Inheritance of one gene Questions in English

(B) The representation $Rr$ denotes the genetic constitution of an organism regarding a specific trait. Since it shows the alleles present (one dominant $R$ and one recessive $r$),it provides information about the genotype of the plant. Therefore,$Rr$ is a genotypic representation.

(B) When an individual possesses two different alleles for a specific gene (e.g.,$Tt$),the condition is referred to as $Heterozygous$.
$Heterozygous$ individuals produce two different types of gametes,each carrying one of the two alleles.
In contrast,$Homozygous$ individuals possess two identical alleles (e.g.,$TT$ or $tt$).
Therefore,the correct term for having two different alleles is $Heterozygous$.

(B) test cross is performed between an individual with a dominant phenotype (genotype unknown,$Aa$ or $AA$) and a homozygous recessive individual $(aa)$.
If the parent with the dominant phenotype is heterozygous $(Aa)$,the cross $Aa \times aa$ results in a $1: 1$ ratio of dominant to recessive offspring $(Aa: aa)$.
If the parent were homozygous dominant $(AA)$,the cross $AA \times aa$ would result in $100\%$ dominant offspring $(Aa)$.
Therefore,a $1: 1$ ratio confirms that the parent with the dominant phenotype was heterozygous.

(C) The correct answer is $C$.
In a Mendel's monohybrid cross,the $F_2$ generation exhibits a genotypic ratio of $1:2:1$.
This ratio consists of:
$1$ homozygous dominant $(TT)$,
$2$ heterozygous dominant $(Tt)$,and
$1$ homozygous recessive $(tt)$.

(A) This is a test cross between a heterozygous tall plant $(Tt)$ and a homozygous recessive dwarf plant $(tt)$.
According to the Punnett square:
- The gametes produced by $Tt$ are $T$ and $t$.
- The gametes produced by $tt$ are $t$ and $t$.
- The resulting genotypes are $Tt$ (tall) and $tt$ (dwarf) in a $1:1$ ratio.
Therefore,the phenotypic and genotypic ratio of the progeny is $1:1$.

(B) When a plant with genotype $Rr$ is selfed,the cross is $Rr \times Rr$.
According to the Punnett square,the offspring genotypes are $RR, Rr, Rr,$ and $rr$.
Out of these $4$ combinations,$2$ are heterozygous ($Rr$ and $Rr$).
Therefore,the percentage of heterozygous individuals is $(2/4) \times 100 = 50\%$.

(C) In garden pea plants,red flower color is dominant over white flower color. Let $R$ represent the dominant allele for red flowers and $r$ represent the recessive allele for white flowers.
$P$ generation: $RR$ (Red) $\times$ $rr$ (White)
$F_1$ generation: All offspring are $Rr$ (Red).
When $F_1$ plants $(Rr)$ are self-pollinated $(Rr \times Rr)$:
Using a Punnett square:
| | $R$ | $r$ |
|---|---|---|
| $R$ | $RR$ | $Rr$ |
| $r$ | $Rr$ | $rr$ |
Genotypes: $1 RR : 2 Rr : 1 rr$
Phenotypes: $3$ Red $(RR, Rr, Rr)$ : $1$ White $(rr)$.
Therefore,the phenotypic ratio in the $F_2$ generation is $3:1$.

(B) When self-pollination occurs between two heterozygous plants with genotype $Tt$,the cross is represented as $Tt \times Tt$.
According to the Punnett square:
- The gametes produced by each parent are $T$ and $t$.
- The possible combinations in the offspring are $TT$,$Tt$,$Tt$,and $tt$.
- Thus,the resulting genotypes are $1$ $TT$ (homozygous dominant),$2$ $Tt$ (heterozygous),and $1$ $tt$ (homozygous recessive).
- The genotype ratio is $1:2:1$.

(C) In a monohybrid cross,a homozygous tall plant $(TT)$ is crossed with a dwarf plant $(tt)$.
In the $F_1$ generation,all plants are heterozygous tall $(Tt)$.
When $F_1$ plants $(Tt)$ are self-pollinated to produce the $F_2$ generation,the genotypes are $TT$,$Tt$,$Tt$,and $tt$.
This results in a genotypic ratio of $1$ homozygous tall $(TT)$ : $2$ heterozygous tall $(Tt)$ : $1$ homozygous dwarf $(tt)$.

(D) The dwarf pea plant is genetically homozygous recessive $(tt)$.
Gibberellic acid treatment only affects the phenotype (height) by promoting stem elongation but does not change the genotype $(tt)$.
When this treated plant $(tt)$ is crossed with a pure tall plant $(TT)$,
the cross is $TT \times tt$.
The resulting offspring in the $F_1$ generation will all have the genotype $Tt$.
According to Mendel's Law of Dominance,the $T$ allele is dominant over the $t$ allele.
Therefore,all offspring will exhibit the tall phenotype.

(A) Let the red flower allele be $R$ and the white flower allele be $r$. The hybrid between red $(RR)$ and white $(rr)$ strains is $Rr$ (red-flowered).
When this hybrid $(Rr)$ is crossed back with the pure red-flowered strain $(RR)$:
$Rr \times RR \rightarrow RR, RR, Rr, Rr$.
All offspring possess at least one dominant $R$ allele,resulting in $100\%$ red-flowered progeny.
Therefore,the correct option is $A$.

(B) When an individual possesses two different alleles for a specific gene (a contrasting character),the condition is known as heterozygous.
For example,in a genotype $Tt$,the individual carries both the dominant allele $T$ and the recessive allele $t$.
Such organisms do not breed true upon self-fertilization because they produce different types of gametes.

(B) $1$. The trait for smooth seeds is dominant over the trait for crinkled (sugary) seeds.
$2$. Let the dominant allele for smooth seeds be $S$ and the recessive allele for crinkled seeds be $s$.
$3$. The cross between a mutant (homozygous recessive,$ss$) and a normal plant (homozygous dominant,$SS$) produces an $F_1$ generation with the genotype $Ss$,which expresses the dominant smooth phenotype.
$4$. When the $F_1$ $(Ss)$ is allowed to self-pollinate $(Ss \times Ss)$,the offspring follow the Mendelian monohybrid ratio.
$5$. The Punnett square for $Ss \times Ss$ results in genotypes: $1 SS : 2 Ss : 1 ss$.
$6$. Phenotypically,this corresponds to $3$ smooth seeds ($SS$ and $Ss$) and $1$ crinkled (sugary) seed $(ss)$.
$7$. Therefore,the ratio is $3$ smooth : $1$ sugary.

(A) To determine the genotypic ratio of the cross $Aa BB \times aa BB$,we perform a Punnett square analysis:
$1$. Identify the gametes produced by each parent:
- Parent $1$ $(Aa BB)$ produces gametes: $AB$ and $aB$.
- Parent $2$ $(aa BB)$ produces gametes: $aB$ and $aB$.
$2$. Perform the cross:
- $(AB) \times (aB) = Aa BB$
- $(aB) \times (aB) = aa BB$
$3$. Resulting genotypes in the $F_1$ generation are $Aa BB$ and $aa BB$ in a ratio of $1:1$.
Therefore,the correct genotypic ratio is $1 Aa BB : 1 aa BB$.

(D) The cross between a heterozygous individual $(Aa)$ and a homozygous recessive individual $(aa)$ is known as a test cross.
According to Mendelian genetics,the Punnett square for this cross $(Aa \times aa)$ results in gametes $A$ and $a$ from the male,and $a$ from the female.
This produces offspring with genotypes $Aa$ and $aa$ in a $1:1$ ratio.
For $1000$ total offspring,the expected distribution is $500\ Aa$ and $500\ aa$.
Option $(d)$ $481\ Aa : 519\ aa$ represents the closest observed experimental result to the theoretical $1:1$ ratio.

(C) According to Mendel's law of dominance,when a homozygous dominant plant (red-flowered,$RR$) is crossed with a homozygous recessive plant (white-flowered,$rr$),the $F_1$ generation offspring will all be heterozygous $(Rr)$.
Since the allele for red color is dominant over the allele for white color,all $F_1$ offspring will exhibit the red-flowered phenotype.

(C) The observed ratio of $73.97\%$ tall and $26.03\%$ dwarf is approximately $3:1$.
According to Mendel's law of segregation,a phenotypic ratio of $3:1$ in the $F_2$ generation is obtained when both parents are heterozygous (hybrid) for the trait.
Cross: $Tt \times Tt \rightarrow TT, Tt, Tt, tt$.
Genotypes: $1$ $TT$ (pure tall),$2$ $Tt$ (hybrid tall),$1$ $tt$ (pure dwarf).
Phenotypes: $3$ tall $(75\%)$ and $1$ dwarf $(25\%)$,which matches the experimental data.

(C) The correct answer is $(C)$.
In Mendel's experiment,the parents are true-breeding,meaning they are homozygous. Let the genotype for red flowers be $RR$ and for white flowers be $rr$.
The $F_1$ generation obtained from the cross $RR \times rr$ results in all heterozygous $Rr$ offspring,which show the dominant red phenotype.
When $F_1$ individuals $(Rr)$ are self-crossed $(Rr \times Rr)$,the $F_2$ generation genotypes are $RR$ $(1)$,$Rr$ $(2)$,and $rr$ $(1)$.
Thus,the phenotypic ratio of red-flowered plants to white-flowered plants is $3 : 1$.

(B) In a Mendelian monohybrid cross,the parents are homozygous for a trait (e.g.,$RR$ and $rr$).
The $F_1$ generation consists of heterozygous individuals $(Rr)$.
During gametogenesis,the alleles segregate,resulting in the formation of two types of gametes: $R$ and $r$.
Therefore,the correct answer is $B$ (Two types).

(B) In a typical monohybrid cross between two heterozygous parents $(Aa \times Aa)$,the $F_2$ generation consists of genotypes in the ratio $1 AA : 2 Aa : 1 aa$.
Here,$AA$ and $aa$ are homozygous,while $Aa$ is heterozygous.
Total homozygous individuals = $1 + 1 = 2$ parts.
Total heterozygous individuals = $2$ parts.
Therefore,the ratio of homozygous to heterozygous individuals is $2 : 2$,which simplifies to $1 : 1$.
Given a total of $120$ plants,the number of homozygous plants is $120 \times (1/2) = 60$ and the number of heterozygous plants is $120 \times (1/2) = 60$.
Thus,the ratio is $60 : 60$.

(D) cross between a heterozygous dominant individual $(Aa)$ and a homozygous recessive individual $(aa)$ is known as a test cross.
In this cross,the gametes produced by the heterozygous parent are $A$ and $a$,while the homozygous recessive parent produces only $a$ gametes.
The resulting offspring genotypes are $Aa$ and $aa$ in a $1:1$ ratio.
Therefore,the phenotypic and genotypic ratio of the progeny is $1:1$.

(B) In this cross,the offspring are $50\%$ tall and $50\%$ dwarf.
This indicates that the tall parent must be heterozygous $(Tt)$ and the dwarf parent must be homozygous recessive $(tt)$.
The cross is: $Tt \times tt \rightarrow Tt$ (tall),$tt$ (dwarf).
This specific type of cross,where an individual of unknown genotype is crossed with a homozygous recessive individual,is known as a test cross.

(C) When a cross is performed between a homozygous dominant parent $(AA)$ and a homozygous recessive parent $(aa)$,the gametes produced by $AA$ are $A$ and the gametes produced by $aa$ are $a$.
Upon fertilization,the fusion of these gametes results in the ${F_1}$ generation with the genotype $Aa$.
Since $A$ is the dominant allele,the phenotype of the ${F_1}$ progeny will express the dominant trait $(A)$.

(A) Let the allele for brown eye colour be $B$ and the allele for blue eye colour be $b$. Since brown is dominant,$B$ is dominant over $b$.
The man is homozygous for brown eye colour,so his genotype is $BB$.
The lady is heterozygous for brown eye colour,so her genotype is $Bb$.
Cross: $BB \times Bb$
Gametes produced by the man: $B, B$
Gametes produced by the lady: $B, b$
Offspring genotypes: $BB, Bb, BB, Bb$.
All offspring will have at least one dominant allele $B$,meaning all children will have brown eyes.
Therefore,the correct option is $A$.

(B) Let the allele for brown eyes be $B$ (dominant) and blue eyes be $b$ (recessive).
$1$. The woman is blue-eyed, so her genotype must be $bb$.
$2$. The man is brown-eyed, so he must have at least one $B$ allele. Since his mother was blue-eyed $(bb)$, he must have inherited one $b$ allele from her. Therefore, his genotype is $Bb$.
$3$. The cross is $bb$ (woman) $\times$ $Bb$ (man).
$4$. The possible offspring genotypes are $Bb$ (brown-eyed) and $bb$ (blue-eyed) in a $1:1$ ratio.
$5$. Thus, $50\%$ of the children are expected to be brown-eyed and $50\%$ are expected to be blue-eyed.

(B) The tobacco plant is heterozygous for albinism,so its genotype is $Aa$.
Self-pollination of the heterozygous plant $(Aa \times Aa)$ results in the following $F_1$ generation genotypes:
$AA$ (homozygous dominant),$Aa$ (heterozygous),$Aa$ (heterozygous),and $aa$ (homozygous recessive).
The ratio of genotypes is $1:2:1$.
The parental genotype is $Aa$,which accounts for $2$ out of $4$ parts,or $50\%$ of the total offspring.
Therefore,out of $1200$ seedlings,the number of seedlings with the parental genotype is $1200 \times 0.5 = 600$.

(C) In humans,the trait for curly hair is dominant $(C)$ and the trait for straight hair is recessive $(c)$.
Assuming the mother is homozygous dominant $(CC)$ and the father is homozygous recessive $(cc)$,all offspring will be heterozygous $(Cc)$ and exhibit curly hair.
However,if the mother is heterozygous $(Cc)$ and the father is homozygous recessive $(cc)$,the cross is $Cc \times cc$.
The resulting genotypes are $Cc$ (curly) and $cc$ (straight) in a $1:1$ ratio.
For $8$ children,the expected distribution is $4$ curly-haired and $4$ straight-haired,resulting in a ratio of $4:4$.

(C) This is a classic example of a test cross.
Let the dominant allele for black colour be $X$ and the recessive allele for white colour be $x$.
The genotype of the heterozygous black dog is $Xx$.
The genotype of the recessive homozygous white bitch is $xx$.
When these are crossed $(Xx \times xx)$,the gametes produced are $X, x$ from the dog and $x$ from the bitch.
The resulting offspring genotypes are $Xx$ (black) and $xx$ (white) in a ratio of $1:1$.

(A) In a monohybrid cross,Mendel observed that the $F_1$ generation expressed only one of the parental traits (dominant).
When $F_1$ hybrids were self-pollinated,the $F_2$ generation showed both parental traits in a phenotypic ratio of $3:1$.
This $3:1$ ratio is a direct result of the law of dominance,where the dominant allele masks the expression of the recessive allele in the heterozygous condition.
Therefore,this phenomenon is explained by complete dominance.

(C) cross between an $F_1$ heterozygote (e.g.,$Tt$) and a recessive parent (e.g.,$tt$) is known as a test cross.
In a test cross,the $F_1$ heterozygote produces two types of gametes: $T$ and $t$.
The recessive parent produces only one type of gamete: $t$.
Performing a Punnett square: $(T, t) \times (t)$ results in offspring with genotypes $Tt$ and $tt$ in a $1:1$ ratio.
Therefore,the phenotypic and genotypic ratio is $1:1$.

(A) Let the dominant allele for red flowers be represented by $R$ and the recessive allele for white flowers be represented by $r$.
Since the red-flowered plant is homozygous,its genotype is $RR$.
The white-flowered plant is recessive,so its genotype must be $rr$.
When these two plants are crossed $(RR \times rr)$,the gametes produced by the red plant are $R$ and the gametes produced by the white plant are $r$.
The resulting $F_1$ generation genotype is $Rr$.
Since $R$ is dominant over $r$,all $F_1$ offspring will have the phenotype of red flowers.

(B) Gibberellin is a plant growth hormone that can induce stem elongation,making a genetically dwarf plant appear tall phenotypically. However,the genotype of the dwarf plant remains homozygous recessive $(tt)$.
When this phenotypically tall (but genotypically $tt$) plant is crossed with a pure tall plant (genotype $TT$),the cross is $tt \times TT$.
The gametes produced by the $tt$ plant are all $t$,and the gametes produced by the $TT$ plant are all $T$.
The resulting $F_1$ generation will have the genotype $Tt$.
Since $T$ (tall) is dominant over $t$ (dwarf),all offspring in the $F_1$ generation will be tall.

(C) The formula for the number of different genotypes produced by a self-cross of an organism with $n$ heterozygous gene pairs is $3^n$.
In the given genotype $AaBbCcDdEeFfGg$,we count the number of heterozygous pairs $(n)$:
$Aa$ $(1)$,$Bb$ $(2)$,$Cc$ $(3)$,$Dd$ $(4)$,$Ee$ $(5)$,$Ff$ $(6)$,$Gg$ $(7)$.
Thus,$n = 7$.
The number of different genotypes is $3^n = 3^7$.

(C) The ratio of tall to dwarf plants is $45:15$,which simplifies to $3:1$.
This $3:1$ phenotypic ratio is characteristic of a monohybrid cross between two heterozygous parents $(Tt \times Tt)$.
According to Mendel's law of segregation,when two heterozygous individuals $(Tt)$ are crossed,the offspring genotypes are $1 TT : 2 Tt : 1 tt$,resulting in a phenotypic ratio of $3$ tall plants to $1$ dwarf plant.

(A) In a dihybrid cross,Mendel studied the inheritance of two pairs of contrasting traits simultaneously.
According to the Law of Independent Assortment,the inheritance of one pair of traits is independent of the other.
Therefore,for each individual trait,the $F_2$ generation follows the same phenotypic ratio as observed in a monohybrid cross.
The phenotypic ratio for a monohybrid cross in the $F_2$ generation is $3:1$ (dominant to recessive).
Thus,for any single pair of contrasting traits in a dihybrid cross,the ratio remains $3:1$.

(A) Let the gene for tallness be represented by $T$ (dominant) and the gene for dwarfness be represented by $t$ (recessive).
$1$. The genotype of a heterozygous tall plant is $Tt$.
$2$. The genotype of a dwarf plant is $tt$.
$3$. The cross is $Tt \times tt$.
$4$. The gametes produced by $Tt$ are $T$ and $t$,while the gametes produced by $tt$ are $t$ and $t$.
$5$. Punnett square analysis:
- $T \times t = Tt$ (Tall)
- $T \times t = Tt$ (Tall)
- $t \times t = tt$ (Dwarf)
- $t \times t = tt$ (Dwarf)
$6$. The offspring ratio is $2$ Tall $(Tt)$ : $2$ Dwarf $(tt)$,which simplifies to $1:1$.
$7$. Therefore,the proportion of dwarf plants is $2/4 = 50\%$.

(B) cross between two homozygous parents that differ in only one gene locus is called a $Monohybrid$ $cross$.
In a $Monohybrid$ $cross$,we study the inheritance of only one pair of contrasting traits (e.g.,tall vs. dwarf plants).
$Back$ $cross$ is a cross between an $F_1$ hybrid and one of its parents.
$Dihybrid$ $cross$ involves the study of two pairs of contrasting traits.
$Trihybrid$ $cross$ involves the study of three pairs of contrasting traits.

(B) test cross is defined as a cross between an individual of unknown genotype (or a hybrid) and a homozygous recessive parent.
In this case,the $F_1$ hybrid $Tt$ is crossed with the homozygous recessive parent $tt$.
This cross is used to determine the genotype of the $F_1$ individual.
The resulting offspring ratio will be $1:1$ $(Tt:tt)$.

(A) $1$. The $F_2$ ratio of $60:20$ simplifies to $3:1$. This indicates a Mendelian monohybrid cross where red is dominant over white.
$2$. Let $R$ be the allele for red flowers and $r$ be the allele for white flowers.
$3$. The heterozygous red-flowered plant has the genotype $Rr$.
$4$. Self-pollination of $Rr$ $(Rr \times Rr)$ results in the following Punnett square: $RR, Rr, Rr, rr$.
$5$. The phenotypic ratio is $3$ red $(RR, Rr, Rr)$ to $1$ white $(rr)$,which is $3:1$.
$6$. Among the options provided,$72:24$ simplifies to $3:1$ $(72/24 = 3)$,which matches the expected phenotypic ratio.

(C) In pea plants,red flower color $(R)$ is dominant over white flower color $(r)$.
Since the offspring show a phenotypic ratio of $3:1$ ($75\%$ red and $25\%$ white),this indicates a typical monohybrid cross between two heterozygous parents $(Rr \times Rr)$.
According to Mendel's law of segregation,the cross $Rr \times Rr$ results in genotypes $RR$ $(25\%)$,$Rr$ $(50\%)$,and $rr$ $(25\%)$.
Thus,the phenotypic ratio is $75\%$ red ($RR$ and $Rr$) and $25\%$ white $(rr)$.
Therefore,both parent plants must be heterozygous $(Rr)$.

(B) In a monohybrid cross,Mendel crossed two pure-breeding plants (e.g.,tall $TT$ and dwarf $tt$) to obtain the $F_1$ generation,which consisted of all heterozygous tall plants $(Tt)$.
When these $F_1$ plants were self-pollinated,the $F_2$ generation was produced.
The gametes produced by $F_1$ $(Tt)$ are $T$ and $t$.
Using a Punnett square,the resulting genotypes are $TT$,$Tt$,$Tt$,and $tt$.
Thus,the genotypic ratio is $1$ $(TT)$ : $2$ $(Tt)$ : $1$ $(tt)$,which is $1:2:1$.

(C) Let the allele for brown eyes be $B$ (dominant) and the allele for blue eyes be $b$ (recessive).
$1$. The woman has blue eyes,so her genotype must be $bb$.
$2$. The man has brown eyes,so he must have at least one $B$ allele. Since his mother had blue eyes $(bb)$,he must have inherited one $b$ allele from her. Thus,his genotype is $Bb$.
$3$. The cross is $bb \times Bb$.
$4$. The possible offspring genotypes are $Bb$ (brown eyes) and $bb$ (blue eyes) in a $1:1$ ratio.
$5$. Therefore,$50\%$ of the children will have brown eyes and $50\%$ will have blue eyes.

(C) When a cross is performed between a homozygous dominant parent $(AA)$ and a homozygous recessive parent $(aa)$,the gametes produced by $AA$ are $A$,and the gametes produced by $aa$ are $a$.
Upon fertilization,the $F_1$ generation receives one allele from each parent,resulting in the genotype $Aa$.
Since the allele $A$ is dominant over $a$,the phenotype of the $F_1$ progeny will express the dominant trait,which is $A$.

(C) pure tall plant has the genotype $TT$,while a hybrid tall plant has the genotype $Tt$.
When a $TT$ plant undergoes self-pollination,all offspring are tall $(TT)$.
When a $Tt$ (hybrid) plant undergoes self-pollination,according to Mendel's laws,offspring are produced in a $3:1$ ratio of tall to dwarf plants.
Therefore,if all offspring are tall after self-pollination,the parent is a pure tall plant. If dwarf plants are also produced after self-pollination,the parent is a hybrid tall plant.

(D) $1$. The total number of plants produced is $140 + 40 = 180$ (Note: The problem mentions $200$ seeds,but $180$ germinated).
$2$. The ratio of tall to dwarf plants is $140:40$,which simplifies to $7:2$ or approximately $3:1$.
$3$. $A$ phenotypic ratio of $3:1$ in the $F_1$ generation is characteristic of a monohybrid cross between two heterozygous parents $(Tt \times Tt)$.
$4$. In a cross between two heterozygous parents $(Tt \times Tt)$,the expected genotypic ratio is $1TT : 2Tt : 1tt$.
$5$. Since both tall ($TT$ and $Tt$) and dwarf $(tt)$ plants are produced,the parent must be heterozygous $(Tt)$ for the height trait.

(D) Let the gene for tallness be represented by $T$ and the gene for dwarfness by $t$.
- $A$ heterozygous tall plant has the genotype $Tt$.
- $A$ homozygous dwarf plant has the genotype $tt$.
- The cross is $Tt \times tt$.
- The gametes produced by $Tt$ are $T$ and $t$,and by $tt$ are $t$ and $t$.
- The Punnett square results are:
- $T \times t = Tt$ (Tall)
- $T \times t = Tt$ (Tall)
- $t \times t = tt$ (Dwarf)
- $t \times t = tt$ (Dwarf)
- The resulting genotypes are $50\% Tt$ (Tall) and $50\% tt$ (Dwarf).
- Therefore,the percentage of dwarf plants in the progeny is $50\%$.

(A) Let the allele for red flowers be $R$ (dominant) and white flowers be $r$ (recessive).
The heterozygous red-flowered plant has the genotype $Rr$.
The pure-breeding red-flowered parent has the genotype $RR$.
When these are crossed $(Rr \times RR)$:
Gametes from $Rr$ are $R$ and $r$.
Gametes from $RR$ are $R$ and $R$.
The resulting genotypes in the offspring are $RR$ and $Rr$.
Both $RR$ (homozygous dominant) and $Rr$ (heterozygous) express the red phenotype.
Therefore,$100\%$ of the offspring will be red-flowered plants.

(A) In a monohybrid cross,a homozygous tall plant $(TT)$ is crossed with a homozygous dwarf plant $(tt)$.
According to Mendel's law of dominance,the $F_1$ generation will result in plants with the genotype $Tt$.
Since $T$ (tall) is dominant over $t$ (dwarf),all plants in the $F_1$ generation will be phenotypically tall but genotypically heterozygous $(Tt)$.

(A) Let the allele for yellow seeds be $Y$ and the allele for green seeds be $y$. Since yellow is dominant,the genotype for a heterozygous yellow-seeded plant is $Yy$. The genotype for a green-seeded plant (recessive) is $yy$.
Performing the cross: $Yy \times yy$.
The gametes produced are $Y$ and $y$ from the heterozygous parent,and $y$ from the green-seeded parent.
The resulting offspring genotypes are $Yy$ (yellow) and $yy$ (green) in a $1:1$ ratio.
Therefore,the expected ratio of yellow to green-seeded plants is $1:1$.