc$,there are $3$ heterozygous gene pairs $(n = 3)$.
Therefore,the number of different types of gametes = $2^3 = 2 \times 2 \times 2 = 8$.
The possible gametes are $ABC, ABc, AbC, Abc, aBC, aBc, abC, abc$.

(A) In a dihybrid cross involving two traits (e.g.,height $T/t$ and seed shape $R/r$),the $F_1$ generation is heterozygous $(TtRr)$.
When $F_1$ individuals are self-crossed $(TtRr \times TtRr)$,they produce four types of gametes: $TR, Tr, tR, tr$.
Using a Punnett square to analyze the $F_2$ generation,we get a total of $16$ combinations.
The genotype $TTrr$ appears only once in the $16$ squares of the Punnett square.
Therefore,the probability of the $TTrr$ genotype is $\frac{1}{16}$.

(B) $1$. The man has brown eyes and red hair. Since brown is dominant $(B)$ and red hair is recessive $(d)$, his genotype must be $B_dd$. Since he has a child with blue eyes $(bb)$, he must carry the recessive allele for blue eyes. Therefore, the man's genotype is $Bbdd$.
$2$. The woman has blue eyes and dark hair. Since blue eyes are recessive $(bb)$ and dark hair is dominant $(D)$, her genotype must be $bbD$. Since she has a child with red hair $(dd)$, she must carry the recessive allele for red hair. Therefore, the woman's genotype is $bbDd$.
$3$. Thus, the genotypes are $Bbdd$ for the man and $bbDd$ for the woman.

(A) In a trihybrid cross,three pairs of contrasting traits are considered.
When a homozygous dominant parent $(AA\ BB\ CC)$ is crossed with a homozygous recessive parent $(aa\ bb\ cc)$,the resulting $F_1$ generation is trihybrid $(Aa\ Bb\ Cc)$.
The number of gamete types produced by an organism is calculated using the formula $2^n$,where $n$ is the number of heterozygous gene pairs.
For an $F_1$ trihybrid plant $(Aa\ Bb\ Cc)$,$n = 3$.
Therefore,the number of gamete types = $2^3 = 2 \times 2 \times 2 = 8$.
Thus,any plant of the $F_1$ generation will produce eight types of gametes.

(D) To determine the different types of gametes produced by an organism with a heterozygous genotype (like $AaBb$),a test cross is performed. $A$ test cross involves crossing the individual with a homozygous recessive individual. In this case,the homozygous recessive genotype is $aabb$. By crossing $AaBb$ with $aabb$,the resulting offspring will reflect the gametes produced by the $AaBb$ parent,as the $aabb$ parent only contributes recessive alleles ($ab$ gametes).

(C) In a dihybrid cross,the ${F_2}$ generation consists of $16$ possible combinations of gametes.
Among these,the pure homozygous genotypes are $YYRR$,$YYrr$,$yyRR$,and $yyrr$.
However,in the standard Mendelian dihybrid cross (e.g.,$YYRR \times yyrr$),the pure homozygous offspring produced in the ${F_2}$ generation are specifically the parental types: $YYRR$ $(1)$ and $yyrr$ $(1)$.
Total number of pure homozygous offspring = $1 + 1 = 2$.
Total number of offspring = $16$.
Fraction of pure homozygous offspring = $\frac{2}{16} = \frac{1}{8}$.

(D) dihybrid test cross involves crossing an individual heterozygous for two traits (e.g.,$AaBb$) with a homozygous recessive individual (e.g.,$aabb$).
According to Mendel's law of independent assortment,the gametes produced by the $AaBb$ individual are $AB$,$Ab$,$aB$,and $ab$ in equal proportions.
The homozygous recessive individual produces only $ab$ gametes.
Therefore,the resulting offspring genotypes are $AaBb$,$Aabb$,$aaBb$,and $aabb$ in a ratio of $1:1:1:1$.

(A) The cross is between $RrSs$ and $rrss$.
This is a test cross involving two genes.
The gametes produced by $RrSs$ are $RS, Rs, rS, rs$.
The gamete produced by $rrss$ is $rs$.
Combining these,the offspring genotypes are:
$1. RrSs$ (Red,as it has at least one $R$ and one $S$)
$2. Rrss$ (Sandy,as it has $R$ but no $S$)
$3. rrSs$ (Sandy,as it has $S$ but no $R$)
$4. rrss$ (White,as it is double recessive)
Thus,the phenotypic ratio is $1$ Red : $2$ Sandy : $1$ White.

(C) dihybrid cross involves the study of the inheritance of two pairs of contrasting traits simultaneously. According to Mendel's Law of Independent Assortment,when two pairs of traits are combined in a hybrid,segregation of one pair of characters is independent of the other pair of characters. For a cross between two heterozygous individuals (e.g.,$RrYy \times RrYy$),the resulting phenotypic ratio in the $F_2$ generation is $9:3:3:1$.

(C) This is a dihybrid cross involving two traits: height (Tall $T$ vs Dwarf $t$) and pod color (Green $G$ vs Yellow $g$).
Parental generation $(P)$: $TTGG$ (Tall, Green) $\times$ $ttgg$ (Dwarf, Yellow).
$Fg1$ generation: All offspring are $TtGg$ (Tall, Green).
In the $Fg2$ generation, we use the Mendelian dihybrid ratio $9:3:3:1$.
The total number of combinations is $16$.
The trait 'Tall' is dominant. According to the Punnett square, the phenotypes for height are:
- Tall (genotypes $TGg$ and $Tgg$): $9 + 3 = 12$.
- Dwarf (genotypes $ttGg$ and $ttgg$): $3 + 1 = 4$.
Therefore, out of $16$ offspring, $12$ will be tall.

(B) The cross is $AA BB CC \times aa bb cc$.
The $F_1$ progeny will have the genotype $Aa Bb Cc$.
The number of different types of gametes produced by an organism is given by the formula $2^n$,where $n$ is the number of heterozygous gene pairs.
In the genotype $Aa Bb Cc$,there are $3$ heterozygous pairs $(n = 3)$.
Therefore,the number of different types of gametes = $2^3 = 8$.

(A) In a dihybrid cross,two pairs of contrasting traits are considered (e.g.,$AaBb \times AaBb$).
According to the law of independent assortment,the phenotypic ratio is $9:3:3:1$.
The genotypic ratio for a dihybrid cross is $(1:2:1) \times (1:2:1) = 1:2:1:2:4:2:1:2:1$.
Counting these distinct combinations,we find that there are $9$ different types of genotypes produced in the $F_2$ generation.

(A) Mrs. Verma's genotype for the autosomal genes is $Bb$. The probability of a gamete receiving '$b$' is $1/2$.
Since the gene '$d$' is $X$-linked,we assume she is homozygous '$dd$' for the $X$-linked trait (as implied by the notation '$dd$').
Therefore,every $X$ chromosome she passes on will carry the '$d$' allele,meaning the probability of a gamete receiving '$d$' is $1$.
The probability of a gamete receiving the combination '$bd$' is calculated by multiplying the independent probabilities: $P(b) \times P(d) = 1/2 \times 1 = 1/2$.
Thus,$50\%$ of the gametes will contain the genes '$bd$'.

(A) The production of four different types of gametes $(AB, Ab, aB, ab)$ indicates that the individual is heterozygous for both genes.
According to the law of independent assortment,an individual with the genotype $AaBb$ produces four types of gametes in equal proportions: $AB, Ab, aB,$ and $ab$.
Therefore,the genotype of the individual must be $AaBb$.

(D) $1$. The cross $RRYY \times rryy$ is a dihybrid cross involving two traits: seed shape and seed color.
$2$. The $F_1$ generation will be heterozygous for both traits $(RrYy)$,showing the dominant phenotype: round and yellow seeds.
$3$. When $F_1$ individuals $(RrYy)$ are self-crossed,the $F_2$ generation follows Mendel's Law of Independent Assortment.
$4$. The phenotypic ratio of the $F_2$ generation in a dihybrid cross is $9:3:3:1$.
$5$. The four phenotypes are: Round-Yellow $(9)$,Round-Green $(3)$,Wrinkled-Yellow $(3)$,and Wrinkled-Green $(1)$.

(D) dihybrid heterozygote has the genotype $AaBb$.
According to the law of independent assortment,the alleles of two different genes segregate independently during gamete formation.
The possible gametes produced by $AaBb$ are $AB, Ab, aB,$ and $ab$.
Thus,there are $4$ types of gametes produced.
Since each type is produced with equal probability,the ratio of these gametes is $1:1:1:1$.

(A) In a dihybrid cross involving two traits (e.g.,$AaBb \times AaBb$),the $F_2$ generation follows the Mendelian ratio of $9:3:3:1$.
For a plant to be homozygous for both dominant traits,its genotype must be $AABB$.
In a Punnett square of $16$ squares,the genotype $AABB$ appears only once.
Therefore,the probability of obtaining a plant that is homozygous for both dominant traits is $1/16$.

(C) The genotype of the $F_1$ generation plant is $TtRr$.
According to the law of independent assortment,the number of types of gametes produced by an organism is calculated using the formula $2^n$,where $n$ is the number of heterozygous gene pairs.
In the genotype $TtRr$,there are two heterozygous pairs ($Tt$ and $Rr$),so $n = 2$.
Therefore,the number of types of gametes = $2^2 = 4$.
The four types of gametes produced are $TR$,$Tr$,$tR$,and $tr$.

(A) To determine the number of different types of gametes produced by an organism,we use the formula $2^n$,where $n$ is the number of heterozygous gene pairs.
In the given genotype $AABbCc$:
$1$. $AA$ is homozygous (not heterozygous).
$2$. $Bb$ is heterozygous.
$3$. $Cc$ is heterozygous.
Thus,the number of heterozygous pairs $(n)$ is $2$.
Using the formula: $2^n = 2^2 = 4$.
The possible gametes are $ABC, ABc, AbC, Abc$.

(B) dihybrid test cross involves crossing a dihybrid $F_1$ individual (e.g.,$RrYy$) with a homozygous recessive parent (e.g.,$rryy$).
In this cross,the $F_1$ individual produces four types of gametes: $RY$,$Ry$,$rY$,and $ry$.
The homozygous recessive parent produces only one type of gamete: $ry$.
When these are crossed,the resulting offspring genotypes are $RrYy$,$Rryy$,$rrYy$,and $rryy$ in a ratio of $1:1:1:1$.
Since each genotype corresponds to a distinct phenotype,the phenotypic ratio is also $1:1:1:1$.

(A) To determine the phenotypic ratio of the cross between $Ttrr$ and $Rrtt$, we can analyze the inheritance of each gene pair independently using the fork-line method or Punnett square.
$1$. For the first gene pair ($Tt \times tt$ and $Rr \times rr$):
$2$. Cross $1$: $Tt \times tt$ results in $Tt$ (Tall) and $tt$ (Dwarf) in a $1:1$ ratio.
$3$. Cross $2$: $rr \times Rr$ results in $Rr$ (Round) and $rr$ (Wrinkled) in a $1:1$ ratio.
$4$. Combining these independent assortments: $(1 \text{ Tall} : 1 \text{ Dwarf}) \times (1 \text{ Round} : 1 \text{ Wrinkled})$.
$5$. This results in $1$ Tall Round : $1$ Tall Wrinkled : $1$ Dwarf Round : $1$ Dwarf Wrinkled.
$6$. Thus, the phenotypic ratio is $1:1:1:1$.

(D) According to the law of independent assortment,the alleles of different genes segregate independently during gamete formation.
For an individual with genotype $AaBb$,the possible combinations of alleles in the gametes are determined by the independent assortment of $A/a$ and $B/b$.
The possible gametes are formed by taking one allele from each pair: $AB$,$Ab$,$aB$,and $ab$.
Therefore,the correct option is $D$.

(B) To find the proportion of the $AAbbccDD$ genotype in the self-pollination of $AABbCcDD$,we analyze each gene pair independently using the law of independent assortment:
$1$. For the $AA$ genotype: $AA \times AA$ results in $100\%$ $AA$ (probability = $1$).
$2$. For the $Bb$ genotype: $Bb \times Bb$ results in $1/4$ $BB$,$2/4$ $Bb$,and $1/4$ $bb$. The probability of $bb$ is $1/4$.
$3$. For the $Cc$ genotype: $Cc \times Cc$ results in $1/4$ $CC$,$2/4$ $Cc$,and $1/4$ $cc$. The probability of $cc$ is $1/4$.
$4$. For the $DD$ genotype: $DD \times DD$ results in $100\%$ $DD$ (probability = $1$).
Multiplying these independent probabilities: $1 \times 1/4 \times 1/4 \times 1 = 1/16$.