Q.How was it concluded that genes are located on chromosomes?

Showing the 12 most recent of 13 on this concept.

1. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Red eyed plain thorax Drosophila flies were crossed with pink eyed, striped thorax flies. F$_1$ generation flies were test crossed with recessive flies. F$_2$ generation result is as follows. Red eyed plain thorax = 80; Red eye striped thorax = 16, Pink eye plain thorax = 12; Pink eye striped thorax = 92. Find out the percentage of recombinants. (A) 12% (B) 16% (C) 14% (D) 86%

   ›Reveal solutionSolution

   This is a two-factor cross where the parental phenotypes are red-eyed plain thorax and pink-eyed striped thorax. The recombinants are red-eyed striped thorax and pink-eyed plain thorax. Their total is 28 out of 200 offspring, giving 14% recombinants.

   The question is about linkage and recombination in Drosophila. When two genes are on the same chromosome, they tend to be inherited together unless crossing over separates them. The percentage of recombinants (offspring with new combinations of traits) directly tells us the recombination frequency, which is a measure of how far apart the genes are on the chromosome.

   Here, we have two traits: eye colour (red vs pink) and thorax pattern (plain vs striped). The F$_1$ flies were test crossed — meaning crossed with a double recessive fly. In a test cross, the offspring phenotypes directly reflect the gametes produced by the F$_1$ parent. So the numbers we see are the gamete frequencies from the F$_1$.

   1. Identify the parental and recombinant types.

      The original parents were red-eyed plain thorax and pink-eyed striped thorax. So the F$_1$ inherited one chromosome with red + plain and the other with pink + striped. In the test cross, the most frequent offspring will be the ones that got a non-recombinant gamete from the F$_1$ — that is, red plain and pink striped. The less frequent ones are recombinants: red striped and pink plain.

   2. Count the offspring.

      • Red plain: 80
      • Pink striped: 92
      • Red striped: 16
      • Pink plain: 12 Total = $80 + 92 + 16 + 12 = 200$.

   3. Sum the recombinants.

      Recombinants = red striped + pink plain = $16 + 12 = 28$.

   4. Calculate the recombination percentage. …

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2. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.In maize coloured (C) and full endosperm (F) is dominant over colourless

   (c) and shrunken

   (f) endosperm. F$_1$ generation was subjected to testcross. It produced four phenotypes in the following percentage. Coloured full = 48%, Coloured shrunken = 5% Colourless full = 7%, Colourless shrunken = 40% Find out the distance between two non-allelic genes (A) 48 units (B) 5 units (C) 7 units (D) 12 units

   ›Reveal solutionSolution

   The distance between two linked genes is determined by the frequency of recombination between them. By identifying the recombinant offspring from the testcross data, we find the recombination frequency to be 12%, which corresponds to a distance of 12 map units.

   Concept and Intuition

   When genes are located on the same chromosome, they are said to be linked. Linked genes tend to be inherited together because they are physically connected. However, during meiosis, a process called crossing over can occur between homologous chromosomes. This exchange of genetic material can lead to new combinations of alleles on the chromosomes, resulting in recombinant offspring.

   The frequency of these recombination events is directly related to the physical distance between the genes on the chromosome. Genes that are far apart on a chromosome are more likely to experience crossing over between them than genes that are close together. This principle allows us to map the relative positions of genes on a chromosome.

   The recombination frequency is calculated as the percentage of recombinant offspring produced in a cross. This percentage is then directly converted into map units (also known as centimorgans, cM), where 1% recombination frequency equals 1 map unit.

   A testcross is a crucial tool for determining recombination frequency. It involves crossing an individual heterozygous for the genes in question (e.g., $CcFf$) with a homozygous recessive individual ($ccff$). The homozygous recessive parent contributes only recessive alleles to the offspring, meaning the phenotype of the offspring directly reflects the gamete contributed by the heterozygous parent. This simplifies the analysis, as we can easily distinguish parental and recombinant gametes from the observed offspring phenotypes.

   Step-by-Step Solution

   1. Identify the alleles and parental genotypes:

      • Coloured (C) is dominant over colourless (c).
      • Full endosperm (F) is dominant over shrunken (f).
      • The F$_1$ generation was subjected to a testcross. This implies the F$_1$ individuals are heterozygous for both genes. Typically, an F$_1$ generation is produced by crossing a homozygous dominant parent with a homozygous recessive parent.
        • Parental cross: $CCFF \times ccff$
        • F$_1$ genotype: $CcFf$
      • The testcross is between the F$_1$ individual ($CcFf$) and a homozygous recessive individual ($ccff$).

   2. Determine the types of gametes produced by the F$_1$ parent and the corresponding offspring phenotypes:

      In the testcross ($CcFf \times ccff$), the homozygous recessive parent ($ccff$) produces only one type of gamete: $cf$. Therefore, the phenotype of the offspring directly reveals the gamete contributed by the F$_1$ parent ($CcFf$).

      • Parental gametes: These are the gametes that carry the same combination of alleles as the original parents (before crossing over). From the initial cross ($CCFF \times ccff$), the F$_1$ individual received $CF$ from one parent and $cf$ from the other. So, the parental gametes produced by the F$_1$ are $CF$ and $cf$.

        • $CF$ gamete from F$_1$ + $cf$ gamete from testcross parent $\rightarrow$ $CcFf$ (Coloured full)
        • $cf$ gamete from F$_1$ + $cf$ gamete from testcross parent $\rightarrow$ $ccff$ (Colourless shrunken)

      • Recombinant gametes: These are the gametes formed due to crossing over, resulting in new combinations of alleles. The recombinant gametes produced by the F$_1$ are $Cf$ and $cF$.

        • $Cf$ gamete from F$_1$ + $cf$ gamete from testcross parent $\rightarrow$ $Ccff$ (Coloured shrunken)
        • $cF$ gamete from F$_1$ + $cf$ gamete from testcross parent $\rightarrow$ $ccFf$ (Colourless full)

   3. Classify the observed offspring phenotypes as parental or recombinant: …

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3. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Study the following and choose the correct statements I. Cumulative effect of two or more genes on a single phenotypic trait is known as polygenetic inheritance II. Karyotype of Turner syndrome is AA+XXY III. The longest gene codes for the muscle protein dystrophin IV. Highest number of genes are located in $21^{\text{st}}$ chromosome (A) I, II (B) III, IV (C) I, III (D) II, IV

   ›Reveal solutionSolution

   This question tests your knowledge of fundamental genetic concepts. We will evaluate each statement individually: polygenic inheritance, karyotype of Turner syndrome, the longest human gene, and gene density on chromosomes. Statements I and III are correct.

   The correct option is (C).

   Understanding genetics requires precise definitions and factual recall. We will examine each statement to determine its accuracy based on established biological principles.

   1. Evaluate Statement I: Cumulative effect of two or more genes on a single phenotypic trait is known as polygenetic inheritance.

      • Polygenic inheritance, also known as quantitative inheritance, describes a situation where a single phenotypic trait is determined by the additive or cumulative effect of multiple genes. These genes often have small, individual effects that combine to produce a wide range of phenotypes, typically showing continuous variation (e.g., human height, skin color, intelligence).
      • Therefore, this statement accurately defines polygenic inheritance.
      • Statement I is correct.

   2. Evaluate Statement II: Karyotype of Turner syndrome is AA+XXY.

      • Turner syndrome is a chromosomal disorder affecting females, characterized by the presence of only one X chromosome instead of the usual two. The typical karyotype for Turner syndrome is $45, \text{X}$ or $44 + \text{XO}$, meaning there are 44 autosomes and one X chromosome.
      • The karyotype AA+XXY (or $47, \text{XXY}$) describes Klinefelter syndrome, which affects males and involves an extra X chromosome.
      • Statement II is incorrect.

   3. Evaluate Statement III: The longest gene codes for the muscle protein dystrophin. …

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4. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.Match the following List – I List – II A. ABO blood groupings I. Frame – shift mutations B. Starch synthesis in pea seeds II. Sickle cell anemia C. Point mutation III. Pleotrophy D. Deletion of DNA base pairs IV. Co-dominance The correct answer is (A) A-IV, B-I, C-II, D-III (B) A-IV, B-II, C-III, D-I (C) A-III, B-IV, C-II, D-I (D) A-IV, B-III, C-II, D-I

   ›Reveal solutionSolution

   Match each genetic phenomenon with its molecular basis: ABO blood groups show co-dominance, starch synthesis involves pleiotropy, point mutations cause sickle cell anemia, and base-pair deletions produce frame-shift mutations. The answer is (D).

   The question tests your understanding of how different genetic mechanisms manifest as observable traits. Each item in List-I represents a classic example from genetics, and you need to connect it to the underlying molecular phenomenon in List-II.

   Let me work through each pairing by examining what makes each example distinctive:

   Analyzing each match

   1. ABO blood groupings → Co-dominance (IV)

      The ABO system involves three alleles: $I^A$, $I^B$, and $i$. When someone inherits both $I^A$ and $I^B$, both alleles are fully expressed simultaneously, producing AB blood type with both A and B antigens on red blood cells. This is the textbook definition of co-dominance—neither allele is recessive to the other, and both phenotypes appear together.

   2. Starch synthesis in pea seeds → Pleiotropy (III)

      A single gene controlling starch synthesis affects multiple, seemingly unrelated traits. Wrinkled peas lack the enzyme that converts sugar to starch, so they accumulate more sugar (which draws in water by osmosis), swell more during development, then shrink and wrinkle when they dry. The same gene thus affects seed shape, water content, sugar concentration, and starch levels—one gene, many phenotypic effects. That is pleiotropy.

   Tip

   Pleiotropy means "many turnings"—one gene turns the dial on multiple traits simultaneously because its product participates in several biological pathways.

   3. Point mutation → Sickle cell anemia (II)

      Sickle cell anemia results from a single nucleotide substitution in the β-globin gene: the sixth codon changes from GAG (glutamic acid) to GTG (valine). This single base change—a point mutation—alters hemoglobin structure, causing red blood cells to deform under low oxygen. It is the canonical example of how one "point" change in DNA can have profound consequences.

   4. Deletion of DNA base pairs → Frame-shift mutations (I) …

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5. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.It is an example for monosomy (A) Klinefelter syndrome (B) Turner syndrome (C) Down syndrome (D) Edward syndrome

   ›Reveal solutionSolution

   Monosomy means the loss of one chromosome from a homologous pair (2n − 1). Turner syndrome is the classic example, with karyotype 45,X.

   Monosomy refers to a chromosomal abnormality where an individual has only one copy of a particular chromosome instead of the normal two. The total chromosome count becomes 2n − 1 (45 in humans instead of 46). This happens when one chromosome from a homologous pair is missing.

   To identify which syndrome represents monosomy, we need to examine the chromosomal composition of each option:

   1. Klinefelter syndrome (47,XXY) involves an extra X chromosome in males. The karyotype is 47,XXY, making it a case of trisomy (2n + 1), not monosomy. Affected individuals have 47 chromosomes total.

   2. Turner syndrome (45,X) occurs when a female has only one X chromosome instead of two. The karyotype is 45,X — exactly one chromosome short of the normal 46. This is the textbook example of monosomy in humans, specifically monosomy of the sex chromosome.

   3. Down syndrome (47,XX+21 or 47,XY+21) results from an extra copy of chromosome 21. With 47 chromosomes total, this is trisomy 21, the most common autosomal trisomy.

   4. Edward syndrome (47,XX+18 or 47,XY+18) similarly involves an extra chromosome 18, giving a total of 47 chromosomes. This is trisomy 18. …

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6. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.Existence of deleterious genes in a population is called (A) Gene flow (B) Genetic drift (C) Genetic load (D) Sewall Wright effect

   ›Reveal solutionSolution

   The term for the presence of harmful genes in a population is genetic load, which measures the reduction in fitness due to deleterious alleles.

   The question asks for the specific term that describes the existence of deleterious (harmful) genes in a population. This is a concept from population genetics, which studies how allele frequencies change over time. Let's break down each option to see why only one fits.

   1. Gene flow is the transfer of alleles between populations through migration. It can introduce new genes, but it doesn't specifically refer to the presence of harmful ones — it's a process, not a state.

   2. Genetic drift is the random change in allele frequencies due to chance events, especially in small populations. The Sewall Wright effect (option D) is actually another name for genetic drift, named after the population geneticist who studied it. Both describe a mechanism of change, not the existence of deleterious genes.

   3. Genetic load is the correct term. It quantifies the burden of harmful alleles in a population's gene pool. These alleles reduce the average fitness of individuals compared to the best possible genotype. The "load" is the difference between the actual fitness of the population and what it would be if every individual had the optimal genotype. …

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7. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.If sex index ratio in Drosophila is 0.33, the sexual phenotype of it is (A) Metamale (B) Metafemale (C) Female (D) Intersex

   ›Reveal solutionSolution

   The sex index ratio (X : A) in Drosophila determines sexual phenotype; a ratio of 0.33 (1X : 3A) corresponds to a metamale, so the correct option is (A).

   ---

   The key concept here is the X : A ratio (number of X chromosomes divided by the number of autosomal sets) in Drosophila melanogaster. Unlike mammals, where the Y chromosome determines maleness, in fruit flies the balance between X chromosomes and autosomes sets the sex. A ratio of 1.0 gives a female, 0.5 gives a male, and values outside this range produce extreme phenotypes.

   A sex index of 0.33 means there is 1 X chromosome for every 3 sets of autosomes (1X : 3A). This is below the normal male ratio of 0.5, so the fly is hyper-masculine — a metamale (also called "supermale"). These flies are sterile and have reduced viability.

   Let’s walk through the reasoning step by step.

   1. Recall the standard X : A ratios in Drosophila

      • Normal female: 2X : 2A → ratio = 1.0
      • Normal male: 1X : 2A → ratio = 0.5
      • Metafemale: 3X : 2A → ratio = 1.5 (or 2X : 1A → ratio = 2.0)
      • Metamale: 1X : 3A → ratio ≈ 0.33 (or 1X : 4A → 0.25)
      • Intersex: ratios between 0.5 and 1.0 (e.g., 2X : 3A → 0.67) produce mosaic or intermediate traits.

   2. Interpret the given ratio

      The problem states the sex index is 0.33. This is exactly 1/3, which corresponds to 1 X chromosome and 3 sets of autosomes (1X : 3A). Since the normal male ratio is 0.5, a lower ratio pushes development further toward maleness, resulting in a metamale.

   3. Eliminate other options …

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8. TG EAPCET 2023Set ap-2023-05-10-AN1 markMCQ
   Q.Study the following and identify the wrong combination I) $F_1$ hybrid \hspace{1cm} Tt \hspace{1cm} Dominant II) $RRyy$ \hspace{1cm} Wrinkled yellow \hspace{1cm} $2/16$ III) Allele \hspace{1cm} Alternate forms of same gene \hspace{1cm} unit of inheritance IV) Chromosome \hspace{1cm} Cell division \hspace{1cm} Two genomes (A) II, IV (B) I, III (C) I, IV (D) II, III

   ›Reveal solutionSolution

   We need to identify the incorrect statements regarding genetic concepts. Statement II incorrectly describes the phenotype and frequency of the $RRyy$ genotype, and Statement IV incorrectly states that a single chromosome contains "two genomes." Therefore, the wrong combinations are II and IV.

   Understanding the fundamental terms in genetics is crucial for solving this problem. We will evaluate each statement based on the definitions and principles of Mendelian genetics and chromosome structure.

   1. Evaluate Statement I: $F_1$ hybrid \hspace{1cm} Tt \hspace{1cm} Dominant

      • An $F_1$ hybrid typically refers to the first filial generation resulting from a cross between two true-breeding (homozygous) parents that differ in a trait. For example, if a true-breeding dominant parent (TT) is crossed with a true-breeding recessive parent (tt), the $F_1$ generation will all have the genotype Tt.
      • The genotype Tt is heterozygous.
      • According to the principle of dominance, if 'T' is the dominant allele and 't' is the recessive allele, an individual with genotype Tt will express the dominant phenotype.
      • Thus, this statement is correct.

   2. Evaluate Statement II: $RRyy$ \hspace{1cm} Wrinkled yellow \hspace{1cm} $2/16$

      • Let's assume standard Mendelian dihybrid cross conventions, where 'R' (Round) is dominant over 'r' (wrinkled), and 'Y' (Yellow) is dominant over 'y' (green).
      • Phenotype of $RRyy$:
        • $RR$ corresponds to the Round phenotype.
        • $yy$ corresponds to the green phenotype.
        • Therefore, the genotype $RRyy$ results in a Round green phenotype.
        • The statement says "Wrinkled yellow," which is incorrect.
      • Frequency of $RRyy$:
        • In a dihybrid cross between two $RrYy$ individuals ($RrYy \times RrYy$), the gametes produced are RY, Ry, rY, ry, each with a frequency of $1/4$.
        • The probability of obtaining $RR$ is $1/4$ (from $Rr \times Rr$).
        • The probability of obtaining $yy$ is $1/4$ (from $Yy \times Yy$).
        • The probability of obtaining $RRyy$ is the product of these individual probabilities: $(1/4) \times (1/4) = 1/16$.
        • The statement says "$2/16$," which is incorrect.
      • Since both the phenotype and the frequency are incorrectly stated, this statement is wrong.

   3. Evaluate Statement III: Allele \hspace{1cm} Alternate forms of same gene \hspace{1cm} unit of inheritance

      • An allele is indeed an alternate form of a gene. For example, the gene for pea seed color can have an allele for yellow (Y) and an allele for green (y).
      • Genes, and by extension their alleles, are considered the fundamental units of inheritance, as they carry the genetic information passed from parents to offspring.
      • Thus, this statement is correct. …

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9. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
   Q.The Drosophila with the Karyotype AAA+XX is (A) Intersex (B) Female (C) Metafemale (D) Metamale

   ›Reveal solutionSolution

   In Drosophila, sex is determined by the X:A ratio (X chromosomes to autosome sets). A karyotype of AAA+XX gives a ratio of 2:3 = 0.67, which falls below the threshold for normal females and produces a metafemale (superfemale).

   Sex determination in Drosophila does not follow the simple XX/XY rule that mammals use. Instead, it depends on the balance between X chromosomes and sets of autosomes, discovered by Calvin Bridges. This X:A ratio acts as a developmental switch that determines sexual phenotype.

   The key thresholds are:

   • X:A ≥ 1.0 → metafemale (superfemale)
   • X:A = 1.0 → normal female
   • 0.5 < X:A < 1.0 → intersex
   • X:A = 0.5 → normal male
   • X:A ≤ 0.5 → metamale (supermale)

   $\text{X:A ratio} = \frac{\text{number of X chromosomes}}{\text{number of autosome sets}}$

   Now let's work through the given karyotype:

   1. Count the X chromosomes: The karyotype shows XX, so there are 2 X chromosomes.

   2. Count the autosome sets: The notation AAA means three complete sets of autosomes, so there are 3 autosome sets (triploid for autosomes).

   3. Calculate the X:A ratio:

   $\text{X:A} = \frac{2}{3} = 0.67$

   4. Interpret the ratio: A ratio of 0.67 is greater than 0.5 but less than 1.0. Wait—this would normally suggest intersex. However, we need to reconsider what "metafemale" means in the Drosophila literature.

      Actually, the classical definition is:

      • When X:A > 1.0, you get a metafemale (more X relative to autosomes than normal)
      • When X:A = 1.0, normal female
      • When 0.5 < X:A < 1.0, intersex

      But let me recalculate: with AAA+XX, if we have three sets of autosomes and only two X chromosomes, the ratio is indeed 2/3 ≈ 0.67, which should be intersex by the standard scale. …

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10. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
    Q.Chromosome with the fewest genes in human beings is (A) X - chromosome (B) Y - chromosome (C) Chromosome - 1 (D) Chromosome - 21

    ›Reveal solutionSolution

    The Y-chromosome is the smallest human chromosome and carries the fewest number of genes — far fewer than the X, chromosome 1, or chromosome 21.

    The question asks which human chromosome has the fewest genes. This is a fact-based genetics question, but the reasoning behind it is rooted in chromosome size and function.

    Human chromosomes vary dramatically in size and gene density. Chromosome 1 is the largest, containing over 2,000 genes. Chromosome 21 is the smallest autosome (non-sex chromosome), with about 200–300 genes. The X-chromosome is large, carrying around 800–900 genes. The Y-chromosome, however, is tiny — only about 60 million base pairs — and most of its length is heterochromatin (non-coding DNA). It carries only about 50–70 protein-coding genes, making it the gene-poorest chromosome by far.

    Let’s walk through the options:

    1. Option (A): X-chromosome — This is a large chromosome (about 155 million base pairs) with roughly 800–900 genes. It’s not the smallest in gene count.

    2. Option (B): Y-chromosome — This is the smallest human chromosome in both size and gene content. It has only about 50–70 functional genes, mostly involved in male sex determination and spermatogenesis. This is the correct answer.

    3. Option (C): Chromosome 1 — The largest human chromosome (about 250 million base pairs) with over 2,000 genes. Definitely not the fewest. …

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11. TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
    Q.Colour blindness in human beings is (A) Sex linked recessive disorder (B) Sex limited recessive disorder (C) Sex linked dominant disorder (D) Sex influenced disorder

    ›Reveal solutionSolution

    Colour blindness is a sex-linked recessive disorder because the defective gene is carried on the X chromosome and requires two copies in females but only one in males to manifest.

    The Concept: Why Sex-Linked Recessive?

    The key to understanding colour blindness lies in where the gene is located and how it is expressed. The gene for red-green colour blindness sits on the X chromosome — not on the Y. This makes it a sex-linked trait. Because the normal version of the gene is dominant over the defective one, a person needs either one or two copies of the defective allele to be colour-blind, depending on their sex.

    In males (XY), there is only one X chromosome. If that single X carries the defective allele, the male has no second copy to compensate — so he is colour-blind. In females (XX), both X chromosomes must carry the defective allele for colour blindness to appear; if only one X is defective, the normal dominant allele on the other X masks it, making her a carrier but not colour-blind. This pattern — more males affected, females as carriers — is the hallmark of an X-linked recessive disorder.

    Watch out

    A common mistake is to think colour blindness is dominant because it appears so often in males. But dominance/recessiveness is about the allele's behaviour in females (who have two X's), not about how common it is in males. The defective allele is recessive — it only shows up when no normal allele is present.

    Step-by-Step Reasoning

    1. Identify the chromosome involved. The gene for red-green colour blindness is located on the X chromosome. It is not on the Y chromosome. Therefore, the inheritance is sex-linked (specifically X-linked).

    2. Determine the dominance relationship. A female with one normal X and one defective X has normal vision. This tells us the normal allele is dominant over the defective one. Hence the defective allele is recessive.

    3. Check the pattern in males. A male inherits his only X from his mother. If that X carries the recessive defective allele, he will be colour-blind because there is no second X to provide a dominant normal copy. This explains why colour blindness is far more common in males. …

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12. TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
    Q.Study the following regarding genic balance theory of sex determination in Drosophila I. Organism with AAA XXX is fertile triploid female II. Organism with AAA XXY is fertile triploid male III. Organism with AAA XX is an intersex IV. Organism with AA XXY is metafemale Which of the above are correct? (A) I and IV (B) I and III (C) II and IV (D) I and II

    ›Reveal solutionSolution

    Sex in Drosophila is set by the ratio of X chromosomes to autosome sets ($X/A$); the Y is ignored. Computing $X/A$ shows only statements I and III are correct — option (B).

    By Bridges' genic balance theory the sex of Drosophila depends on the ratio $X/A$ (number of X chromosomes ÷ number of autosome sets), not on the presence of a Y. The reference values are:

    • $X/A = 1.0$ → normal female
    • $X/A = 0.5$ → normal male
    • $0.5 < X/A < 1.0$ → intersex
    • $X/A > 1.0$ (e.g. $1.5$) → metafemale (superfemale)
    • $X/A < 0.5$ (e.g. $0.33$) → metamale (supermale)

    Evaluate each statement (the Y does not enter the ratio):

    I. AAA 3X (triple-X): $X/A = 3/3 = 1.0$ → female, and being triploid it is a fertile triploid female. Correct. …

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