Heredity and Variation

159 soru

Soru 21Soru

A reciprocal translocation between non-homologous chromosomes alters the arrangement of genes on the affected chromosomes without changing the total count of chromosomes in a diploid cell, whereas meiotic non-disjunction causes aneuploidy by altering the total chromosome number in the resulting gametes.

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Cevap: True

Cevap

The statement is True.
Reciprocal translocation represents a structural chromosomal mutation where segments of non-homologous chromosomes are exchanged without altering the total chromosome count (2n2n). Non-disjunction, on the other hand, is a nuclear division error where chromosomes fail to segregate properly during Anaphase I or II of meiosis, producing gametes with missing or extra whole chromosomes, which leads to numerical aneuploidy (such as Down syndrome, 47,+2147, +21).

Adım Adım Çözüm

1
Analyze the nature of reciprocal translocation
Reciprocal translocation involves non-homologous chromosomes swapping segments. This rearranges gene loci (a structural alteration) but leaves the total count of chromosomes intact.
Structural chromosomal aberrations modify chromosome architecture rather than chromosome complement number.
2
Analyze the mechanism and outcome of meiotic non-disjunction
Non-disjunction occurs when homologous chromosomes fail to segregate in Anaphase I or sister chromatids fail to separate in Anaphase II, yielding abnormal gametes (n+1n+1 or n1n-1).
Failure of proper spindle separation directly alters gametic chromosome counts, causing numerical aberrations (aneuploidy) upon fertilization.
3
Evaluate the complete comparative statement
The distinction drawn between structural alteration (translocation preserving number) and numerical alteration (non-disjunction causing aneuploidy) is scientifically accurate.
Both biological concepts are correctly defined and contrasted.

Anahtar Kavram

Distinction between structural chromosomal aberrations (e.g., translocation) and numerical chromosomal aberrations (e.g., aneuploidy via non-disjunction)
Soru 22Soru

Match each type of gene mutation or chromosomal aberration on the left with its corresponding description on the right.

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Öğeler

Substitution mutation
Frameshift mutation
Aneuploidy
Polyploidy

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Cevap

Substitution mutation matches replacement of a single nucleotide base; Frameshift mutation matches addition or deletion altering the reading frame; Aneuploidy matches gain or loss of specific individual chromosomes (2n+12n+1 or 2n12n-1); Polyploidy matches possession of extra complete chromosome sets (3n3n or 4n4n).
Substitution mutation corresponds to swapping a single nucleotide base for another. Frameshift mutation occurs when additions or deletions of nucleotides shift the triplet codon reading sequence. Aneuploidy describes the loss or gain of individual chromosomes, while polyploidy describes organisms having additional full sets of chromosomes.

Adım Adım Çözüm

1
Distinguish between gene mutations and numerical chromosomal aberrations.
Substitution and frameshift are point gene mutations, while aneuploidy and polyploidy involve changes in chromosome number.
Gene mutations affect nucleotide sequences within a gene, whereas numerical aberrations alter chromosome counts.
2
Identify the mechanisms of point mutations.
Substitution replaces a single base, whereas frameshift adds or deletes bases shifting downstream codons.
Codons are read in triplets, so non-multiple-of-three insertions or deletions disrupt all subsequent amino acid translation.
3
Distinguish between aneuploidy and polyploidy.
Aneuploidy affects individual chromosome counts (2n±12n \pm 1), whereas polyploidy involves entire extra sets of chromosomes (3n,4n3n, 4n).
Non-disjunction of a single chromosome pair causes aneuploidy, whereas failure of spindle formation during total cell division leads to polyploidy.

Anahtar Kavram

Gene Mutations vs Chromosomal Numerical Aberrations
Soru 23Soru

In monohybrid inheritance experiments involving guinea pig coat color, where the allele for black fur (BB) is completely dominant over the allele for white fur (bb), match each specified parental cross or test cross scenario on the left with its corresponding phenotypic or genotypic outcome in the offspring on the right.

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Öğeler

Cross between a heterozygous black guinea pig (BbBb) and a white guinea pig (bbbb)
Cross between two heterozygous black guinea pigs (Bb×BbBb \times Bb) yielding a total of 160 offspring
A test cross of a dominant black guinea pig that produces 100% black offspring across multiple litters
Cross between pure-breeding black (BBBB) and pure-breeding white (bbbb) parents to produce the F1F_1 generation

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Cevap

The correct pairings match each monohybrid inheritance scenario to its precise Mendelian ratio or outcome: (1) Heterozygous black crossed with white matches the 1:1 genotypic and phenotypic ratio; (2) Two heterozygous black parents producing 160 offspring matches 120 black and 40 white offspring (3:1 ratio); (3) A test cross producing 100% dominant offspring confirms a homozygous dominant parent (BB); and (4) Crossing pure-breeding parents yields 100% uniform heterozygous F1 offspring.
Each cross directly demonstrates a fundamental aspect of Mendel's First Law. Heterozygote ×\times homozygous recessive gives a 1:11:1 ratio; two heterozygotes produce a 3:13:1 phenotypic ratio (120:40 out of 160); a test cross yielding zero recessive offspring confirms homozygous dominance; and contrasting pure lines produce uniform F1F_1 heterozygotes.

Adım Adım Çözüm

1
Analyze the cross between BbBb and bbbb
Gametes: Parent 1 produces BB and bb in equal proportion (1:11:1), Parent 2 produces only bb. Offspring genotypes: 12Bb\frac{1}{2} Bb (black fur) and 12bb\frac{1}{2} bb (white fur).
Mendel's Law of Segregation dictates that alleles segregate during gamete formation so that each gamete carries only one allele for each gene.
2
Calculate expected offspring numbers for Bb×BbBb \times Bb with total N=160N = 160
Monohybrid phenotypic ratio is 33 dominant : 11 recessive. Dominant count =34×160=120= \frac{3}{4} \times 160 = 120; Recessive count =14×160=40= \frac{1}{4} \times 160 = 40.
A monohybrid cross of two heterozygotes generates a genotypic ratio of 1BB:2Bb:1bb1 BB : 2 Bb : 1 bb, which simplifies to a 3:13:1 phenotypic ratio under complete dominance.
3
Evaluate the test cross of an unknown black parent with a recessive bbbb parent
If the parent were BbBb, white offspring would appear in approximately 50%50\% of cases. Since 100%100\% of offspring are black, the unknown parent must be homozygous dominant (BBBB).
A test cross uses a known homozygous recessive individual to reveal whether an organism expressing a dominant trait is homozygous or heterozygous.
4
Determine the outcome of crossing homozygous contrasting parents (BB×bbBB \times bb)
All F1F_1 offspring receive BB from the black parent and bb from the white parent, making 100%100\% of the F1F_1 generation heterozygous (BbBb).
This illustrates Mendel's principle of uniformity in the F1F_1 generation when crossing pure lines.

Anahtar Kavram

Mendel's Law of Segregation and Monohybrid Inheritance Ratios
Soru 24Soru

During gametogenesis, non-disjunction of chromosome 21 occurs specifically during Meiosis I. If all resulting gametes are subsequently fertilized by normal haploid gametes, what percentage of the produced zygotes will have trisomy 21?

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Cevap: 50%

Cevap

50% of the resulting zygotes will exhibit trisomy 21.
When non-disjunction of a chromosome pair takes place during Meiosis I, the pair fails to segregate. Consequently, one daughter cell receives both homologous chromosomes while the other receives none. Following normal chromatid separation in Meiosis II, half of the gametes carry an extra chromosome (n+1n+1) and half are deficient by one chromosome (n1n-1). Fertilization of the n+1n+1 gametes with normal haploid gametes (nn) produces trisomic (2n+12n+1) zygotes. Therefore, 50% of the zygotes will have trisomy 21.

Adım Adım Çözüm

1
Analyze the meiotic stage where non-disjunction occurs
In Meiosis I, non-disjunction means homologous chromosomes fail to separate into distinct daughter cells.
Understanding the exact stage of meiotic failure determines the chromosome distribution in the daughter cells after Meiosis I.
2
Determine the chromosomal content of the resulting four gametes after Meiosis II
Meiosis I yields one cell with an extra chromosome (n+1n+1) and one cell missing that chromosome (n1n-1). Normal chromatid separation in Meiosis II creates two (n+1)(n+1) gametes and two (n1)(n-1) gametes.
All four gametes produced are genetically abnormal after a Meiosis I non-disjunction event.
3
Calculate the proportion of trisomic zygotes post-fertilization
Fertilization of the two (n+1)(n+1) gametes by normal haploid (nn) gametes gives (2n+1)(2n+1), which is trisomy 21. Two out of four total zygotes (50%) will be trisomic (and 50% will be monosomic).
Trisomy requires the combination of an (n+1)(n+1) gamete with a normal nn gamete.

Anahtar Kavram

Chromosomal Non-disjunction and Aneuploidy
Soru 25Soru

In fruit flies (*Drosophila melanogaster*), grey body color (BB) is dominant over black body color (bb), and normal wing length (VV) is dominant over vestigial wings (vv). If two heterozygous grey-bodied, normal-winged flies (BbVvBbVv) are crossed and produce a total of 320 offspring, how many of these offspring are expected to exhibit a black body with normal wings?

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Cevap: 60

Cevap

60 offspring are expected to have a black body with normal wings.
In a dihybrid cross of two heterozygous individuals (BbVv×BbVvBbVv \times BbVv), the independent assortment of alleles yields an F2 phenotypic ratio of 9:3:3:19 : 3 : 3 : 1. The proportion of offspring displaying the recessive trait for the first gene (black body, bbbb) and the dominant trait for the second gene (normal wings, V_V\_) is 316\frac{3}{16}. Multiplying 316\frac{3}{16} by the total offspring count of 320320 gives exactly 6060.

Adım Adım Çözüm

1
Determine parental genotypes and set up the dihybrid cross.
Parental cross is BbVv×BbVvBbVv \times BbVv, where both parents are heterozygous for both traits.
Mendel's Second Law states that alleles of different genes assort independently during gamete formation.
2
Identify the expected phenotypic ratio for the offspring.
The phenotypic ratio for a dihybrid cross between two double heterozygotes (BbVv×BbVvBbVv \times BbVv) is 9:3:3:19 : 3 : 3 : 1.
- Grey body, normal wings (B_V_B\_V\_) = 916\frac{9}{16}
- Grey body, vestigial wings (B_vvB\_vv) = 316\frac{3}{16}
- Black body, normal wings (bbV_bbV\_) = 316\frac{3}{16}
- Black body, vestigial wings (bbvvbbvv) = 116\frac{1}{16}
The probability of recessive body color (bb=14bb = \frac{1}{4}) combined with dominant wing length (V_=34V\_ = \frac{3}{4}) is calculated as 14×34=316\frac{1}{4} \times \frac{3}{4} = \frac{3}{16}.
3
Calculate the expected number of offspring with black body and normal wings.
316×320=60\frac{3}{16} \times 320 = 60.
Multiplying the expected phenotypic fraction by the total offspring count yields the specific expected count.

Anahtar Kavram

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Ratios
Tahmini Süre:1m 30s
Soru 26Soru

Match each human phenotypic feature listed on the left with its characteristic variation pattern and underlying genetic mechanism on the right.

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Öğeler

Human skin pigmentation spectrum
ABO blood group classification
Presence or absence of a widow's peak hairline
Total dermal ridge count in fingerprints

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Cevap

Human skin pigmentation matches continuous variation governed by polygenic inheritance with environmental modification. ABO blood group classification matches discontinuous variation controlled by multiple alleles at a single locus. Widow's peak hairline matches discontinuous variation governed by simple monogenic Mendelian inheritance. Total dermal ridge count matches continuous variation determined predominantly by additive polygenes with minimal environmental influence.
Continuous traits (such as skin pigmentation and fingerprint ridge counts) form a continuous spectrum of phenotypic values controlled by polygenes, differing in their environmental sensitivity. Discontinuous traits (such as ABO blood groups and widow's peak) form distinct qualitative classes without intermediate phenotypes, controlled by single loci with either multiple alleles or simple Mendelian dominance.

Adım Adım Çözüm

1
Analyze the phenotypic distribution pattern for each feature to classify it as either continuous (quantitative spectrum) or discontinuous (discrete qualitative categories).
Skin pigmentation and dermal ridge count are continuous traits; ABO blood groups and widow's peak are discontinuous traits.
Continuous traits show intermediate gradation across a population, whereas discontinuous traits exhibit distinct non-overlapping phenotypes.
2
Examine the underlying genetic architecture (monogenic vs. polygenic vs. multiple alleles) for each trait.
Skin pigmentation and ridge count involve polygenes; ABO blood group involves multiple alleles at one locus (IA,IB,iI^A, I^B, i); widow's peak involves a single monogenic locus.
Polygenic inheritance produces continuous variation through additive effects of multiple genes, whereas single loci yield distinct phenotypic classes.
3
Evaluate the impact of environmental factors on the expression of each continuous trait.
Skin tone varies significantly with solar radiation exposure (high environmental effect), while finger ridge counts are genetically fixed early in development with minimal environmental susceptibility.
Environmental modification alters phenotypic expression of certain polygenic traits without altering underlying genotype.

Anahtar Kavram

Polygenic vs. Monogenic Inheritance and Environmental Influence on Continuous and Discontinuous Variation
Soru 27Soru

While the ABO blood group system in human populations is governed by three multiple alleles (IAI^A, IBI^B, and ii), any given diploid individual can inherit a maximum of two alleles, with IAI^A and IBI^B exhibiting codominance with each other while both display complete dominance over the allele ii.

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Cevap: True

Cevap

True
The statement accurately distinguishes between population-level multiple alleles and individual diploid genetic constitution, while correctly identifying the codominant relationship between IAI^A and IBI^B alongside their complete dominance over the recessive ii allele.

Adım Adım Çözüm

1
Analyze allele distribution in populations versus diploid individuals.
Multiple alleles refer to the presence of more than two allele options in a population gene pool (IAI^A, IBI^B, ii), but diploid organisms possess only two locus copies (one maternal, one paternal).
Meiotic segregation and diploid chromosome organization restrict an individual to possessing a maximum of two alleles for any autosomal gene.
2
Examine the phenotypic expression pattern of heterozygous genotype IAIBI^A I^B.
Both antigen A and antigen B are independently and fully expressed on red blood cell surfaces, defining codominance.
Codominance occurs when neither allele masks the other, leading to simultaneous, distinct expression of both allele products.
3
Evaluate the interaction between alleles IAI^A / IBI^B and allele ii.
Genotypes IAiI^A i and IBiI^B i result in blood group A and blood group B phenotypes respectively, as allele ii produces a non-functional enzyme.
Allele ii is recessive, so IAI^A and IBI^B exhibit complete dominance over ii.

Anahtar Kavram

Multiple Alleles and Codominance in ABO Blood Groups
Tahmini Süre:1m 30s
Soru 28Soru

In maize plants, the allele for smooth kernels (SS) is completely dominant over the allele for wrinkled kernels (ss). According to Mendel's Law of Segregation, a monohybrid cross between two heterozygous smooth-kernel maize plants (Ss×SsSs \times Ss) produces offspring with a genotypic ratio of 3:13:1.

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Cevap: False

Cevap

The statement is False. A monohybrid cross between two heterozygous parents yields a genotypic ratio of 1:2:11:2:1 (1 SS:2 Ss:1 ss1\ SS : 2\ Ss : 1\ ss) and a phenotypic ratio of 3:13:1 (3 smooth : 1 wrinkled).
Evaluating the statement as False is correct because Mendel's Law of Segregation dictates that crossing two monohybrids (Ss×SsSs \times Ss) yields three distinct genotypes (SSSS, SsSs, and ssss) in a 1:2:11:2:1 ratio, while the 3:13:1 ratio applies only to the physical phenotypes.

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1
Determine the parental genotypes and gametes produced during meiosis according to Mendel's First Law.
Both parents are heterozygous (SsSs). Each parent produces gametes with either the SS allele (50%50\%) or the ss allele (50%50\%).
Mendel's Law of Segregation states that allele pairs separate during gamete formation so that each gamete carries only one allele for each gene.
2
Construct a Punnett square for the monohybrid cross (Ss×SsSs \times Ss).
Offspring genotypes are 1 SS1\ SS, 2 Ss2\ Ss, and 1 ss1\ ss.
Combining the gametes gives 25% SS25\%\ SS, 50% Ss50\%\ Ss, and 25% ss25\%\ ss.
3
Compare the resulting genotypic ratio with the ratio stated in the prompt.
The genotypic ratio is 1:2:11:2:1, whereas the phenotypic ratio (smooth to wrinkled) is 3:13:1.
The prompt incorrectly asserts that 3:13:1 is the genotypic ratio, confusing genotypic outcome with phenotypic outcome.

Anahtar Kavram

Mendel's Law of Segregation and Monohybrid Cross Ratios
Tahmini Süre:1m 0s
Soru 29Soru

In genetic studies of pea plants, the gene governing flower placement exists as two structural variants: axial (AA) and terminal (aa). Which of the following terms specifically describes these alternative forms of a single gene that occupy identical locations on homologous chromosomes?

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Cevap: Alleles

Cevap

Alleles are the alternative forms of a gene occupying identical positions on homologous chromosomes.
The term alleles specifically denotes alternative molecular sequence forms of the same gene that occupy corresponding loci on homologous chromosomes and determine contrasting traits.

Adım Adım Çözüm

1
Identify the biological entity described in the stem.
The stem describes alternative molecular forms of a gene (axial AA vs terminal aa) controlling the same trait.
Genes often exist in multiple structural variations that lead to contrasting characteristics.
2
Differentiate between gene location and gene variant.
The physical position on a chromosome is a locus, whereas the contrasting gene versions located at that locus are alleles.
Distinguishing between structural variants (alleles) and positional coordinates (loci) is essential in basic genetics.

Anahtar Kavram

Alleles and Gene Loci
Tahmini Süre:1m 0s
Soru 30Soru

Match each fundamental genetics term on the left with its precise biological definition on the right.

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Öğeler

Locus
Allele
Genotype
Phenotype

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Cevap

Locus matches the specific physical position of a gene on a chromosome; Allele matches an alternative functional form of a gene; Genotype matches the specific allelic composition of an organism; Phenotype matches the observable physical or physiological characteristics of an organism.
Each genetic term matches its corresponding definition: Locus is the physical gene location on a chromosome, Allele is a gene variant, Genotype is the allelic composition, and Phenotype is the observable structural or functional trait.

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1
Identify the chromosomal position term.
Locus corresponds to the fixed physical site of a gene on a chromosome.
Chromosomal positioning terminology distinguishes the locus from the gene itself.
2
Distinguish gene variants.
Allele corresponds to alternative sequence versions of the same gene.
Genes exist in variant molecular forms called alleles.
3
Separate internal genetic makeup from outward appearance.
Genotype is the inherited genetic constitution, while Phenotype is the observable physical feature.
Genotype provides the instructions; phenotype is the expressed physical manifestation.

Anahtar Kavram

Basic Genetics Terminology
Tahmini Süre:1m 30s
Soru 31Soru

In pea plants, the allele for axial flower position (AA) is completely dominant over the allele for terminal flower position (aa). If two heterozygous axial-flowered plants (AaAa) are crossed, what is the expected genotypic ratio (AA:Aa:aaAA : Aa : aa) among their offspring according to Mendel's First Law?

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Cevap: 1:2:11 : 2 : 1

Cevap

The expected genotypic ratio among the offspring is 1:2:11 : 2 : 1 (1AA:2Aa:1aa1\,AA : 2\,Aa : 1\,aa).
In a monohybrid cross between two heterozygous individuals (Aa×AaAa \times Aa), Mendel's Law of Segregation dictates that alleles segregate into gametes independently. Punnett square analysis gives 1/4AA1/4\,AA, 2/4Aa2/4\,Aa, and 1/4aa1/4\,aa, producing an exact genotypic ratio of 1:2:11 : 2 : 1.

Adım Adım Çözüm

1
Identify parental genotypes and alleles produced
Both parents have the genotype AaAa. Gametes produced by each parent are AA and aa in equal proportions (1/2A1/2\,A and 1/2a1/2\,a).
Mendel's Law of Segregation states that allele pairs separate during gamete formation so that each gamete carries only one allele for each gene.
2
Construct a Punnett square for the cross Aa×AaAa \times Aa
Offspring combinations are: AAAA (1/41/4), AaAa (2/42/4), and aaaa (1/41/4).
Random fertilization combines gametes from both parents to form offspring genotypes.
3
Determine the genotypic ratio
The ratio of genotypes is 1AA:2Aa:1aa1\,AA : 2\,Aa : 1\,aa, which simplifies to 1:2:11 : 2 : 1.
The question specifically asks for the ratio of genotypes (AA:Aa:aaAA : Aa : aa), not physical appearances (phenotypes).

Anahtar Kavram

Mendel's Law of Segregation and Monohybrid Cross Genotypic Ratio
Tahmini Süre:1m 0s
Soru 32Soru

In tomato plants (*Solanum lycopersicum*), purple stem (PP) is dominant over green stem (pp), and cut leaf (CC) is dominant over potato leaf (cc). A plant heterozygous for both traits (PpCcPpCc) is crossed with a homozygous recessive plant (ppccppcc). What proportion of the offspring is expected to have purple stems and potato leaves?

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Cevap: 1/4 (25%)

Cevap

1/4 (25%)
In a dihybrid test cross between a double heterozygote (PpCcPpCc) and a double homozygous recessive individual (ppccppcc), the heterozygous parent produces four gamete types (PCPC, PcPc, pCpC, pcpc) in equal proportions (1/41/4 each). When combined with the recessive parent's pcpc gamete, four phenotypic classes result with an expected frequency of 1/41/4 (25%25\%) each. Therefore, the proportion of offspring showing purple stems and potato leaves (PpccPpcc) is 1/41/4.

Adım Adım Çözüm

1
Identify parental genotypes and determine gametes produced
The heterozygous parent (PpCcPpCc) produces four types of gametes in equal proportions: PCPC, PcPc, pCpC, and pcpc (1/41/4 each). The homozygous recessive parent (ppccppcc) produces only one type of gamete: pcpc (100%100\%).
Mendel's Law of Independent Assortment states that alleles for different traits segregate independently during gamete formation.
2
Determine offspring genotypes and phenotypes
Combining gametes yields four offspring genotypes: PpCcPpCc (purple stem, cut leaf), PpccPpcc (purple stem, potato leaf), ppCcppCc (green stem, cut leaf), and ppccppcc (green stem, potato leaf), each occurring at a frequency of 1/4.
A test cross matches each distinct gamete from the heterozygous parent with a recessive allele pair from the tester parent.
3
Calculate the expected proportion for the targeted phenotype
The proportion of offspring displaying purple stems (P_) and potato leaves (cccc), represented by genotype PpccPpcc, is 1/4 or 25%.
Exactly one of the four equally probable phenotypic outcomes exhibits the dominant stem color and recessive leaf form.

Anahtar Kavram

Dihybrid Test Cross Ratio
Tahmini Süre:1m 30s
Soru 33Soru

In tomato plants, the allele for red fruit (RR) is completely dominant over the allele for yellow fruit (rr). If a heterozygous red-fruited plant is self-pollinated and produces 400400 offspring in the F2F_2 generation, how many of these plants are expected to be heterozygous red-fruited?

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Cevap: 200200

Cevap

The expected number of heterozygous red-fruited plants is 200.
Crossing two heterozygous individuals (Rr×RrRr \times Rr) produces genotypes in a 1 RR:2 Rr:1 rr1\ RR : 2\ Rr : 1\ rr ratio according to Mendel's First Law. Therefore, half (50%50\%) of the 400400 total offspring are expected to have the heterozygous genotype (RrRr), which equals 200200 plants.

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1
Determine the parental genotypes and set up the monohybrid cross.
The self-pollinated plant is heterozygous (RrRr), so the cross is Rr×RrRr \times Rr.
Mendel's First Law (Law of Segregation) states that paired alleles segregate during gamete formation so each gamete carries only one allele.
2
Calculate the genotypic ratio of the F2 generation.
The offspring genotypic ratio is 1 RR:2 Rr:1 rr1\ RR : 2\ Rr : 1\ rr, meaning 1/41/4 homozygous dominant, 2/42/4 (or 1/21/2) heterozygous, and 1/41/4 homozygous recessive.
Random combination of parental gametes (RR and rr) yields probabilities of 25% RR25\%\ RR, 50% Rr50\%\ Rr, and 25% rr25\%\ rr.
3
Compute the expected count of heterozygous offspring out of 400 total plants.
Heterozygous count = 24×400=200\frac{2}{4} \times 400 = 200 plants.
Multiplying the probability of the heterozygous genotype (1/21/2) by the total number of offspring gives the expected number.

Anahtar Kavram

Mendel's Law of Segregation and Monohybrid Genotypic Ratios
Soru 34Soru

In guinea pigs (*Cavia porcellus*), black coat color (BB) is dominant over white coat color (bb), and short hair (SS) is dominant over long hair (ss). If two heterozygous guinea pigs (BbSsBbSs) are mated and produce a total of 400400 offspring, how many offspring are expected to display the black coat and long hair phenotype?

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Cevap: 75

Cevap

75 offspring are expected to display the black coat and long hair phenotype.
In a dihybrid cross between two individuals heterozygous for both traits (BbSs×BbSsBbSs \times BbSs), independent assortment produces an F2F_2 phenotypic ratio of 9:3:3:19:3:3:1. The phenotype with one dominant and one recessive trait (black coat and long hair, B_ssB\_ss) represents 316\frac{3}{16} of the total offspring population. Out of 400400 offspring, the expected count is calculated as 316×400=75\frac{3}{16} \times 400 = 75.

Adım Adım Çözüm

1
Determine the phenotypic ratio from the dihybrid cross BbSs×BbSsBbSs \times BbSs.
The expected ratio is 9:3:3:19 : 3 : 3 : 1 for (black coat, short hair) : (black coat, long hair) : (white coat, short hair) : (white coat, long hair).
Mendel's Law of Independent Assortment dictates that the two gene pairs segregate independently during gamete formation.
2
Determine the proportion of offspring showing the black coat and long hair phenotype (B_ssB\_ss).
The proportion is 316\frac{3}{16}.
The probability of inheriting the dominant coat trait (B_B\_) is 34\frac{3}{4}, and the probability of inheriting the recessive hair length trait (ssss) is 14\frac{1}{4}. Combined probability = 34×14=316\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}.
3
Multiply the proportion by the total population size to find the expected number of offspring.
316×400=75\frac{3}{16} \times 400 = 75.
Applying the theoretical phenotypic probability to the sample size of 400 yields the expected count.

Anahtar Kavram

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Ratios
Tahmini Süre:1m 30s
Soru 35Soru

Match each dihybrid parental cross genotype of independently assorting genes with its expected offspring phenotypic ratio.

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Öğeler

AaBb×AaBbAaBb \times AaBb
AaBb×aabbAaBb \times aabb
AaBb×AabbAaBb \times Aabb
AABb×aaBbAABb \times aaBb

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Cevap

The correct matches pair AaBb×AaBbAaBb \times AaBb with 9:3:3:19:3:3:1, AaBb×aabbAaBb \times aabb with 1:1:1:11:1:1:1, AaBb×AabbAaBb \times Aabb with 3:3:1:13:3:1:1, and AABb×aaBbAABb \times aaBb with 3:13:1.
Each parent cross is accurately matched to its offspring phenotypic distribution by applying the product rule of probability to independently assorting gene pairs: AaBb×AaBbAaBb \times AaBb produces 9:3:3:19:3:3:1, AaBb×aabbAaBb \times aabb produces 1:1:1:11:1:1:1, AaBb×AabbAaBb \times Aabb produces 3:3:1:13:3:1:1, and AABb×aaBbAABb \times aaBb produces 3:13:1.

Adım Adım Çözüm

1
Analyze the dihybrid self-cross AaBb×AaBbAaBb \times AaBb
Combining independent monohybrid crosses (3 dominant:1 recessive)×(3 dominant:1 recessive)(3 \text{ dominant} : 1 \text{ recessive}) \times (3 \text{ dominant} : 1 \text{ recessive}) produces the phenotypic ratio 9:3:3:19 : 3 : 3 : 1.
According to Mendel's Law of Independent Assortment, the segregation of alleles for one gene occurs independently of the segregation of alleles for another gene.
2
Analyze the dihybrid test cross AaBb×aabbAaBb \times aabb
The heterozygous parent produces four distinct gamete types (AB,Ab,aB,abAB, Ab, aB, ab) in equal frequencies of 25%25\% each, while the homozygous recessive parent produces only abab gametes, yielding a 1:1:1:11:1:1:1 phenotypic ratio.
Test cross phenotypic ratios directly mirror the gametic frequencies produced by the heterozygous parent.
3
Analyze the cross AaBb×AabbAaBb \times Aabb
For gene A (Aa×AaAa \times Aa), expected probabilities are 34\frac{3}{4} dominant (A_A\_) and 14\frac{1}{4} recessive (aaaa). For gene B (Bb×bbBb \times bb), expected probabilities are 12\frac{1}{2} dominant (B_B\_) and 12\frac{1}{2} recessive (bbbb). Multiplying probabilities gives 38A_B_:38A_bb:18aaB_:18aabb\frac{3}{8} A\_B\_ : \frac{3}{8} A\_bb : \frac{1}{8} aaB\_ : \frac{1}{8} aabb, simplified as 3:3:1:13:3:1:1.
Applying the product rule of probability for independent genetic events.
4
Analyze the cross AABb×aaBbAABb \times aaBb
For gene A (AA×aaAA \times aa), 100%100\% of offspring express the dominant phenotype (AaAa). For gene B (Bb×BbBb \times Bb), 34\frac{3}{4} express the dominant phenotype (B_B\_) and 14\frac{1}{4} express the recessive phenotype (bbbb). The total offspring ratio simplifies to 3:13:1.
Since trait A displays zero phenotypic variation in the progeny, the overall phenotypic ratio is determined solely by the segregation of trait B.

Anahtar Kavram

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Ratios
Soru 36Soru

A man with blood group A, whose mother had blood group O, marries a woman with blood group B, whose father had blood group O. What is the probability that their first child will have blood group O?

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Cevap: 25%

Cevap

The probability that their first child will have blood group O is 25%.
Both parents inherited a recessive allele ii from their respective blood group O parents, giving them heterozygous genotypes (IAiI^A i and IBiI^B i). Crossing these genotypes produces four offspring genotypes (IAIBI^A I^B, IAiI^A i, IBiI^B i, iiii) in equal proportions, resulting in a 25% (1 in 4) chance for a child with blood group O (iiii).

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1
Determine parental genotypes from their phenotypes and family histories.
The man has blood group A (IAIAI^A I^A or IAiI^A i). Because his mother was blood group O (iiii), he inherited an ii allele, giving him genotype IAiI^A i. The woman has blood group B (IBIBI^B I^B or IBiI^B i) and her father was blood group O (iiii), giving her genotype IBiI^B i.
An individual with blood group O has genotype iiii and must pass a recessive ii allele to all offspring.
2
Perform a genetic cross between the two parents (IAi×IBiI^A i \times I^B i).
The offspring genotypes formed are IAIBI^A I^B (blood group AB), IAiI^A i (blood group A), IBiI^B i (blood group B), and iiii (blood group O) in a 1:1:1:1 ratio.
Punnett square analysis of two heterozygous individuals yields four equally likely genotype combinations.
3
Calculate the probability of obtaining genotype iiii (blood group O).
1 out of 4 possible outcomes is genotype iiii, which equals 25% or 14\frac{1}{4}.
Blood group O is expressed only when the homozygous recessive genotype iiii is present.

Anahtar Kavram

Multiple Alleles and Codominance in ABO Blood Group Inheritance
Tahmini Süre:1m 30s
Soru 37Soru

Match each non-Mendelian genetic concept listed on the left with its corresponding biological example or defining feature on the right.

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Öğeler

Incomplete Dominance
Codominance
Multiple Alleles

Eşleşmeler

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Cevap

Incomplete Dominance matches with pink flower color in snapdragons; Codominance matches with simultaneous expression of IAI^A and IBI^B antigens in AB blood group; Multiple Alleles matches with the existence of IAI^A, IBI^B, and ii gene variants in a population.
Matching Incomplete Dominance to pink snapdragons correctly identifies intermediate phenotypic blending. Matching Codominance to blood type AB correctly identifies joint expression of two functional alleles. Matching Multiple Alleles to the ABO gene locus (IAI^A, IBI^B, ii) correctly identifies gene loci that have more than two alternative allele variants within a population.

Adım Adım Çözüm

1
Analyze Incomplete Dominance
Incomplete dominance involves a blending of phenotypes in heterozygous offspring.
Crossing red and white homozygous snapdragons produces pink heterozygous offspring because the single functional dominant allele produces insufficient pigment for full red expression.
2
Analyze Codominance
Codominance involves full, distinct, non-blended phenotypic expression of both inherited alleles.
Heterozygotes carrying both IAI^A and IBI^B alleles produce both A and B glycoprotein antigens on red blood cell surfaces equally.
3
Analyze Multiple Alleles
Multiple alleles describes a population-level genetic phenomenon where a single locus has more than two allele options.
While any diploid individual inherits only two alleles, the human population possesses three distinct alleles (IAI^A, IBI^B, and ii) governing the ABO blood group locus.

Anahtar Kavram

Patterns of Non-Mendelian Inheritance (Incomplete Dominance, Codominance, Multiple Alleles)
Soru 38Soru

A woman with normal vision whose father was color-blind marries a man with normal vision. What is the probability that any male child born to this couple will be color-blind?

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Cevap: 50%50\%

Cevap

The probability that any male child born to this couple will be color-blind is 50%50\%.
The mother is a carrier (XCXcX^C X^c) because she inherited the color-blindness allele (XcX^c) from her affected father. When crossed with a normal male (XCYX^C Y), sons inherit their Y chromosome from the father and one of the mother's two X chromosomes. Therefore, there is a 50%50\% chance that a son receives the XcX^c allele and manifests color blindness.

Adım Adım Çözüm

1
Determine the parental genotypes.
The woman's father was color-blind (XcYX^c Y), so she inherited the XcX^c allele. Being phenotypically normal, her genotype is XCXcX^C X^c (heterozygous carrier). The man has normal vision, so his genotype is XCYX^C Y.
Red-green color blindness is an X-linked recessive trait.
2
Construct a Punnett square for the cross XCXc×XCYX^C X^c \times X^C Y.
The possible offspring genotypes are XCXCX^C X^C (normal female), XCXcX^C X^c (carrier female), XCYX^C Y (normal male), and XcYX^c Y (color-blind male).
Each parent contributes one sex chromosome to offspring.
3
Calculate the probability specific to male offspring.
The male offspring genotypes are XCYX^C Y and XcYX^c Y. Out of these 22 possible male outcomes, 11 is color-blind (XcYX^c Y). The probability among male children is 12=50%\frac{1}{2} = 50\%.
The question asks specifically for the probability among sons, requiring evaluation within the male subset only.

Anahtar Kavram

Sex-Linked Inheritance and X-Linked Recessive Traits
Tahmini Süre:1m 30s
Soru 39Soru

Match each sex determination system or sex-linked inheritance pattern on the left with its correct biological characteristic or organism example on the right.

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Öğeler

XX-XO System
ZZ-ZW System
X-linked Recessive Inheritance
Y-linked (Holandric) Inheritance

Eşleşmeler

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Cevap

The correct pairings are: XX-XO System matches grasshoppers with XOXO males; ZZ-ZW System matches birds with heterogametic ZWZW females; X-linked Recessive Inheritance matches traits passed from carrier mothers to sons; Y-linked Inheritance matches traits passed exclusively from an affected father to all sons.
Each mechanism accurately corresponds to its characteristic biological model: XX-XO occurs in grasshoppers (XOXO males); ZZ-ZW occurs in birds where females are heterogametic (ZWZW); X-linked recessive traits pass from carrier mothers to sons; and Y-linked traits pass exclusively from fathers to sons.

Adım Adım Çözüm

1
Identify the chromosomal structure of the XX-XO system.
Determine that males lack a Y chromosome (XOXO) while females have two X chromosomes (XXXX), which is characteristic of grasshoppers.
Distinguishing insect sex determination from mammalian systems.
2
Identify the heterogametic sex in the ZZ-ZW system.
Recognize that in birds, females are heterogametic (ZWZW) and males are homogametic (ZZZZ).
Reversing the standard male-heterogametic concept seen in humans.
3
Compare transmission pathways of X-linked versus Y-linked traits.
X-linked recessive conditions are passed from carrier females to sons, whereas Y-linked traits are passed strictly along the male line (father to son).
Applying rules of sex-linked pedigree analysis.

Anahtar Kavram

Sex Determination Systems and Sex-Linked Inheritance Patterns
Soru 40Soru

During a crop breeding experiment, two individual maize plants both exhibit resistance to a fungal pathogen. Genetic sequencing reveals that Plant 1 has the genetic constitution RRRR, whereas Plant 2 has the genetic constitution RrRr. Which of the following statements correctly compares the genetic characteristics of these two plants?

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Cevap: They express the same phenotype but possess different genotypes.

Cevap

They express the same phenotype but possess different genotypes.
Phenotype refers to the observable physical characteristics of an organism, while genotype refers to its specific genetic composition. Because allele RR is dominant over allele rr, both RRRR and RrRr produce the disease-resistant phenotype. However, their genotypes differ because one plant is homozygous dominant (RRRR) while the other is heterozygous (RrRr).

Adım Adım Çözüm

1
Identify the phenotypic expression of both plants.
Both Plant 1 (RRRR) and Plant 2 (RrRr) exhibit resistance to the fungal pathogen, meaning they have identical phenotypes.
Phenotype is the observable physical or physiological trait of an organism produced by the interaction of its genotype with the environment.
2
Analyze the genotypic constitution of each plant.
Plant 1 is homozygous dominant (RRRR) and Plant 2 is heterozygous (RrRr).
Genotype refers to the actual allele combination present at a specific gene locus.
3
Compare the phenotype and genotype between the two plants.
The two plants share the same observable phenotype (disease resistance) despite having different genotypes (RRRR versus RrRr).
The dominant allele (RR) completely masks the presence of the recessive allele (rr) in the heterozygous state.

Anahtar Kavram

Phenotype vs. Genotype and Allelic Dominance
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