401. Match each sex-determination system in Column I with the factor in Column II that provides the alternative developmental possibilities relevant to sex determination. Each Column II entry is used once.
| Column I | Column II |
|---|
| P. Human \(XX/XY\) | 1. X-bearing versus Y-bearing sperm from the male |
| Q. \(XX/XO\) | 2. X-bearing versus sex-chromosome-free sperm from the male |
| R. Bird \(ZZ/ZW\) | 3. Z-bearing versus W-bearing ova from the female |
| S. Honey bee haplodiploidy | 4. Fertilised diploid versus unfertilised haploid development |
ⓐ. P-2, Q-1, R-4, S-3
ⓑ. P-1, Q-2, R-3, S-4
ⓒ. P-1, Q-3, R-2, S-4
ⓓ. P-3, Q-2, R-1, S-4
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: Humans have an \(XX/XY\) system: females produce only X-bearing ova, while males produce X-bearing and Y-bearing sperm, so P maps to 1. In an \(XX/XO\) system, the female is again homogametic, whereas the male produces sperm that either carry X or lack a sex chromosome, giving Q-2. Birds use the \(ZZ/ZW\) system, in which the male is homogametic and the female is heterogametic; Z-bearing and W-bearing ova provide the alternatives, so R-3. Honey bees use haplodiploidy rather than an alternative sex-chromosome-gamete system. Fertilised diploid eggs develop as females and unfertilised haploid eggs develop as males, giving S-4. The four mappings distinguish male heterogamety, female heterogamety and ploidy-based sex determination without treating these mechanisms as interchangeable.
402. In humans, suppose only \(Y\)-bearing sperm participate in fertilisation. In birds, suppose only \(W\)-bearing ova participate in fertilisation. Under the usual \(XX/XY\) and \(ZZ/ZW\) systems, respectively, which offspring outcomes are expected?
ⓐ. Human females and bird males
ⓑ. Human females and bird females
ⓒ. Human males and bird males
ⓓ. Human males and bird females
Correct Answer: Human males and bird females
Explanation: A human ovum carries X. If only Y-bearing sperm fertilise those ova, every resulting zygote has the combination \(XY\) and is male under the simple human chromosomal model. In birds, the normal male is \(ZZ\) and contributes Z-bearing sperm. If only W-bearing ova from the heterogametic \(ZW\) female participate, each fertilisation produces \(ZW\), which is female. The two cases illustrate opposite locations of heterogamety. The human male supplies the variable X or Y sperm, while the avian female supplies the variable Z or W ovum. Restricting the heterogametic parent to one gamete class removes one of the two usual sex-chromosome outcomes and shifts the offspring entirely toward the sex associated with that retained gamete class. The same conclusion is supported by the fact that in an \(XX/XY\) system the male is heterogametic, while in a \(ZZ/ZW\) system the female is heterogametic; the heterogametic sex produces two sex-chromosome gamete types.
403. A human couple and a bird pair each produce two offspring. Assume independent fertilisations and equal production of the two relevant gamete classes by the heterogametic parent. What is the probability that each pair produces one male and one female, in either order?
ⓐ. Human: \(\frac{1}{4}\); bird: \(\frac{1}{4}\)
ⓑ. Human: \(\frac{1}{2}\); bird: \(\frac{1}{4}\)
ⓒ. Human: \(\frac{1}{4}\); bird: \(\frac{1}{2}\)
ⓓ. Human: \(\frac{1}{2}\); bird: \(\frac{1}{2}\)
Correct Answer: Human: \(\frac{1}{2}\); bird: \(\frac{1}{2}\)
Explanation: In each system, the stated assumptions give a probability of \(\frac{1}{2}\) for each sex at every independent fertilisation. The chromosomal mechanism differs between the organisms: in humans the \(XY\) male produces X- or Y-bearing sperm, while in birds the \(ZW\) female produces Z- or W-bearing ova. Nevertheless, the two-offspring probability calculation has the same structure. One male and one female can occur in two orders. The probability of male then female is \(\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\), and female then male also has probability \(\frac{1}{4}\). Adding the mutually exclusive sequences gives \(\frac{1}{2}\). Equal numerical probability does not imply identical biological mechanisms; the heterogametic parent differs between humans and birds. Male heterogamety in an \(XX/XY\) system differs from female heterogamety in a \(ZZ/ZW\) system; previous births do not alter the chromosomal probability for a later independent conception.
404. A haploid honey bee male carries allele \(M\). He fathers a diploid daughter with a queen that carries allele \(m\) at the corresponding locus. The daughter is \(Mm\) and later produces an unfertilised egg that develops into a son. Assuming ordinary segregation in the daughter, what is the probability that this grandson receives allele \(M\)?
ⓐ. \(1\)
ⓑ. \(\frac{1}{2}\)
ⓒ. \(\frac{1}{4}\)
ⓓ. \(0\)
Correct Answer: \(\frac{1}{2}\)
Explanation: The original haploid male carries only allele \(M\) at the locus, so sperm produced by him transmit \(M\). His diploid daughter receives \(M\) from him and \(m\) from the queen, giving genotype \(Mm\). The daughter's son develops from an unfertilised egg, so he receives his genetic material entirely through that maternal egg and has no father. During the daughter's gamete formation, \(M\) and \(m\) segregate into different eggs with expected probabilities of \(\frac{1}{2}\) each. The probability that the grandson receives the allele originally contributed by his maternal grandfather is \(\frac{1}{2}\). This problem combines the unusual honey bee ancestry pattern with ordinary segregation in a diploid female rather than treating haplodiploidy as an exception to allele segregation. A diploid queen carries homologous allele pairs, whereas a haploid male carries only one allele at each locus; allele probabilities differ between the diploid maternal and haploid paternal contributions.
405. Assess the statements listed below about mutation, recombination and inheritance.
I. Recombination can place existing alleles into a new linked combination.
II. Mutation can create an altered genetic state that did not previously exist.
III. A mutation confined to a non-reproductive somatic lineage is automatically transmitted to offspring.
IV. A mutation entering a gamete-forming lineage can contribute to inherited variation.
ⓐ. I and III only
ⓑ. II and III only
ⓒ. I, II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: Recombination changes associations among alleles that already exist, particularly when linked loci are considered. Mutation differs as it can alter genetic material itself and create a new allelic or chromosomal state. Heritability depends on where the mutation occurs. A genetic alteration restricted to a somatic cell lineage does not automatically enter gametes and is not necessarily transmitted through sexual reproduction. If a mutation occurs in a lineage that contributes to gamete formation, the altered genetic material can be passed to descendants when a gamete carrying it participates in fertilisation. The statements distinguish three separate issues: creation of new genetic states, rearrangement of existing states, and access of a genetic alteration to the reproductive lineage. Only the third statement incorrectly treats every mutation as automatically heritable.
406. Select the sequence that best connects a heritable point mutation with a later family phenotype.
ⓐ. Base change → altered allele → gametic transmission → altered product → phenotype
ⓑ. Phenotype → gamete formation → base-pair alteration → chromosome gain → changed allele
ⓒ. Independent assortment → altered protein → mutation disappears → phenotype
ⓓ. Chromosome segregation → phenotype → point mutation → fertilisation
Correct Answer: Base change → altered allele → gametic transmission → altered product → phenotype
Explanation: A heritable point mutation begins with alteration of genetic material at the base-pair level. That change creates an altered allele. For the mutation to affect a descendant, the altered allele must enter a gamete and be transmitted through fertilisation. Once present in the descendant's genotype, the allele can influence the structure or amount of the relevant gene product, which can then produce a cellular or organismal phenotype. The sequence keeps genetic origin, inheritance and expression in their correct causal order. A phenotype does not precede the mutation that causes it, and independent assortment does not create the altered sequence itself. The reasoning links mutation as a source of variation with ordinary inheritance of the resulting allele and with downstream genotype-to-phenotype expression. The biological boundary to keep in view is that classifying a mutation requires identifying what changed in the genetic material before linking that change to a phenotypic consequence.
407. A pedigree shows an affected woman and an unaffected man with two children, one affected and one unaffected. Under a simple autosomal dominant model in which unaffected individuals are \(aa\), what can be inferred about the affected mother's genotype?
ⓐ. She must be \(AA\) because a dominant phenotype is always homozygous.
ⓑ. She must be \(Aa\) because the unaffected child must receive \(a\) from both parents.
ⓒ. She must be \(aa\) because she transmitted \(a\) to the unaffected child.
ⓓ. Her genotype cannot be narrowed because a dominant allele never reveals zygosity.
Correct Answer: She must be \(Aa\) because the unaffected child must receive \(a\) from both parents.
Explanation: In a simple autosomal dominant model, an unaffected person has genotype \(aa\). The unaffected father can therefore contribute only \(a\) alleles. Their unaffected child is also \(aa\), so the child must have received one \(a\) allele from the affected mother. That observation rules out \(AA\) for the mother, since an \(AA\) parent could contribute only the dominant allele. The mother is affected, so she cannot be \(aa\); her genotype must be \(Aa\). The affected child is also compatible with this inference because that child can receive the dominant allele from the mother and \(a\) from the father. Pedigree reasoning here depends on using the unaffected offspring to expose the hidden recessive allele carried by a dominant-phenotype parent.
408. A woman is unaffected but has an affected brother with an autosomal recessive disorder. Her parents are known heterozygous carriers. She has a child with a man who is also known to be a carrier. Given that the woman is unaffected, what is the probability that their child is affected?
ⓐ. \(\frac{1}{4}\)
ⓑ. \(\frac{1}{8}\)
ⓒ. \(\frac{1}{3}\)
ⓓ. \(\frac{1}{6}\)
Correct Answer: \(\frac{1}{6}\)
Explanation: The woman's parents are \(Aa\times Aa\), so their offspring probabilities are \(\frac{1}{4}\) \(AA\), \(\frac{1}{2}\) \(Aa\), and \(\frac{1}{4}\) \(aa\). She is known to be unaffected, so \(aa\) is excluded. Among the unaffected possibilities, the probability that she is a carrier is \(\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}\). Her partner is known to be \(Aa\). If she is also a carrier, an \(Aa\times Aa\) cross produces an affected \(aa\) child with probability \(\frac{1}{4}\). The combined probability is: \[ \frac{2}{3}\times\frac{1}{4}=\frac{1}{6} \] The calculation requires pedigree-based conditional genotype inference before the reproductive-risk calculation is performed. 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.
409. An affected individual with an autosomal recessive disorder has children with an unaffected person whose genotype is unknown. Their first child is affected. What can now be concluded about the unaffected parent under the simple model?
ⓐ. The unaffected parent must be a heterozygous carrier.
ⓑ. The unaffected parent must be homozygous normal.
ⓒ. The unaffected parent must also be affected genetically but not phenotypically.
ⓓ. The genotype remains completely undetermined.
Correct Answer: The unaffected parent must be a heterozygous carrier.
Explanation: The affected parent has the homozygous recessive genotype and contributes the recessive allele to every child. An affected child must receive a second recessive allele from the other parent. The other parent is stated to be phenotypically unaffected, so that parent cannot be homozygous recessive under the simple complete-recessive model. The only genotype that is both unaffected and capable of transmitting the recessive allele is the heterozygous carrier state. The affected child reveals the hidden allele in the unaffected parent. This inference is analogous to the logic of a test cross: a recessive offspring can expose an allele that is not visible in the parent's phenotype. The conclusion follows from the offspring genotype requirement rather than from the frequency of the disorder in the wider population.
410. Study the disorder information below.
| Case | Primary genetic or biochemical feature |
|---|
| P | Reduced synthesis of beta-globin |
| Q | Glutamic acid replaced by valine in beta-globin |
| R | Impaired conversion of phenylalanine to tyrosine |
| S | Defect in a protein of the blood-clotting cascade |
Which matching is correct?
ⓐ. P = sickle-cell anaemia; Q = beta-thalassemia; R = haemophilia; S = phenylketonuria
ⓑ. P = phenylketonuria; Q = sickle-cell anaemia; R = beta-thalassemia; S = haemophilia
ⓒ. P = beta-thalassemia; Q = sickle-cell anaemia; R = phenylketonuria; S = haemophilia
ⓓ. P = haemophilia; Q = phenylketonuria; R = sickle-cell anaemia; S = beta-thalassemia
Correct Answer: P = beta-thalassemia; Q = sickle-cell anaemia; R = phenylketonuria; S = haemophilia
Explanation: Beta-thalassemia is primarily a quantitative globin defect, so reduced synthesis of beta-globin identifies P. Sickle-cell anaemia is a qualitative beta-globin disorder in which the described mutation replaces glutamic acid with valine, identifying Q. Phenylketonuria involves impaired phenylalanine metabolism given that normal conversion of phenylalanine to tyrosine is disrupted, identifying R. Haemophilia affects the blood-clotting process through deficiency of a required clotting protein, identifying S. The four disorders cannot be separated reliably merely by calling them inherited conditions. Their inheritance patterns and biological mechanisms differ. Mechanism-level distinctions separate an autosomal quantitative globin disorder, an autosomal qualitative globin disorder, an autosomal metabolic disorder and an X-linked clotting disorder. The same conclusion is supported by the fact that thalassemia is a quantitative globin-synthesis defect: alpha-thalassemia involves alpha-globin genes on chromosome \(16\), whereas beta-thalassemia involves \(\mathrm{HBB}\) on chromosome \(11\).
411. A gamete contains two copies of the X chromosome and a normal haploid autosomal set. It fertilises a normal Y-bearing sperm. Which zygotic condition follows from these chromosome contributions?
ⓐ. \(45,XO\)
ⓑ. \(46,XY\)
ⓒ. \(47,XXY\)
ⓓ. \(47,+21\)
Correct Answer: \(47,XXY\)
Explanation: The abnormal gamete contributes two X chromosomes, while the normal sperm contributes Y. The resulting sex-chromosome complement is \(XXY\). Since the autosomal sets are otherwise described as normal, the zygote contains one chromosome more than the usual human diploid total, giving \(47\) chromosomes. The complement \(47,XXY\) is associated with Klinefelter's syndrome. The abnormality is a sex-chromosome aneuploidy rather than an autosomal trisomy. An \(XO\) zygote would require only one X and no second sex chromosome, while trisomy \(21\) specifically requires an extra copy of chromosome \(21\). The problem is solved by adding the sex chromosomes contributed by the two gametes and then identifying the resulting aneuploid complement.
412. Assertion: Down's syndrome and Klinefelter's syndrome can both involve \(47\) chromosomes but are not the same chromosomal disorder.
Reason: Aneuploidy must be interpreted according to which chromosome is gained or lost, not only by the total chromosome count.
ⓐ. 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: Down's syndrome usually involves trisomy \(21\), so the extra chromosome is an autosome. Klinefelter's syndrome has the complement \(47,XXY\), so the extra chromosome is a sex chromosome. Both individuals can have a total of \(47\) chromosomes while possessing different karyotypic abnormalities and different phenotypic consequences. The Reason identifies the critical principle: numerical chromosome analysis requires chromosome identity as well as total count. Merely observing \(47\) chromosomes is not sufficient to name the syndrome. The same logic also distinguishes these conditions from Turner's syndrome, where one X chromosome is absent and the total is \(45\). Aneuploidy is defined by gain or loss of particular chromosomes, so the specific chromosome involved is biologically decisive.
413. Match each observation in Column I with the most appropriate genetic relation in Column II. Each Column II entry is used once.
| Column I | Column II |
|---|
| P. One gene influences several phenotypic effects | 1. Pleiotropy |
| Q. Several genes contribute additively to one character | 2. Polygenic inheritance |
| R. Parental allele combinations greatly exceed recombinant combinations | 3. Linkage |
| S. One chromosome is present in three copies | 4. Trisomy |
ⓐ. P-2, Q-1, R-4, S-3
ⓑ. P-3, Q-2, R-1, S-4
ⓒ. P-1, Q-3, R-4, S-2
ⓓ. P-1, Q-2, R-3, S-4
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: P describes the one-gene-to-many-effects relationship that defines pleiotropy. Q reverses that direction: several genes contribute to one character, identifying polygenic inheritance. R describes a two-locus transmission pattern in which original parental combinations are preserved more often than recombinant combinations, which is characteristic of linkage. S describes the presence of three copies of one chromosome, defining trisomy, a form of aneuploidy. These four observations operate at different biological levels: gene-to-phenotype relationship, multilocus control of a trait, physical association of genes on chromosomes, and chromosome-number abnormality. The mapping requires identifying the precise genetic relation being tested rather than treating all non-simple inheritance patterns as equivalent departures from a monohybrid dominant-recessive model. This reasoning is consistent with the principle that recombination frequency is the recombinant fraction in an informative cross and serves as a relative measure of gene separation, not an exact physical base-pair distance.
414. An inheritance study produces four findings: a single locus shows ordinary segregation, two loci show a parental excess in a test cross, one descendant contains a newly altered allele, and a family member has a karyotype with an extra chromosome \(21\). Which interpretation best integrates the findings?
ⓐ. All four observations can be explained by failure of the Law of Segregation.
ⓑ. The findings represent segregation, linkage, mutation and trisomy \(21\), respectively.
ⓒ. The parental excess proves mutation, while the altered allele proves independent assortment.
ⓓ. The extra chromosome \(21\) converts the linked loci into polygenic inheritance.
Correct Answer: The findings represent segregation, linkage, mutation and trisomy \(21\), respectively.
Explanation: The first observation shows that alleles at one locus separate normally during gamete formation. The second concerns the relationship between two loci: a parental excess in an informative test cross supports linkage rather than failure of one-locus segregation. The third introduces a different source of variation, given that an altered allele requires mutation of genetic material rather than simple reshuffling of existing alleles. The fourth occurs at the chromosome-number level: an additional chromosome \(21\) is trisomy \(21\), the chromosomal basis of Down's syndrome. The observations should not be forced into one explanatory mechanism. Inheritance can simultaneously involve normal segregation, non-independent transmission of linked genes, origin of new genetic variation through mutation and chromosome-number abnormalities arising from segregation errors.
415. Review the statements below about genetic and phenotypic novelty.
I. An offspring phenotype absent in both parents does not by itself prove that a new mutation occurred.
II. Independent assortment can produce new combinations of parental alleles.
III. Recombination can create non-parental combinations without creating new alleles.
IV. Only mutation can generate any phenotype not visible in the parents.
ⓐ. I and IV only
ⓑ. II, III and IV only
ⓒ. I, II and III only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: Offspring can differ from both parents even when every allele they carry was already present in the parental genotypes. Segregation and independent assortment can produce multilocus combinations that neither parent happened to express, while recombination can rearrange existing linked alleles into non-parental associations. Such outcomes represent new combinations rather than new allelic states. Mutation is different since it alters genetic material itself and can create an allele that was not previously present. The fourth statement is consequently too strong: phenotypic novelty is not equivalent to mutational novelty. A recessive phenotype hidden in heterozygous parents, a recombinant combination of linked alleles, or a new assortment of independently segregating genes can all appear in offspring without requiring mutation of the underlying alleles.
416. Two plants with genotype \(AaBb\) show the same dominant phenotype for both characters. They produce an \(aabb\) offspring even though neither parent shows either recessive phenotype. No new allele is detected. Which explanation is sufficient if the loci assort independently?
ⓐ. Carrier parents can produce \(aabb\) through segregation plus independent assortment.
ⓑ. The dominant alleles were converted into recessive alleles during fertilisation.
ⓒ. A chromosome-number mutation produced both recessive phenotypes simultaneously.
ⓓ. The appearance of \(aabb\) proves that the parental plants were not diploid.
Correct Answer: Carrier parents can produce \(aabb\) through segregation plus independent assortment.
Explanation: Each \(AaBb\) parent contains both recessive alleles even though complete dominance prevents those alleles from appearing in the parental phenotype. Segregation allows \(a\) to separate from \(A\) and \(b\) from \(B\) during gamete formation. With independent assortment, a gamete carrying both \(a\) and \(b\) can be produced. If each parent contributes such a gamete, the offspring becomes \(aabb\) and expresses both recessive phenotypes. No mutation is needed since the alleles were already present in the parental genotypes. The result illustrates how sexual reproduction can uncover hidden recessive variation and create a phenotypic combination absent from both parents simply by redistributing existing hereditary material. 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.
417. Test-cross and sequence data from three offspring groups are shown below.
| Group | Genetic observation | Sequence observation |
|---|
| P | Four classes occur near equal frequency | No new alleles detected |
| Q | Two parental classes greatly exceed two recombinant classes | No new alleles detected |
| R | An allele differs in sequence from both parental alleles | New altered sequence detected |
Which interpretation best matches P, Q and R?
ⓐ. P = linkage; Q = mutation; R = independent assortment
ⓑ. P = mutation; Q = polyploidy; R = recombination
ⓒ. P = co-dominance; Q = incomplete dominance; R = linkage
ⓓ. P = independent assortment; Q = linkage; R = mutation
Correct Answer: P = independent assortment; Q = linkage; R = mutation
Explanation: Group P contains four test-cross classes in roughly equal proportions, which is consistent with a heterozygote producing four gametic combinations at similar frequencies under effective independent assortment. Group Q contains all four classes but shows a strong parental excess. That combination of parental predominance and smaller recombinant classes is characteristic of linked genes with some recombination. Group R is fundamentally different: the allele itself has an altered sequence that was absent from both parents. Rearranging existing alleles cannot explain that observation, so mutation is required. The table separates three levels of genetic evidence: unbiased combination of existing alleles, biased transmission of physically associated loci with occasional recombination, and genuine alteration of genetic material. A mutation changes genetic material, whereas recombination rearranges alleles that already exist; new heritable sequence states require alteration of genetic material rather than assortment alone.
418. In the cross \(AaBbCc\times AaBbCc\), assume complete dominance and independent assortment at all three loci. What is the probability that an offspring shows recessive phenotypes for exactly two of the three characters?
ⓐ. \(\frac{3}{64}\)
ⓑ. \(\frac{9}{64}\)
ⓒ. \(\frac{27}{64}\)
ⓓ. \(\frac{1}{8}\)
Correct Answer: \(\frac{9}{64}\)
Explanation: At each locus in a heterozygous self-cross, the probability of the recessive phenotype is \(\frac{1}{4}\), while the probability of the dominant phenotype is \(\frac{3}{4}\). Exactly two recessive phenotypes can occur in three distinct arrangements: recessive at \(A\) and \(B\) but dominant at \(C\); recessive at \(A\) and \(C\) but dominant at \(B\); or recessive at \(B\) and \(C\) but dominant at \(A\). Each specific arrangement has probability: \[ \frac{1}{4}\times\frac{1}{4}\times\frac{3}{4}=\frac{3}{64} \] There are three mutually exclusive arrangements, so the total probability is: \[ 3\times\frac{3}{64}=\frac{9}{64} \] The calculation depends on independent assortment among the three loci. 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.
419. Assertion: Linkage can alter the joint distribution of two genes without preventing segregation of the alleles at either gene.
Reason: Homologous chromosomes can separate normally while genes located on the same chromosome tend to remain associated unless recombination separates them.
ⓐ. 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: Segregation concerns separation of alternative alleles at a locus, while linkage concerns the relationship between different loci carried on the same chromosome. A linked \(AB/ab\) heterozygote can still segregate \(A\) from \(a\) and \(B\) from \(b\) through normal separation of homologous chromosomes. However, the alleles on the same physical chromosome tend to travel together, producing more parental than recombinant combinations. Recombination can break those original associations and generate non-parental types. The Reason directly explains why linkage changes two-locus assortment without requiring failure of one-locus segregation. This distinction is essential when a progeny ratio deviates from independent-assortment expectations but still shows ordinary transmission of each individual allele pair. Linkage modifies the joint frequencies of allele combinations without preventing segregation at the individual loci; deviation from the independent-assortment expectation provides the evidence for physical association.
420. A linked heterozygote has the arrangement \(AB/ab\) and shows a total recombination frequency of \(20\%\). Assume reciprocal parental classes are equal and reciprocal recombinant classes are equal. What is the expected probability of an \(AB\) gamete?
ⓐ. \(10\%\)
ⓑ. \(20\%\)
ⓒ. \(40\%\)
ⓓ. \(80\%\)
Correct Answer: \(40\%\)
Explanation: A recombination frequency of \(20\%\) means that the two recombinant gamete classes together account for \(20\%\) of gametes. The remaining \(80\%\) are parental. In the arrangement \(AB/ab\), the parental gametes are \(AB\) and \(ab\). Since the stated conditions specify that reciprocal parental classes occur equally, each parental class represents half of the \(80\%\) parental total: \[ \frac{80\%}{2}=40\% \] The recombinant classes \(Ab\) and \(aB\) would each represent \(10\%\). The important step is to distinguish total recombinant frequency from the frequency of a single recombinant or parental class. Recombination percentage is first used to partition all gametes into recombinant and parental groups, after which the stated symmetry is applied within each group. Linked genes tend to preserve parental allele combinations more often than recombinant combinations in an informative cross; reciprocal recombinant classes together make up the total recombination percentage.
