Principles of Inheritance and Variation MCQs with Answers – Part 2 (Class 12 Biology)

Explanation: Production of both \(R\) and \(r\) gametes requires the parent to carry both alleles at the seed-shape locus, so it must be \(Rr\) there. Similarly, production of both \(Y\) and \(y\) requires genotype \(Yy\) at the colour locus. Combining those deductions gives \(RrYy\). Independent assortment then allows each allele at one locus to combine with either allele at the other, generating four gamete types in equal expected proportions. A parent homozygous at either locus would produce fewer allele combinations. This problem reverses the usual direction of Mendelian reasoning: instead of predicting gametes from a known genotype, the observed gamete set is used to reconstruct the diploid genotype that could generate it.

Explanation: Gametes carry one allele from each gene pair, so an \(RrYy\) parent contributes either \(R\) or \(r\) and either \(Y\) or \(y\) to a given gamete. When the two loci assort independently, the four combinations occur with equal expected probability. Segregation describes separation within each allelic pair, while independent assortment describes the relation between the two separate segregation events. A gamete cannot normally be labelled \(RrYy\), since that notation retains both alleles at both loci and represents a diploid genotype. The valid statements collectively show that dihybrid gamete formation depends on ordinary segregation at each locus plus independence in how the two locus-specific outcomes combine. The biological boundary to keep in view is that in the ABO system, \(I^A\) and \(I^B\) are co-dominant to each other, while \(i\) is recessive to both; a phenotype may therefore correspond to more than one genotype.

Explanation: The plant is heterozygous only at the seed-shape locus. It can contribute \(R\) or \(r\) with expected probabilities of \(\frac{1}{2}\) each. At the colour locus it is \(YY\), so every gamete necessarily receives \(Y\). The only gamete types are consequently \(RY\) and \(rY\), expected in a \(1:1\) relation. The expected number of \(rY\) gametes is \(400\times\frac{1}{2}=200\). The reduction from four gamete classes to two comes from homozygosity at one locus, not from failure of independent assortment. Independent assortment can generate alternative combinations only where alternative alleles are actually available to segregate from the parent. The relevant genetic probability follows once one recognises that for an \(RrYy\) heterozygote under independent assortment, the four gamete types \(RY\), \(Ry\), \(rY\), and \(ry\) have equal expected frequencies. An \(RrYy\) heterozygote forms four gamete types in equal expected proportions when the loci assort independently; linkage would change the joint gamete frequencies even though alleles at each locus still segregate.

Explanation: The total number of gametes is \(1000\), so equal representation of four classes would give an expectation of about \(250\) of each type. The observed counts lie very close to that value. An \(RrYy\) parent has two possible alleles at each locus, and independent assortment predicts all four combinations with probability \(\frac{1}{4}\) each. Small departures from exactly equal counts are compatible with sampling variation and do not by themselves contradict the expectation. The important evidence is the near-equal recovery of both parental and newly combined gamete types. This pattern is what one expects when segregation occurs normally at both loci and the outcome at one locus does not systematically constrain the outcome at the other. An \(RrYy\) heterozygote forms four gamete types in equal expected proportions when the loci assort independently; conditional information must be applied after identifying which one-locus outcomes remain possible.

Explanation: Wrinkled green offspring must be recessive for both characters, so their genotype is \(rryy\). In the self-cross \(RrYy\times RrYy\), the probability of obtaining \(rr\) at the seed-shape locus is \(\frac{1}{4}\), and the probability of obtaining \(yy\) at the independently assorting colour locus is also \(\frac{1}{4}\). The joint probability is \(\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}\). Together with the other three phenotypic fractions, this completes the familiar \(9:3:3:1\) dihybrid relation. The smallest class represents the offspring that must receive a recessive allele from both parents at both independently segregating loci. An \(RrYy\) heterozygote forms four gamete types in equal expected proportions when the loci assort independently; the familiar combined ratios arise from multiplication of biologically independent segregation events. An \(RrYy\) heterozygote forms four gamete types in equal expected proportions when the loci assort independently; this allows a complex cross to be reduced to two simpler one-locus decisions.

Explanation: Each locus in \(RrYy\times RrYy\) behaves as a monohybrid cross under the stated assumptions. At either locus, the dominant phenotype has probability \(\frac{3}{4}\) and the recessive phenotype has probability \(\frac{1}{4}\). Independent assortment allows these probabilities to be multiplied across loci. The four combinations become \(\frac{9}{16}\) dominant-dominant, \(\frac{3}{16}\) dominant-recessive, \(\frac{3}{16}\) recessive-dominant, and \(\frac{1}{16}\) recessive-recessive. These fractions correspond to \(9:3:3:1\). The Reason supplies the probability structure that generates the asserted phenotypic ratio, linking independent one-locus segregation patterns into the combined two-gene outcome. The same conclusion is supported by the fact that for an \(RrYy\) heterozygote under independent assortment, the four gamete types \(RY\), \(Ry\), \(rY\), and \(ry\) have equal expected frequencies.

Explanation: A wrinkled phenotype requires the recessive seed-shape genotype \(rr\), so both alleles at that locus are fixed by the phenotype. Yellow seed colour requires at least one dominant \(Y\) allele, meaning the colour genotype may be either \(YY\) or \(Yy\). Combining these locus-specific conditions gives \(rrYY\) or \(rrYy\). This example illustrates why two-gene phenotype interpretation should be decomposed into separate decisions for each locus before the results are combined. One phenotype can uniquely determine a recessive homozygous genotype at one locus while leaving two possible genotypes at the second locus. Correct dihybrid reasoning keeps those constraints separate until the final multilocus genotype is assembled.

Explanation: At the seed-shape locus, \(Rr\times Rr\) gives a probability of \(\frac{1}{4}\) for \(rr\). Independently, \(Yy\times Yy\) gives a probability of \(\frac{1}{4}\) for \(yy\). The requested offspring must satisfy both genotype conditions at the same time. Multiplying the independent probabilities gives \(\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}\). Biologically, this is the double-recessive class, corresponding to the wrinkled green phenotype. The result also explains why this phenotype is the least frequent class in the standard \(9:3:3:1\) distribution. Both recessive homozygous conditions must coincide within the same zygote. Combined phenotype classes in a two-gene cross depend on both one-locus genotype probabilities and the dominance relation at each locus; the familiar combined ratios arise from multiplication of biologically independent segregation events. Combined genotype probabilities in a two-gene cross can be obtained locus by locus; the probability of the double-recessive genotype is the product of the two recessive-genotype probabilities.

Explanation: The total is \(1600\) offspring. A \(9:3:3:1\) expectation predicts \(900\) round yellow, \(300\) round green, \(300\) wrinkled yellow, and \(100\) wrinkled green individuals. The observed counts are very close to those values. Such agreement is consistent with complete dominance at each locus and independent assortment between the two gene pairs under the simple Mendelian model. The data should be interpreted as approximate rather than demanding exact numerical equality, since biological samples can show random departures from theoretical proportions. The four-class pattern is especially informative as the two single-recessive classes occur at similar intermediate frequencies while the double-recessive class remains much less frequent. This reasoning is consistent with the principle that for an \(RrYy\) heterozygote under independent assortment, the four gamete types \(RY\), \(Ry\), \(rY\), and \(ry\) have equal expected frequencies.

Explanation: Segregation and independent assortment address related but distinct levels of inheritance. Segregation concerns the two alleles of a single gene pair separating so that individual gametes receive one allele. Independent assortment asks whether the segregation outcome of one gene pair occurs independently of the segregation outcome of another gene pair. A pair of linked genes can still show allele segregation at each locus while displaying non-independent combinations across the two loci. This makes the third statement biologically important: failure of independent assortment does not automatically imply failure of segregation. Treating the two laws as synonyms removes the distinction needed to understand why monohybrid inheritance may remain Mendelian even when a dihybrid ratio departs from \(9:3:3:1\). The biological boundary to keep in view is that a dihybrid outcome is built from two one-locus segregation events, and combined phenotype classes depend on both allele segregation and the dominance relation at each locus.

Explanation: The \(rryy\) tester contributes only \(ry\) gametes, so variation among the offspring directly reflects the gamete types supplied by the \(RrYy\) parent. An approximately \(1:1:1:1\) distribution of the four test-cross phenotypes indicates that \(RY\), \(Ry\), \(rY\), and \(ry\) gametes are being produced at roughly equal frequencies. This is the expected pattern when the two heterozygous loci assort independently. The test cross is especially informative since the recessive tester does not conceal which alleles arrived from the dihybrid parent. Offspring phenotype classes can thus serve as a readable record of the gametic combinations produced by that parent. An \(RrYy\) heterozygote forms four gamete types in equal expected proportions when the loci assort independently; the product rule is valid only when the stated independence assumption applies.

Explanation: The process begins during gamete formation, when \(R\) separates from \(r\) and \(Y\) separates from \(y\). Under independent assortment, these segregation events generate four expected gamete types. Selfing brings gametes from two such parents together, producing multiple zygotic genotype combinations. Only after those genotypes are established can dominance be applied at each locus to classify offspring into round yellow, round green, wrinkled yellow, and wrinkled green phenotypes. Grouping the genotypes gives the \(9:3:3:1\) expectation. The order matters biologically: phenotype ratios arise from allele segregation, gamete combination, fertilisation, and genotype expression rather than directing those processes themselves. A useful conceptual check is that for an \(RrYy\) heterozygote under independent assortment, the four gamete types \(RY\), \(Ry\), \(rY\), and \(ry\) have equal expected frequencies.

Explanation: Round seeds require the dominant seed-shape phenotype, which occurs with probability \(\frac{3}{4}\) in \(Rr\times Rr\). Green seeds require the recessive colour genotype \(yy\), which occurs with probability \(\frac{1}{4}\) in \(Yy\times Yy\). With independent assortment, the two conditions can be combined by multiplication: \(\frac{3}{4}\times\frac{1}{4}=\frac{3}{16}\). This class contains more than one possible genotype at the shape locus, since round offspring may be \(RR\) or \(Rr\), but the colour locus must be \(yy\). The calculation connects phenotype conditions to the correct locus-specific probabilities before combining them into a two-character outcome. Combined phenotype classes in a two-gene cross depend on both one-locus genotype probabilities and the dominance relation at each locus; the familiar combined ratios arise from multiplication of biologically independent segregation events. Combined phenotype classes in a two-gene cross depend on both one-locus genotype probabilities and the dominance relation at each locus; conditional information must be applied after identifying which one-locus outcomes remain possible.

Explanation: Under complete dominance, each heterozygous monohybrid self-cross produces dominant and recessive phenotypes in an expected \(3:1\) relation. When two such loci assort independently, the phenotype probabilities combine multiplicatively. Dominant at both loci corresponds to \(\frac{3}{4}\times\frac{3}{4}=\frac{9}{16}\); dominant at the first and recessive at the second gives \(\frac{3}{16}\); the reciprocal combination also gives \(\frac{3}{16}\); and recessive at both gives \(\frac{1}{16}\). These four fractions form \(9:3:3:1\). The multiplication represents joint occurrence of independent locus-specific conditions and is not merely an arithmetic shortcut detached from the underlying segregation events.

Explanation: Linkage concerns the association of genes on the same chromosome and can make some allele combinations occur more often than others. Segregation at each locus can still take place, so individual gametes continue to receive one allele for each gene. What may fail is the assumption that the two segregation events combine independently to yield equal frequencies of all four gamete types from a double heterozygote. Once those gametic frequencies become unequal, the standard \(9:3:3:1\) dihybrid expectation need not appear. This boundary is important when applying Mendel's Law of Independent Assortment: the ratio follows from independence under the stated model and should not be imposed automatically on every pair of genes.

Explanation: The cross should first be separated into its two locus-specific components. At the \(T\) locus, \(Tt\times Tt\) gives the dominant phenotype with probability \(\frac{3}{4}\). At the \(Y\) locus, \(Yy\times yy\) gives the dominant phenotype only when the first parent contributes \(Y\), so that probability is \(\frac{1}{2}\). The two loci are stated to assort independently, allowing the joint probability to be calculated as \(\frac{3}{4}\times\frac{1}{2}=\frac{3}{8}\). This approach avoids constructing a large Punnett square and keeps genotype and phenotype requirements distinct. Each locus is solved according to its own parental genotypes before the two required outcomes are combined. Combined phenotype probabilities in a two-gene cross can be obtained locus by locus; the requested joint phenotype follows from multiplying the dominant-phenotype probability at each locus under the stated independence assumption.

Explanation: The second parent has genotype \(Ttyy\). At the \(Y\) locus, this parent can contribute only \(y\), since it carries no \(Y\) allele. Every offspring must consequently receive at least one \(y\) from that parent. Genotype \(YY\) is impossible in the progeny, regardless of what happens at the independently assorting \(T\) locus. Genotypes \(Yy\) are possible when the \(TtYy\) parent contributes \(Y\). Among the listed choices, only \(TTYY\) requires two \(Y\) alleles and is excluded by the parental genotype. The problem shows why multilocus crosses should be decomposed locus by locus: impossibility at one locus is sufficient to eliminate an entire combined genotype.

Explanation: The first parent is heterozygous at both loci and, with independent assortment, can produce \(TY\), \(Ty\), \(tY\), and \(ty\) gametes. The second parent is \(Ttyy\), so it can contribute \(Ty\) or \(ty\). Since both parents can supply \(y\), offspring with genotype \(yy\) are possible. The genotype \(YY\), however, cannot occur since the second parent possesses no \(Y\) allele to contribute. This statement set requires separate analysis of the two loci before considering the combined cross. Heterozygosity at the \(T\) locus affects both parents, while the fixed \(yy\) state of the second parent imposes a strict constraint on colour-locus offspring genotypes. Segregation during gamete formation is the basis for the genotype probabilities recovered after fertilisation; a test cross becomes informative when a dominant phenotype could correspond to more than one genotype.

Explanation: Genotype \(ttyy\) is homozygous recessive at both loci. To obtain \(tt\), each parent must contribute \(t\), and to obtain \(yy\), each must contribute \(y\). The required gamete from each parent is consequently \(ty\). The first parent \(TtYy\) can produce \(ty\) among its four independently assorting gamete types, while the second parent \(Ttyy\) can also produce \(ty\). Their fusion supplies \(t+t\) and \(y+y\), generating the double-recessive genotype. This backward inference demonstrates the connection between offspring genotype and the exact haploid contributions that must have united during fertilisation. It is the same segregation logic used prospectively, applied in reverse.

Explanation: At the \(T\) locus, \(Tt\times Tt\) gives dominant and recessive phenotypes in a \(3:1\) relation. At the \(Y\) locus, \(Yy\times yy\) gives dominant and recessive phenotypes in a \(1:1\) relation. Combining these independent probabilities gives \(\frac{3}{8}\) dominant-dominant, \(\frac{3}{8}\) dominant-recessive, \(\frac{1}{8}\) recessive-dominant, and \(\frac{1}{8}\) recessive-recessive, which corresponds to \(3:3:1:1\). Out of \(200\) offspring, the theoretical counts would be \(75\), \(75\), \(25\), and \(25\), closely matching the observations. The altered ratio arises from different parental genotypes at the two loci, not from a failure of independent assortment. A useful conceptual check is that independent assortment allows the two loci to be solved separately and their probabilities combined by the product rule; linkage would invalidate that independence assumption.