501. A pedigree shows unaffected parents with two affected children, one male and one female. A second family shows an affected father, unaffected mother, all daughters carrying the allele and no sons inheriting it directly from the father. Which pair of inheritance interpretations is most consistent with the two families?
ⓐ. First: autosomal dominant; second: autosomal recessive
ⓑ. First: X-linked recessive; second: Y-linked
ⓒ. First: autosomal recessive; second: X-linked recessive
ⓓ. First: X-linked dominant; second: autosomal dominant
Correct Answer: First: autosomal recessive; second: X-linked recessive
Explanation: In the first family, two unaffected parents produce affected offspring of both sexes. Under a simple model, this is readily explained if both parents are heterozygous carriers of an autosomal recessive allele. The affected daughter is especially useful as an X-linked recessive explanation would ordinarily require an affected father, which is not present. The second family shows the characteristic chromosome route of an X-linked recessive allele from an affected father. He passes his mutant X to all daughters and his Y to sons, so daughters can carry the allele while sons do not inherit it directly from him. The two pedigrees illustrate how similar recessive expression can arise through different chromosomal locations and transmission routes.
502. A carrier woman \(X^HX^h\) and a normal man \(X^HY\) have three independent sons. What is the probability that at least one son has haemophilia?
ⓐ. \(\frac{7}{8}\)
ⓑ. \(\frac{1}{8}\)
ⓒ. \(\frac{3}{8}\)
ⓓ. \(\frac{1}{2}\)
Correct Answer: \(\frac{7}{8}\)
Explanation: Each son receives Y from the normal father and one X chromosome from the carrier mother. The mother transmits \(X^H\) or \(X^h\) with equal expected probability, so each son independently has probability \(\frac{1}{2}\) of haemophilia and \(\frac{1}{2}\) of being unaffected. The probability that none of the three sons is affected is: \[ \left(\frac{1}{2}\right)^3=\frac{1}{8} \] The required event is the complement, at least one affected son: \[ 1-\frac{1}{8}=\frac{7}{8} \] Given that the three children are already specified to be sons, no additional probability for male sex is included. Previous outcomes do not change the allele-transmission probability in later independent fertilisations. An affected male transmits his X-linked allele to every daughter but does not pass that X chromosome directly to a son; the son's phenotype can directly reveal which maternal X he received.
503. A patient has reduced synthesis of one globin chain but no stated amino-acid substitution in the produced chain. Another patient produces a structurally altered beta-globin that behaves abnormally under low oxygen tension. Which conceptual contrast best separates the disorders?
ⓐ. Chromosomal versus polygenic disorder
ⓑ. Sex-linked versus autosomal
ⓒ. Dominant versus co-dominant
ⓓ. Globin quantity versus structure
Correct Answer: Globin quantity versus structure
Explanation: Reduced synthesis of a globin chain is the central quantitative feature of thalassemia. The primary problem is the amount of the relevant globin product that is produced. Sickle-cell anaemia differs since the beta-globin product itself has an altered amino-acid sequence. The structurally abnormal haemoglobin can polymerise under low oxygen tension and distort red blood cells, making it a qualitative defect in the globin product. Both disorders can be inherited as autosomal recessive conditions, so their difference is not fundamentally autosomal versus sex-linked. The most informative distinction is amount versus structure: thalassemia reduces production of a globin chain, while sickle-cell anaemia changes the nature and behaviour of beta-globin. Sickle-cell anaemia is primarily a qualitative beta-globin defect, whereas thalassemia principally reduces synthesis of a globin chain; quality of the globin product must be separated from quantity of globin synthesis.
504. A couple are both carriers of an autosomal recessive disorder. They have two children, one affected and one unaffected. What is the probability that their third child is affected?
ⓐ. \(\frac{1}{4}\)
ⓑ. \(\frac{1}{2}\)
ⓒ. \(\frac{1}{8}\)
ⓓ. \(\frac{3}{4}\)
Correct Answer: \(\frac{1}{4}\)
Explanation: If both parents are heterozygous carriers, the cross is \(Aa\times Aa\). Each independent conception has genotype probabilities \(\frac{1}{4}\) \(AA\), \(\frac{1}{2}\) \(Aa\), and \(\frac{1}{4}\) \(aa\). The affected genotype is \(aa\), so the probability for any individual child is \(\frac{1}{4}\). The outcomes of the first two children do not alter the alleles carried by the parents or the segregation probabilities in the next reproductive event. The third child's risk remains \(\frac{1}{4}\). This illustrates why observed family sequence should not be confused with changing genetic probability. Mendelian probabilities describe independent fertilisation events under unchanged parental genotypes rather than a requirement that a family eventually match the theoretical ratio exactly. The numerical relation can be checked by noting that pedigree analysis constrains genotypes from affected and unaffected relatives, allowing inheritance patterns to be inferred without controlled human crosses.
505. Two karyotypes are compared.
| Karyotype | Observation |
|---|
| P | \(47\) chromosomes with three copies of chromosome \(21\) |
| Q | \(47\) chromosomes with \(XXY\) sex chromosomes |
Which statement is correct?
ⓐ. P and Q are genetically identical because both contain \(47\) chromosomes.
ⓑ. P is polyploid, whereas Q is aneuploid.
ⓒ. P is a sex-chromosome disorder, whereas Q is autosomal trisomy.
ⓓ. Both are aneuploid; P affects an autosome, Q a sex chromosome.
Correct Answer: Both are aneuploid; P affects an autosome, Q a sex chromosome.
Explanation: Karyotype P has an additional copy of chromosome \(21\), producing trisomy \(21\), the chromosomal basis of Down's syndrome. Karyotype Q has an additional X chromosome in a male, producing \(47,XXY\), the characteristic complement of Klinefelter's syndrome. Both conditions involve gain of one particular chromosome rather than an entire chromosome set, so both are forms of aneuploidy. Their identical total chromosome counts do not make them the same disorder. The identity of the abnormal chromosome determines the classification and biological consequences. P is an autosomal trisomy, whereas Q is a sex-chromosome aneuploidy. Polyploidy would require complete additional haploid sets rather than one extra chromosome. Down's syndrome involves trisomy \(21\), Klinefelter's syndrome typically involves \(47,XXY\), and Turner's syndrome involves \(45,XO\); sex-chromosome aneuploidy and autosomal trisomy represent different chromosome targets within the same numerical-error category.
506. A family contains a woman affected with an autosomal recessive disorder. Her father is affected and her mother is unaffected. Under the simple model, what must be true of the mother's genotype?
ⓐ. She must be homozygous normal.
ⓑ. She must carry the recessive allele.
ⓒ. She must be affected genetically but not phenotypically.
ⓓ. Her genotype cannot be constrained by the daughter's phenotype.
Correct Answer: She must carry the recessive allele.
Explanation: An affected daughter with an autosomal recessive disorder has the homozygous recessive genotype \(aa\). She must receive one \(a\) allele from each parent. Her affected father is \(aa\) and necessarily contributes \(a\). The mother must supply the second \(a\). Since the mother is phenotypically unaffected, she cannot be \(aa\) under the simple complete-recessive model. She must be \(Aa\), an unaffected carrier. The daughter's genotype exposes an allele that is hidden in the mother's phenotype. This inference is possible given that an autosomal recessive affected individual requires two mutant alleles, so the parental contributions can be traced directly even when one parent does not express the disorder.
507. Evaluate the statements that follow about Down's, Klinefelter's and Turner's syndromes.
I. All three involve aneuploid chromosome complements.
II. Down's syndrome involves an autosomal chromosome.
III. Klinefelter's and Turner's syndromes involve sex chromosomes.
IV. All three necessarily have \(47\) chromosomes.
ⓐ. I and IV only
ⓑ. I, II and III only
ⓒ. II, III and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: Down's syndrome usually involves trisomy \(21\), so an autosome is present in an extra copy and the total is generally \(47\). Klinefelter's syndrome involves the sex-chromosome complement \(47,XXY\), also giving \(47\) chromosomes. Turner's syndrome differs since one X chromosome is missing, giving \(45,XO\). All three are aneuploid since each involves gain or loss of particular chromosomes rather than whole chromosome sets. The fourth statement is false: Turner's syndrome has \(45\), not \(47\), chromosomes. The comparison demonstrates why aneuploidy includes both chromosome gain and chromosome loss and why total chromosome number must be interpreted together with the chromosome involved. Down's syndrome involves trisomy \(21\), Klinefelter's syndrome typically involves \(47,XXY\), and Turner's syndrome involves \(45,XO\); the total chromosome number follows from gain or loss of a complete chromosome, not from dominance at one locus.
508. A woman has genotype \(X^CX^c\) for colour vision and \(I^Ai\) for ABO blood group. Her husband is \(X^cY\) and \(I^Bi\). Assuming independent inheritance, what is the probability that a randomly chosen child is a colour-blind daughter with blood group O?
ⓐ. \(\frac{1}{4}\)
ⓑ. \(\frac{1}{32}\)
ⓒ. \(\frac{1}{8}\)
ⓓ. \(\frac{1}{16}\)
Correct Answer: \(\frac{1}{16}\)
Explanation: A colour-blind daughter must receive \(X^c\) from both parents. The father is \(X^cY\), so every daughter receives his \(X^c\). The child must first be female, with probability \(\frac{1}{2}\), and the carrier mother must contribute \(X^c\), also with probability \(\frac{1}{2}\). Thus the probability of a colour-blind daughter is \(\frac{1}{4}\). At the ABO locus, \(I^Ai\times I^Bi\) produces blood group O, genotype \(ii\), with probability \(\frac{1}{4}\). With independent inheritance: \[ \frac{1}{4}\times\frac{1}{4}=\frac{1}{16} \] The calculation combines sex determination, X-linked recessive inheritance and autosomal multiple-allele inheritance within one offspring. Sex-specific transmission in X-linked inheritance follows directly from which parent supplies the offspring's X chromosome; father-to-son transmission is therefore excluded for the X-linked allele under the simple model. Sex-specific transmission in X-linked inheritance follows directly from which parent supplies the offspring's X chromosome; probabilities should be conditioned on sex only when the problem has already fixed the offspring's sex.
509. A new mutation produces allele \(m^\ast\). In the next generation, \(m^\ast\) segregates from \(m\), and at another independently assorting locus it becomes associated with either \(B\) or \(b\) in different offspring. Which statement correctly assigns the roles of the processes?
ⓐ. Mutation creates \(m^\ast\); segregation and assortment redistribute it among offspring.
ⓑ. Independent assortment created \(m^\ast\), while mutation merely separated it from \(m\).
ⓒ. Segregation altered \(m\) into \(m^\ast\), while dominance produced the different \(B/b\) associations.
ⓓ. Mutation is unnecessary because any new allele can arise through assortment alone.
Correct Answer: Mutation creates \(m^\ast\); segregation and assortment redistribute it among offspring.
Explanation: The appearance of \(m^\ast\) as a genetic state not previously present requires alteration of genetic material, which is the role of mutation. Once \(m^\ast\) exists in a heterozygous individual, ordinary Mendelian processes can determine its later transmission. Segregation can separate \(m^\ast\) from its alternative allele \(m\) into different gametes. At another locus, independent assortment can place the \(m^\ast\)-bearing gamete contribution into offspring genotypes carrying either \(B\) or \(b\). These later processes increase combinational variation but do not recreate the mutation. The sequence distinguishes origin of new allelic variation from redistribution of that variation during sexual reproduction. Mutagens can increase mutation frequency without directing a specific useful mutation; new heritable sequence states require alteration of genetic material rather than assortment alone.
510. A genetics investigation reports the following four observations.
P. A recessive allele hidden in \(F_1\) reappears in \(F_2\).
Q. A linked-gene test cross shows mostly parental types but some recombinants.
R. One individual carries a newly altered allele sequence.
S. Another individual has \(45,XO\).
Which integrated interpretation is most appropriate?
ⓐ. P, Q, R and S all represent failures of Mendelian segregation.
ⓑ. P and Q require mutation, whereas R and S require independent assortment.
ⓒ. P segregation; Q linkage/recombination; R mutation; S sex-chromosome aneuploidy.
ⓓ. P reflects co-dominance, Q reflects polygenic inheritance, R reflects linkage, and S reflects polyploidy.
Correct Answer: P segregation; Q linkage/recombination; R mutation; S sex-chromosome aneuploidy.
Explanation: Observation P shows that the recessive allele remains intact in the heterozygous \(F_1\) and can segregate to reappear in \(F_2\), opposing any idea of permanent blending. Observation Q concerns two linked loci: parental combinations predominate given that of physical association, while recombinant classes demonstrate that recombination can still occur. Observation R requires mutation given that the allele itself has acquired a new genetic state rather than merely a new association with existing alleles. Observation S is \(45,XO\), the chromosome complement associated with Turner's syndrome and classified as sex-chromosome monosomy, a form of aneuploidy. The four observations represent distinct but connected levels of inheritance and variation rather than one universal mechanism. The biological boundary to keep in view is that a normal gamete contributes one member of each chromosome pair; an \(n+1\) gamete can create trisomy after fertilisation, whereas an \(n-1\) gamete can create monosomy.
511. A breeder begins with a sexually reproducing plant population that already varies in seed size. For several generations, only plants producing larger seeds are chosen as parents, and the average seed size of the selected line increases. No mutagenic treatment is used. Which interpretation is most appropriate?
ⓐ. Selection directed specific mutations toward larger seed size in every generation.
ⓑ. Selection propagated existing heritable variation without requiring directed mutations.
ⓒ. Sexual reproduction prevented genetic variation because only one phenotype was selected.
ⓓ. The increase proves that environmental variation alone was inherited unchanged.
Correct Answer: Selection propagated existing heritable variation without requiring directed mutations.
Explanation: Sexual reproduction can generate and reshuffle inherited variation already present in a population. When a breeder repeatedly chooses individuals with a desired phenotype as parents, alleles contributing to that phenotype can become more represented among descendants. The procedure does not require the breeder to direct a particular mutation into existence. Mutation can generate new variation, but selection acts on variation that is available to be inherited. The observation connects heredity, variation and artificial selection without invoking evolutionary mechanisms beyond the supported scope. The critical distinction is between choosing among existing heritable differences and causing a predetermined genetic change merely because a particular phenotype is desired. A mutation changes genetic material, whereas recombination rearranges alleles that already exist; the resulting phenotype depends on the genetic change but does not redefine the type of mutation that occurred.
512. Examine the statements given below about Mendel's experimental strategy.
I. True-breeding parents reduce uncertainty about the parental genotypes being crossed.
II. Large progeny numbers make expected inheritance proportions easier to recognise.
III. Observation of a uniform \(F_1\) generation alone is sufficient to establish the Law of Segregation.
IV. Following subsequent generations allows a phenotype hidden in \(F_1\) to reappear and reveal allele separation.
ⓐ. I and III only
ⓑ. II and III only
ⓒ. I, II and III only
ⓓ. I, II and IV only
Correct Answer: I, II and IV only
Explanation: True-breeding lines provide genetically stable starting material, allowing the parental contributions to be interpreted clearly. Large progeny numbers make theoretical probability patterns more visible by reducing the effect of random sampling fluctuations on observed ratios. A uniform \(F_1\) generation can reveal dominance, but it does not by itself demonstrate segregation since the recessive alternative remains hidden in the heterozygote. Following the cross into later generations is essential: formation of different gamete classes by \(F_1\) individuals allows the recessive phenotype to reappear in \(F_2\). Thus the experimental power came from combining controlled parental material, substantial sample sizes and observations across generations rather than relying only on the first hybrid generation. A useful conceptual check is that ABO family evidence should first restrict the possible parental genotypes, after which the gametes and offspring genotypes can be evaluated.
513. Assertion: Reappearance of the recessive parental phenotype in the \(F_2\) generation argues against permanent blending of hereditary factors in the \(F_1\).
Reason: The two alleles remain discrete in the heterozygous \(F_1\) and separate during gamete formation.
ⓐ. Both are true, and Reason explains the Assertion
ⓑ. Both are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but the stated Reason is false
ⓓ. Assertion is false, but the stated Reason is true
Correct Answer: Both are true, and Reason explains the Assertion
Explanation: In a monohybrid cross, the recessive phenotype may disappear visibly in the \(F_1\) as the heterozygous genotype expresses the dominant allele. Its later reappearance in \(F_2\) shows that the recessive hereditary factor was not destroyed or permanently blended with the dominant factor. Instead, the two alleles remained distinct within the heterozygote. During gamete formation they segregated, allowing some fertilisations to unite two recessive alleles and reconstruct the recessive homozygous genotype. The Reason supplies the mechanism that explains the Assertion. This observation helped distinguish particulate inheritance from a permanent blending model and connects the visible \(F_2\) phenotype directly with the Law of Segregation. Segregation during gamete formation is the basis for the genotype probabilities recovered after fertilisation; this is why a recessive allele can reappear after being unexpressed in a heterozygote.
514. A tall pea plant of unknown genotype is crossed with a known heterozygous tall plant \(Tt\). Among their offspring, a dwarf plant is produced. What can be concluded about the unknown tall parent?
ⓐ. It must be \(TT\) because it is phenotypically tall.
ⓑ. Its genotype remains either \(TT\) or \(Tt\) because the other parent carries \(t\).
ⓒ. It must be \(Tt\): a dwarf offspring requires \(t\) from both parents.
ⓓ. It must be \(tt\) because a dwarf offspring appeared.
Correct Answer: It must be \(Tt\): a dwarf offspring requires \(t\) from both parents.
Explanation: A dwarf offspring has genotype \(tt\). The known heterozygous parent can contribute one \(t\) allele, but the second \(t\) must come from the unknown parent. Since the unknown parent is phenotypically tall, it cannot be \(tt\) under complete dominance. It must carry both \(T\) and \(t\), giving genotype \(Tt\). A \(TT\) parent could contribute only \(T\), making \(tt\) offspring impossible regardless of the allele contributed by the other parent. The offspring phenotype thus reveals an allele hidden in the unknown parent's dominant phenotype. This is the same diagnostic principle that makes recessive progeny particularly informative when inferring the genotype of an individual showing a dominant character.
515. In the cross \(RrYy\times RRYy\), complete dominance and independent assortment apply. An offspring is known to be round and yellow. What is the probability that its genotype is \(RrYy\)?
ⓐ. \(\frac{1}{6}\)
ⓑ. \(\frac{2}{3}\)
ⓒ. \(\frac{1}{3}\)
ⓓ. \(\frac{1}{2}\)
Correct Answer: \(\frac{1}{3}\)
Explanation: At the shape locus, \(Rr\times RR\) produces \(RR\) and \(Rr\) with equal probability, and every offspring is round. Among round offspring, the probability of \(Rr\) remains \(\frac{1}{2}\). At the colour locus, \(Yy\times Yy\) gives \(YY:Yy:yy\) in the ratio \(1:2:1\). Once the offspring is known to be yellow, the \(yy\) class is excluded, so the probability of \(Yy\) is \(\frac{2}{3}\). Independent assortment allows the two conditional genotype requirements to be combined: \[ \frac{1}{2}\times\frac{2}{3}=\frac{1}{3} \] The dominant phenotype at both loci does not uniquely identify a double heterozygote. Independent assortment concerns the joint transmission of alleles at different loci, not the segregation of alleles within one locus; this allows a complex cross to be reduced to two simpler one-locus decisions. Independent assortment concerns the joint transmission of alleles at different loci, not the segregation of alleles within one locus; linkage would change the joint gamete frequencies even though alleles at each locus still segregate.
516. A double heterozygote is test-crossed, producing the following progeny.
| Progeny class | Number |
|---|
| \(AB\) | \(380\) |
| \(ab\) | \(370\) |
| \(Ab\) | \(125\) |
| \(aB\) | \(125\) |
Which interpretation is best supported?
ⓐ. Parental arrangement \(Ab/aB\) with \(50\%\) recombination
ⓑ. Parental arrangement \(AB/ab\) with \(25\%\) recombination
ⓒ. Parental arrangement \(AB/ab\) with \(12.5\%\) recombination
ⓓ. Independent assortment with four equal gamete classes
Correct Answer: Parental arrangement \(AB/ab\) with \(25\%\) recombination
Explanation: In a test cross, the most frequent offspring classes identify the most frequent gametes from the heterozygous parent. The large \(AB\) and \(ab\) classes are parental, establishing the arrangement \(AB/ab\). Recombinant offspring are \(Ab\) and \(aB\), with a combined count of \(125+125=250\). Total progeny equal \(1000\), so the recombination frequency is: \[ \frac{250}{1000}\times100=25\% \] The marked excess of parental over recombinant classes supports linkage rather than independent assortment. The \(12.5\%\) value would incorrectly treat only one reciprocal recombinant class as the entire recombinant fraction. Recombination frequency is calculated from the combined recombinant progeny. The interpretation also depends on the fact that in a test cross, progeny classes reflect gametes from the heterozygote, so an excess of parental combinations over recombinants supports linkage. Recombination frequency is the recombinant fraction of progeny and serves as a relative measure of separation between linked genes; deviation from the independent-assortment expectation provides the evidence for physical association.
517. Two linked gene pairs are studied in equal-sized test-cross populations. Pair P shows \(1.3\%\) recombinant offspring, whereas pair Q shows \(37.2\%\). A graph plots parental percentage on the y-axis for the two pairs. Which pattern is expected?
ⓐ. P and Q must have identical parental percentages because both pairs are linked.
ⓑ. Q should have the higher parental percentage because higher recombination means stronger linkage.
ⓒ. P should have \(1.3\%\) parental progeny, while Q should have \(37.2\%\).
ⓓ. P has more parental types than Q because its recombinant fraction is smaller.
Correct Answer: P has more parental types than Q because its recombinant fraction is smaller.
Explanation: Parental and recombinant classes together account for the progeny in the simplified comparison. Pair P has only \(1.3\%\) recombinants, leaving approximately \(98.7\%\) parental offspring. Pair Q has \(37.2\%\) recombinants, leaving approximately \(62.8\%\) parental offspring. A graph of parental percentage should show a much taller bar for P. Biologically, the very small recombinant fraction indicates tighter linkage for pair P, whereas Q shows weaker linkage and greater relative gene separation. The comparison translates recombinant-frequency information into its complementary parental-frequency pattern rather than merely recalling which numerical value represents stronger linkage. This reasoning is consistent with the principle that linked genes still segregate at their individual loci; linkage changes the joint frequencies of allele combinations by favouring parental combinations over recombinants. Linkage modifies the joint frequencies of allele combinations without preventing segregation at the individual loci; the parental and recombinant classes must therefore be identified before any percentage is interpreted.
518. Two linked gene pairs are compared. Pair \(P-Q\) shows \(6\%\) recombination, whereas pair \(P-R\) shows \(17\%\) recombination. Which conclusion is most consistent with these observations?
ⓐ. \(P-Q\) is more tightly linked than \(P-R\).
ⓑ. \(P-R\) is more tightly linked because it produces more recombinant offspring.
ⓒ. Both pairs have equal linkage because both recombination values are below \(50\%\).
ⓓ. The recombination values cannot provide any information about relative linkage strength.
Correct Answer: \(P-Q\) is more tightly linked than \(P-R\).
Explanation: A lower recombination frequency indicates that parental allele combinations are retained more often and therefore supports stronger linkage. Pair \(P-Q\) shows only \(6\%\) recombination, whereas \(P-R\) shows \(17\%\). Thus, \(P-Q\) is more tightly linked and has a smaller relative separation than \(P-R\) in this comparison. Recombination frequency provides relative information about linkage and gene separation; it does not need to be converted into an additive three-gene map to make this conclusion.
519. In the simplified additive pigmentation model, \(AaBbCc\) is self-crossed. An offspring is known to contain at least \(4\) contributing dominant alleles among the six allele copies. What is the probability that it contains exactly \(5\) contributing dominant alleles?
ⓐ. \(\frac{3}{11}\)
ⓑ. \(\frac{5}{16}\)
ⓒ. \(\frac{3}{32}\)
ⓓ. \(\frac{6}{64}\)
Correct Answer: \(\frac{3}{11}\)
Explanation: At each independently assorting locus in \(Aa\times Aa\), the numbers of contributing dominant alleles are \(2\), \(1\) and \(0\) with probabilities \(\frac{1}{4}\), \(\frac{1}{2}\) and \(\frac{1}{4}\). Across three loci, the six allele transmissions give the binomial distribution for the total number of contributing dominant alleles. The probabilities for totals \(4\), \(5\) and \(6\) are \[ P(4)=\frac{15}{64},\qquad P(5)=\frac{6}{64},\qquad P(6)=\frac{1}{64}. \] The condition "at least \(4\)" restricts the relevant sample space to these outcomes, whose combined probability is \(\frac{22}{64}\). Renormalising within that conditioned set gives \[ P(5\mid \geq 4)=\frac{6/64}{22/64}=\frac{3}{11}. \] The calculation uses allele-count probabilities rather than dominant-phenotype classes; the condition changes the denominator but not the segregation pattern at any locus. The relevant genetic probability follows once one recognises that the additive pigmentation model counts contributing dominant alleles across several loci rather than classifying each locus only as dominant or recessive in phenotype.
520. A graph compares a polygenic pigmentation trait under two environments. The x-axis shows the number of contributing dominant alleles, while the y-axis shows mean pigmentation. In both environments, pigmentation increases as dominant-allele count increases, but all values in Environment Q are shifted downward relative to Environment P. Which conclusion best fits the graph?
ⓐ. Environment Q has converted some dominant alleles into recessive alleles.
ⓑ. The trait is genetic in Environment P but non-genetic in Environment Q.
ⓒ. The additive genetic trend remains, while environment shifts phenotype.
ⓓ. The downward shift proves pleiotropy rather than polygenic inheritance.
Correct Answer: The additive genetic trend remains, while environment shifts phenotype.
Explanation: In both environments, increasing the number of contributing dominant alleles is associated with increasing pigmentation, so the additive genetic component remains evident. The overall downward shift in Environment Q indicates that environmental conditions modify the expressed phenotype even for comparable additive genotypes. This is consistent with a polygenic character whose observed value reflects both multiple genetic contributions and environmental influence. The graph does not show alleles changing identity, nor does it eliminate the genetic relationship in one environment. Pleiotropy would involve one gene affecting several phenotypic traits, whereas the described relation concerns one trait influenced by several loci together with environment. The decisive feature is preservation of the genotype-related trend together with an environment-dependent shift in phenotype. A useful conceptual check is that independent loci allow allele-count probabilities to be combined, but the final phenotype represents the summed genetic contribution rather than a single-locus dominance decision.
