301. Assertion: Polygenic inheritance and pleiotropy describe opposite directions of a gene-to-phenotype relationship.
Reason: Polygenic inheritance involves several genes contributing to one character, whereas pleiotropy involves one gene influencing several phenotypic effects.
ⓐ. Assertion is true, but the stated Reason is false
ⓑ. Both are true, but Reason does not explain Assertion
ⓒ. Assertion is false, but the stated Reason is true
ⓓ. Both are true, and Reason explains the Assertion
Correct Answer: Both are true, and Reason explains the Assertion
Explanation: Polygenic inheritance and pleiotropy can be distinguished by following the direction of genetic influence. In polygenic inheritance, several different genes contribute to one phenotypic character, often generating a graded range of values through additive effects and environmental modification. In pleiotropy, one gene influences multiple phenotypic outcomes as its product participates in a biological process with several downstream consequences. The Reason states exactly why the two concepts can be viewed as opposite structural relationships. Human skin colour illustrates many genes influencing one trait, while phenylketonuria illustrates how alteration of one gene can have several consequences. Neither concept should be confused with multiple allelism, which concerns more than two allelic forms of one gene in a population.
302. Three observations are made.
P. One gene has \(3\) allelic forms in a population.
Q. Three genes contribute additively to one character.
R. One altered gene causes several phenotypic effects.
Which classification is correct?
ⓐ. P = multiple allelism; Q = polygenic inheritance; R = pleiotropy
ⓑ. P = pleiotropy; Q = multiple allelism; R = polygenic inheritance
ⓒ. P = polygenic inheritance; Q = pleiotropy; R = multiple allelism
ⓓ. P = co-dominance; Q = multiple allelism; R = incomplete dominance
Correct Answer: P = multiple allelism; Q = polygenic inheritance; R = pleiotropy
Explanation: Observation P concerns the number of alternative forms available at a single locus in the population, so it represents multiple allelism. Observation Q involves several different genes contributing to one character, which defines polygenic inheritance. Observation R reverses that direction, with one gene producing several phenotypic consequences, and represents pleiotropy. These categories can occur within the same genetic analysis but answer different biological questions. Multiple allelism concerns variation at one gene, polygenic inheritance concerns the number of genes influencing one trait, and pleiotropy concerns the number of phenotypic effects influenced by one gene. The classification is obtained by identifying the underlying genetic relation rather than by merely counting how many variants or phenotypes are mentioned.
303. A heterozygous self-cross has genotype ratio \(1:2:1\). Under complete dominance, the phenotypic ratio is \(3:1\). If the same alleles instead show incomplete dominance while segregation remains unchanged, what phenotypic ratio is expected?
ⓐ. \(9:3:3:1\)
ⓑ. \(1:1\)
ⓒ. \(1:2:1\)
ⓓ. \(3:1\)
Correct Answer: \(1:2:1\)
Explanation: Segregation of the two alleles during gamete formation is unchanged, so the self-cross still produces the genotype ratio \(1\) homozygote:\(2\) heterozygotes:\(1\) opposite homozygote. What changes is the phenotype assigned to the heterozygous class. Under complete dominance, the heterozygote is grouped phenotypically with one homozygote, producing \(3:1\). Under incomplete dominance, the heterozygote has its own intermediate phenotype, so all three genotype classes remain phenotypically distinguishable. The phenotypic ratio consequently becomes \(1:2:1\). This comparison shows that deviation from a \(3:1\) phenotype ratio does not necessarily indicate linkage, mutation or failed segregation. A changed dominance relation alone can alter phenotype frequencies while preserving the underlying Mendelian genotype distribution. The relevant genetic probability follows once one recognises that incomplete dominance and co-dominance alter heterozygous expression without changing the allele segregation expected from a heterozygous self-cross.
304. A genetic system has three alleles in the population. Two of those alleles are simultaneously expressed when they occur together in a heterozygote, while each is dominant over the third allele. Which pair of concepts is required to describe the system completely?
ⓐ. Polygenic inheritance and pleiotropy
ⓑ. Multiple allelism and co-dominance
ⓒ. Incomplete dominance and linkage
ⓓ. Segregation failure and polyploidy
Correct Answer: Multiple allelism and co-dominance
Explanation: The existence of three allelic forms of one gene in the population establishes multiple allelism. The second observation concerns how two particular alleles behave when paired in a heterozygote: both are expressed rather than one masking the other, which defines co-dominance. The ABO blood-group system illustrates precisely this combination, with \(I^A\), \(I^B\) and \(i\) as three population-level alleles and \(I^A\) and \(I^B\) co-dominant in an \(I^AI^B\) individual. Multiple allelism and co-dominance describe different dimensions of the same genetic system. One concerns the number of allelic forms available in the population, while the other concerns heterozygous expression of a particular allele pair. In incomplete dominance or co-dominance, the alleles still segregate normally during gamete formation; comparison with complete dominance must focus on expression of the heterozygote rather than on disappearance of an allele.
305. An \(F_2\) population has three phenotypic classes in an approximate \(1:2:1\) ratio. The heterozygous individuals are visibly intermediate between the two homozygotes. Which explanation best fits the observations?
ⓐ. Strong linkage has converted a dihybrid ratio into three classes.
ⓑ. A chromosome-number mutation has produced three phenotypes.
ⓒ. Incomplete dominance changes phenotype mapping, not segregation.
ⓓ. Co-dominance must be present because three phenotypes occur.
Correct Answer: Incomplete dominance changes phenotype mapping, not segregation.
Explanation: The decisive observation is not simply the \(1:2:1\) ratio but the phenotype of the heterozygote. It lies between the two homozygous phenotypes, which identifies incomplete dominance. Selfing the heterozygote still produces the Mendelian genotype ratio \(1:2:1\), but as all three genotypes have distinguishable phenotypes, the phenotypic ratio also remains \(1:2:1\). The result does not require linkage or a chromosome-number alteration. Co-dominance would involve simultaneous expression of both allelic contributions rather than an intermediate phenotype. This case demonstrates why an unusual phenotypic ratio should first be connected with the actual genotype-to-phenotype relation before being attributed to another genetic process such as linkage.
306. Evaluate the following statements.
I. Incomplete dominance can change an \(F_2\) phenotypic ratio without disrupting segregation.
II. Linkage can alter expected two-gene assortment while segregation at each locus still occurs.
III. Co-dominance requires permanent blending of alleles.
IV. Multiple allelism refers to more than two allelic forms in the population, not in one diploid individual.
ⓐ. I and III only
ⓑ. I, II and IV only
ⓒ. II, III and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: The first statement distinguishes phenotype expression from allele transmission. Incomplete dominance changes how the heterozygous genotype appears but does not prevent the two alleles from segregating. The second statement makes a parallel distinction at two loci: linked genes can show non-independent combinations even though each allele pair segregates normally. The third statement is false given that co-dominance reflects simultaneous expression of both allelic effects, not permanent blending of hereditary material. The fourth statement correctly defines multiple allelism as a population-level property of a locus. Together, these statements show why several inheritance patterns that differ from the simplest Mendelian phenotype ratios should not be treated as failures of Mendelian segregation. Each modifies a different biological relation.
307. In a two-gene cross, locus \(R/r\) shows incomplete dominance, while independently assorting locus \(Y/y\) shows complete dominance. Two double heterozygotes \(RrYy\) are crossed. What is the probability of an offspring having the heterozygous \(Rr\) phenotype and the dominant \(Y\)-controlled phenotype?
ⓐ. \(\frac{3}{8}\)
ⓑ. \(\frac{1}{8}\)
ⓒ. \(\frac{1}{4}\)
ⓓ. \(\frac{9}{16}\)
Correct Answer: \(\frac{3}{8}\)
Explanation: The two loci should be analysed separately before their probabilities are combined. At the incomplete-dominance locus, \(Rr\times Rr\) produces \(Rr\) with probability \(\frac{1}{2}\). Since the heterozygote has its own phenotype, the required \(Rr\) phenotype has exactly that probability. At the independently assorting \(Y\) locus, \(Yy\times Yy\) produces the dominant phenotype with probability \(\frac{3}{4}\). Since both conditions must occur in the same offspring and the loci assort independently, the joint probability is: \[ \frac{1}{2}\times\frac{3}{4}=\frac{3}{8} \] The problem integrates altered dominance at one locus with ordinary complete dominance at another without changing the underlying segregation rules. In incomplete dominance or co-dominance, the alleles still segregate normally during gamete formation; comparison with complete dominance must focus on expression of the heterozygote rather than on disappearance of an allele. In incomplete dominance or co-dominance, the alleles still segregate normally during gamete formation; the altered phenotype ratio should not be mistaken for failure of Mendelian segregation.
308. Which statement correctly distinguishes human chromosomal sex determination from X-linked inheritance?
ⓐ. Both processes refer only to whether an individual has \(XX\) or \(XY\).
ⓑ. X-linked inheritance determines whether a zygote is male or female, while \(XX/XY\) determines disease status.
ⓒ. Carrier status for an X-linked disorder changes an \(XX\) individual into an \(XY\) individual.
ⓓ. \(XX/XY\) determines sex; X-linked inheritance follows alleles on the X chromosome.
Correct Answer: \(XX/XY\) determines sex; X-linked inheritance follows alleles on the X chromosome.
Explanation: Human chromosomal sex determination concerns which sex-chromosome combination is established at fertilisation. An \(X\)-bearing ovum fertilised by an \(X\)-bearing sperm produces \(XX\), while fertilisation by a \(Y\)-bearing sperm produces \(XY\). X-linked inheritance instead concerns which allele is carried at a gene located on the X chromosome and how that allele is transmitted through a family. An \(XX\) female can be a carrier of an X-linked recessive allele without any change in her chromosomal sex. Keeping these processes separate is essential when analysing haemophilia or colour blindness, since sex determines the X-chromosome context in which the allele is expressed but is not itself the disease allele.
309. A male has genotype \(X^hY\), where \(X^h\) carries a recessive haemophilia allele. Which interpretation correctly separates the two pieces of information in this genotype?
ⓐ. \(X^h\) establishes maleness, while Y determines haemophilia.
ⓑ. Y determines male sex; \(X^h\) is the X-linked haemophilia allele.
ⓒ. Both Y and \(X^h\) are required copies of the haemophilia allele.
ⓓ. \(X^hY\) represents an autosomal recessive genotype unrelated to sex chromosomes.
Correct Answer: Y determines male sex; \(X^h\) is the X-linked haemophilia allele.
Explanation: The genotype contains information about two related but distinct biological processes. Presence of Y together with X gives the male \(XY\) chromosomal sex complement. The superscript \(h\) on the X chromosome represents a particular allele of an X-linked gene associated with haemophilia. The Y chromosome does not serve as the haemophilia allele in this notation. As the male has only one X chromosome, the recessive \(X^h\) allele can be expressed without a second X-linked normal allele to mask it. The genotype demonstrates how sex determination and sex-linked inheritance intersect: the chromosome combination establishes chromosomal sex, while the allele carried on one of those chromosomes determines the inherited disorder status.
310. An \(XX\) woman is heterozygous for an X-linked recessive colour-blindness allele and has normal colour vision. Which conclusion is correct?
ⓐ. Carrying the recessive allele changes her chromosomal sex to male.
ⓑ. Her carrier state means that both X chromosomes contain the mutant allele.
ⓒ. Her chromosomal sex and her carrier status are separate genetic properties.
ⓓ. She cannot transmit the mutant allele because it is not phenotypically expressed.
Correct Answer: Her chromosomal sex and her carrier status are separate genetic properties.
Explanation: The woman has an \(XX\) sex-chromosome complement, which establishes her chromosomal sex under the human model. At the colour-vision locus, she carries one normal allele and one recessive mutant allele, making her heterozygous and a carrier. These two descriptions refer to different levels of information. Being a carrier does not alter the number or identity of her sex chromosomes, and lack of phenotypic expression does not prevent the mutant allele from entering her ova during segregation. She can transmit either X-linked allele to offspring. This distinction is essential for pedigree analysis given that carrier status concerns a gene located on the X chromosome, whereas \(XX\) or \(XY\) concerns the chromosome combination involved in sex determination.
311. Match each process in Column I with its most appropriate contribution to inherited variation in Column II. Each Column II entry is used once.
| Column I | Column II |
|---|
| P. Segregation | 1. Produces non-parental combinations of linked alleles |
| Q. Independent assortment | 2. Separates alternative alleles into different gametes |
| R. Recombination | 3. Creates an alteration in genetic material |
| S. Mutation | 4. Combines segregation outcomes of different gene pairs independently |
ⓐ. P-4, Q-2, R-3, S-1
ⓑ. P-1, Q-3, R-4, S-2
ⓒ. P-2, Q-1, R-3, S-4
ⓓ. P-2, Q-4, R-1, S-3
Correct Answer: P-2, Q-4, R-1, S-3
Explanation: Segregation separates the two alleles of one gene pair during gamete formation, allowing different gametes to receive different alternatives. Independent assortment concerns how segregation outcomes of different gene pairs combine when those loci behave independently. Recombination generates non-parental allele combinations, especially important when linked genes are considered. Mutation differs fundamentally since it alters the genetic material itself rather than merely redistributing existing alleles. All four processes can contribute to variation, but they do so through different mechanisms. Segregation and assortment reorganise inherited alternatives, recombination creates new associations among existing alleles, and mutation can introduce a new altered genetic state. Linkage modifies the expected pattern of assortment without eliminating segregation at individual loci.
312. A genetic experiment yields three observations: parental combinations greatly exceed recombinant combinations for two loci; a small number of non-parental combinations still occur; and one offspring contains a newly altered allele not present in either parent. Which interpretation best accounts for all three findings?
ⓐ. Linkage explains parental excess, recombination non-parental types, and mutation the new allele.
ⓑ. Segregation alone explains parental excess, recombinant combinations and creation of the new allele.
ⓒ. Complete dominance explains the parental excess, while polyploidy explains all non-parental classes.
ⓓ. Independent assortment requires the parental excess, and co-dominance creates the new allele.
Correct Answer: Linkage explains parental excess, recombination non-parental types, and mutation the new allele.
Explanation: The three observations require three related but distinct genetic processes. A large excess of parental combinations over recombinant combinations indicates that the loci are linked and tend to remain associated during transmission. The minority non-parental combinations show that recombination can still occur despite linkage. The appearance of a genuinely altered allele that was absent from both parents cannot be explained merely by rearranging existing alleles; it requires mutation of genetic material. This integrated interpretation separates chromosome association, reshuffling and genetic alteration rather than assigning every form of variation to one process. It also captures an important broader principle: linkage modifies assortment, recombination reorganises existing hereditary variants, and mutation supplies new genetic alterations. A useful conceptual check is that mutation changes genetic material, while recombination rearranges existing alleles; these processes can both contribute to variation but do so in different ways.
313. Mendel's experimental design depends on starting with genetically stable parental lines. Suppose plants classified as "pure tall" produce both tall and dwarf offspring when self-pollinated before the planned hybridisation. What is the most important consequence for the inheritance experiment?
ⓐ. The dwarf phenotype must have arisen through directed mutation.
ⓑ. Uncertain parental genotype prevents clean interpretation of later generations.
ⓒ. The plants remain true-breeding because their visible parental phenotype is tall.
ⓓ. The observation proves incomplete dominance rather than complete dominance.
Correct Answer: Uncertain parental genotype prevents clean interpretation of later generations.
Explanation: A true-breeding line should reproduce the same trait consistently through repeated self-pollination. If a supposed pure tall line produces dwarf progeny on selfing, it is not genetically stable for the character being studied. Under a simple tall-dwarf Mendelian model, segregation of dwarf offspring suggests that the parent carries a recessive allele rather than being homozygous for the tall allele. Using such a parent would make the starting genotype uncertain, so uniformity or segregation observed in later generations could no longer be attributed cleanly to a cross between two known homozygous alternatives. Mendel's use of true-breeding lines was important as it fixed the parental allele composition before hybridisation. The problem is experimental ambiguity, not evidence that the recessive phenotype has been newly created or that dominance relations have automatically changed. True-breeding parents provide fixed starting allele combinations for interpreting later generations; this is why a recessive allele can reappear after being unexpressed in a heterozygote.
314. A tall pea plant has produced only tall offspring in several previous self-pollinations, but a later self-pollination yields both tall and dwarf progeny. Under the simple complete-dominance model, which conclusion is most defensible?
ⓐ. The dwarf offspring prove that the tall allele has become recessive.
ⓑ. The plant must be \(TT\) because its earlier progeny were all tall.
ⓒ. The plant must have lost the \(T\) allele during the final self-pollination.
ⓓ. A dwarf offspring shows the tall parent carried \(t\), excluding certain \(TT\).
Correct Answer: A dwarf offspring shows the tall parent carried \(t\), excluding certain \(TT\).
Explanation: A dwarf offspring has genotype \(tt\), so it must receive a \(t\) allele from each gamete involved in its formation. If the plant was self-pollinated, the same parental plant supplied both gametes. The plant must consequently have been capable of producing \(t\)-bearing gametes and is consistent with genotype \(Tt\), not \(TT\). Earlier all-tall progeny do not change the biological meaning of the later recessive offspring. A finite sample from a heterozygous self-cross can, by chance, fail to contain the recessive phenotype even though \(t\) is present. The later observation supplies decisive genotype information given that \(TT\) cannot generate \(tt\) offspring under normal segregation. This illustrates why repeated generations and sufficiently large progeny samples strengthen genetic inference when establishing true-breeding material.
315. From the \(F_2\) of \(Tt\times Tt\), two tall plants are selected independently and crossed with each other. What is the probability that both selected plants are heterozygous and that their first offspring is dwarf?
ⓐ. \(\frac{1}{9}\)
ⓑ. \(\frac{2}{9}\)
ⓒ. \(\frac{1}{6}\)
ⓓ. \(\frac{4}{9}\)
Correct Answer: \(\frac{1}{9}\)
Explanation: In the \(F_2\), the genotype ratio is \(1\,TT:2\,Tt:1\,tt\). Among tall plants, the \(tt\) class is absent, so the relevant conditional composition is \(1\,TT:2\,Tt\). A randomly selected tall plant has probability \(\frac{2}{3}\) of being \(Tt\). For two independently selected tall plants, the probability that both are heterozygous is \(\frac{2}{3}\times\frac{2}{3}=\frac{4}{9}\). Only after establishing that both parents are \(Tt\) do we evaluate their cross. \(Tt\times Tt\) produces a dwarf \(tt\) offspring with probability \(\frac{1}{4}\). Combining the selection stage with the offspring stage gives \(\frac{4}{9}\times\frac{1}{4}=\frac{1}{9}\). The problem requires conditional genotype inference before applying ordinary Mendelian offspring probability. Genotype must be established before phenotype is assigned under the relevant dominance relation; a test cross becomes informative when a dominant phenotype could correspond to more than one genotype.
316. A tall plant selected from the \(F_2\) of \(Tt\times Tt\) is test-crossed with \(tt\). A dwarf offspring is obtained. What can now be concluded about the selected tall plant?
ⓐ. Its genotype could still be either \(TT\) or \(Tt\).
ⓑ. It must have been \(TT\), because it originally showed the dominant phenotype.
ⓒ. It was \(Tt\), since a dwarf offspring requires parental \(t\).
ⓓ. Its genotype cannot be inferred unless the full \(F_2\) population size is known.
Correct Answer: It was \(Tt\), since a dwarf offspring requires parental \(t\).
Explanation: The tester has genotype \(tt\) and can contribute only \(t\) gametes. A dwarf offspring must have genotype \(tt\), so the tested tall parent must also have contributed \(t\). A tall plant carrying \(t\) under complete dominance has genotype \(Tt\). A homozygous dominant \(TT\) parent crossed with \(tt\) can produce only \(Tt\) offspring, all of which are tall. The appearance of even one dwarf offspring identifies the hidden recessive allele in the tall parent. The inference does not depend on the original \(F_2\) population size since the progeny phenotype provides direct evidence about the allele transmitted by the tested individual. This is the diagnostic power of a test cross: the homozygous recessive tester exposes alleles that may be hidden behind a dominant phenotype.
317. In Antirrhinum, \(RR\) is red, \(Rr\) is pink and \(rr\) is white. A random non-white plant is chosen from the progeny of \(Rr\times Rr\) and crossed with a white plant. What is the probability that the first offspring of this new cross is white?
ⓐ. \(\frac{1}{4}\)
ⓑ. \(\frac{1}{3}\)
ⓒ. \(\frac{1}{2}\)
ⓓ. \(\frac{2}{3}\)
Correct Answer: \(\frac{1}{3}\)
Explanation: The original \(Rr\times Rr\) cross gives \(RR:Rr:rr\) in the ratio \(1:2:1\), corresponding to red:pink:white in the same ratio. Once the selected plant is specified to be non-white, the \(rr\) class is excluded. Among the remaining plants, the conditional probabilities are \(\frac{1}{3}\) for \(RR\) and \(\frac{2}{3}\) for \(Rr\). If the chosen plant is \(RR\), crossing with \(rr\) gives only \(Rr\) pink progeny and no white offspring. If it is \(Rr\), the cross \(Rr\times rr\) gives white offspring with probability \(\frac{1}{2}\). The overall probability is \(\frac{2}{3}\times\frac{1}{2}=\frac{1}{3}\). This combines incomplete-dominance phenotype information with conditional genotype selection and a changed second cross. The expected value is obtained from the biological relation that an intermediate heterozygote indicates incomplete dominance, whereas simultaneous expression of both allelic effects indicates co-dominance; both differ from complete dominance at the phenotype-mapping step.
318. Enzyme activity associated with three genotypes is measured as follows.
| Genotype | Relative functional product | Phenotype |
|---|
| \(AA\) | \(100\) units | Normal |
| \(Aa\) | \(52\) units | Normal |
| \(aa\) | \(0\) units | Altered |
The normal phenotype requires at least \(40\) units of functional product. Which interpretation best explains dominance in this system?
ⓐ. One \(A\) allele supplies enough product for a normal phenotype.
ⓑ. Allele \(A\) physically destroys allele \(a\) in the heterozygote.
ⓒ. Genotypes \(AA\) and \(Aa\) must produce exactly equal amounts of product whenever dominance occurs.
ⓓ. Allele \(a\) is absent from \(Aa\) because its product level is below the normal requirement.
Correct Answer: One \(A\) allele supplies enough product for a normal phenotype.
Explanation: The data distinguish biochemical product level from visible phenotype. \(AA\) produces \(100\) units, while \(Aa\) produces only \(52\), so the two genotypes are not biochemically identical. However, both exceed the stated \(40\)-unit requirement for the normal phenotype. A single functional \(A\) allele supplies sufficient product for normal expression, allowing \(Aa\) to resemble \(AA\) phenotypically. The recessive \(a\) allele remains present in the genotype and can still segregate into gametes; dominance does not require its removal or destruction. The table also shows why dominance cannot be defined as equal product output from homozygote and heterozygote. The phenotype depends on whether functional product is sufficient for the biological requirement, not on whether the two genotypes manufacture identical quantities. A heterozygote carries two alleles but contributes only one allele of the pair to each normal gamete; this is why a recessive allele can reappear after being unexpressed in a heterozygote.
319. A homozygous individual producing surface marker P is crossed with a homozygous individual producing surface marker Q. Every \(F_1\) individual displays both P and Q simultaneously. Selfing the \(F_1\) gives three phenotypic classes in a \(1:2:1\) ratio. Which explanation best fits both generations?
ⓐ. Complete dominance with loss of the recessive marker in \(F_1\)
ⓑ. Incomplete dominance producing an intermediate P-Q marker
ⓒ. Polygenic inheritance involving separate genes for P and Q
ⓓ. Co-dominance with normal segregation of the two alleles
Correct Answer: Co-dominance with normal segregation of the two alleles
Explanation: The decisive \(F_1\) observation is simultaneous expression of both parental markers rather than an intermediate state between them. That identifies co-dominance. If the two alleles segregate normally during gamete formation, selfing the heterozygote gives the genotype ratio \(1:2:1\). Since each of the three genotypes has a distinguishable phenotype under co-dominant expression, the phenotypic ratio can also be \(1:2:1\). The outcome does not require blending of markers or a failure of segregation. It also does not require several genes, since the pattern can arise from two alleles of one gene whose effects are both expressed in the heterozygote. The combination of \(F_1\) phenotype and \(F_2\) segregation is essential: the first identifies the expression relation, while the second confirms continued discrete inheritance of the alleles.
320. A woman with blood group A and a man with blood group B have one child with blood group O and another with blood group AB. Which parental genotypes are required under the standard ABO model?
ⓐ. Mother \(I^AI^A\); father \(I^BI^B\)
ⓑ. Mother \(I^AI^A\); father \(I^Bi\)
ⓒ. Mother \(I^Ai\); father \(I^Bi\)
ⓓ. Mother \(I^Ai\); father \(I^BI^B\)
Correct Answer: Mother \(I^Ai\); father \(I^Bi\)
Explanation: The O child has genotype \(ii\), so each parent must contribute an \(i\) allele. A group A parent capable of contributing \(i\) must be \(I^Ai\), and a group B parent capable of contributing \(i\) must be \(I^Bi\). These genotypes also permit an AB child since the A parent can contribute \(I^A\) while the B parent contributes \(I^B\), producing \(I^AI^B\). The two children together identify both hidden parental recessive alleles and the expressed A and B alleles. A homozygous A or homozygous B parent could not produce the O child since such a parent lacks \(i\). The family data are more informative than parental phenotypes alone, which would otherwise leave each parent with two possible genotypes.
