Heredity and Variation

159 questions

Question 81Question

Match each specified plant genotype with the number of genetically distinct gamete types it can produce during meiosis according to Mendel's Law of Independent Assortment.

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Items

YyRrYyRr (Heterozygous at two gene loci)
YYRrYYRr (Homozygous dominant at one locus and heterozygous at another)
yyrryyrr (Homozygous recessive at both gene loci)
YyRrSsYyRrSs (Heterozygous at three independently assorting gene loci)

Matches

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Answer

The correct matches are: YyRrYyRr matches with 4 distinct gamete types; YYRrYYRr matches with 2 distinct gamete types; yyrryyrr matches with 1 distinct gamete type; and YyRrSsYyRrSs matches with 8 distinct gamete types.
According to Mendel's Second Law, alleles of unlinked genes assort independently during meiosis. The number of genetically distinct gametes produced by an organism is given by 2n2^n, where nn is the number of heterozygous gene pairs. Thus, YyRrYyRr has 2 heterozygous pairs giving 4 gametes (222^2), YYRrYYRr has 1 heterozygous pair giving 2 gametes (212^1), yyrryyrr has 0 heterozygous pairs giving 1 gamete (202^0), and YyRrSsYyRrSs has 3 heterozygous pairs giving 8 gametes (232^3).

Step-by-Step Solution

1
Identify the formula for determining the number of distinct gametes.
The formula is 2n2^n, where nn represents the number of heterozygous gene pairs.
Mendel's Law of Independent Assortment states that alleles for different traits segregate independently during gamete formation.
2
Calculate gamete numbers for each given genotype.
For YyRrYyRr, n=2    22=4n=2 \implies 2^2 = 4; for YYRrYYRr, n=1    21=2n=1 \implies 2^1 = 2; for yyrryyrr, n=0    20=1n=0 \implies 2^0 = 1; for YyRrSsYyRrSs, n=3    23=8n=3 \implies 2^3 = 8.
Counting the number of heterozygous pairs (nn) directly determines the variety of gametes produced.

Key Concept

Gamete Genotype Determination and Mendel's Law of Independent Assortment
Estimated Time:45s
Question 82Question

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 a heterozygous black, short-haired guinea pig (BbSsBbSs) is mated with a white, long-haired guinea pig (bbssbbss) and they produce a total of 640 offspring, how many of the offspring are expected to display a black coat and long hair?

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Answer: 160

Answer

160 offspring are expected to have a black coat and long hair.
In a dihybrid testcross involving a double heterozygote (BbSsBbSs) and a homozygous recessive individual (bbssbbss), the offspring phenotypes appear in equal ratios of 1:1:1:1 (25% for each phenotypic class). The black coat, long hair phenotype (BbssBbss) corresponds to 1/4 of the total offspring. Multiplying 1/4 by 640 yields exactly 160 expected offspring.

Step-by-Step Solution

1
Determine the type of genetic cross and parental genotypes.
The cross is a dihybrid testcross between BbSsBbSs and bbssbbss.
One parent is heterozygous for both independently assorting traits (BbSsBbSs), and the other parent is homozygous recessive (bbssbbss).
2
Determine the proportion of offspring expected to have the phenotype black coat and long hair (BbssBbss).
The proportion of BbssBbss offspring is 14\frac{1}{4} (or 25%25\%).
The BbSsBbSs parent produces four types of gametes (BSBS, BsBs, bSbS, bsbs) in equal proportions (14\frac{1}{4} each). Combining BsBs with bsbs yields BbssBbss.
3
Calculate the expected count out of 640 total offspring.
14×640=160\frac{1}{4} \times 640 = 160.
Multiplying the expected phenotypic fraction by the total offspring count yields the absolute expected count.

Key Concept

Dihybrid testcross ratio and probability calculation
Question 83Question

A botanical study recorded two phenotypic traits in a population of *Hibiscus sabdariffa*: total leaf surface area (measured in cm2\text{cm}^2) and petal pigmentation pattern (either solid crimson or white-striped). Leaf surface area exhibited a smooth, bell-shaped frequency distribution curve across the population, with mean values altering significantly when cloned specimens were grown in nutrient-deficient versus nutrient-rich soils. In contrast, petal pigmentation pattern showed strict discrete categories with no intermediate forms, remaining completely unchanged across different soil conditions. Which of the following statements correctly explains the biological mechanisms responsible for the variation observed in these two traits?

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Answer: Leaf surface area exhibits continuous variation controlled by polygenes and modified by environmental conditions, whereas petal pigmentation pattern exhibits discontinuous variation governed by monogenic inheritance and unaffected by the environment.

Answer

Leaf surface area exhibits continuous variation controlled by polygenes and modified by environmental conditions, whereas petal pigmentation pattern exhibits discontinuous variation governed by monogenic inheritance and unaffected by the environment.
The phenotypic trait showing a normal bell-shaped curve and response to soil nutrients (leaf surface area) is a classic example of continuous variation, which is polygenic and modified by environmental conditions. Conversely, the trait displaying clear-cut, non-overlapping phenotypic categories unaffected by environmental shifts (petal pigmentation pattern) represents discontinuous variation, which is governed by monogenic inheritance.

Step-by-Step Solution

1
Analyze the phenotypic distribution and environmental sensitivity of leaf surface area.
Leaf surface area forms a continuous bell-shaped curve and changes phenotypic expression when grown in different soil conditions.
Continuous traits are quantitative, controlled by multiple additive genes (polygenic inheritance), and significantly influenced by environmental factors such as nutrient availability.
2
Analyze the phenotypic distribution and environmental sensitivity of petal pigmentation pattern.
Petal pigmentation shows clear-cut, discrete categories (solid crimson vs. white-striped) without intermediate phenotypes, remaining stable across varied soil environments.
Discontinuous traits are qualitative, controlled by one or a few major genes (monogenic/oligogenic inheritance), and generally unaffected by environmental variation.
3
Synthesize the biological mechanisms for both traits to identify the correct option.
Leaf surface area = Continuous variation (polygenic + environmental impact); Petal pigmentation = Discontinuous variation (monogenic + uninfluenced by environment).
Matching phenotypic features to their corresponding genetic architecture and environmental susceptibility confirms the correct biological interpretation.

Key Concept

Continuous and Discontinuous Variation
Estimated Time:2m 0s
Question 84Question

In humans, normal skin pigmentation is governed by a dominant allele (AA), while albinism is caused by a recessive allele (aa). Match each parental cross combination on the left with its corresponding expected offspring phenotypic and genotypic outcome on the right according to Mendel's First Law.

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Items

Heterozygous normal individual (AaAa) ×\times Heterozygous normal individual (AaAa)
Homozygous normal individual (AAAA) ×\times Albino individual (aaaa)
Heterozygous normal individual (AaAa) ×\times Albino individual (aaaa)
Homozygous normal individual (AAAA) ×\times Heterozygous normal individual (AaAa)

Matches

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Answer

The parental crosses match their expected offspring outcomes as follows: Aa×AaAa \times Aa produces a 3:13:1 phenotypic ratio of normal to albino offspring; AA×aaAA \times aa produces 100%100\% normal heterozygous carriers (AaAa); Aa×aaAa \times aa produces a 1:11:1 ratio of normal to albino offspring; and AA×AaAA \times Aa produces 100%100\% normal offspring with a 1:11:1 ratio of non-carriers (AAAA) to carriers (AaAa).
Each parental cross follows Mendel's Law of Segregation. A cross between two heterozygotes (Aa×AaAa \times Aa) yields a 3:1 phenotypic ratio (3 normal : 1 albino). A cross between homozygous dominant and homozygous recessive (AA×aaAA \times aa) produces 100% normal carriers (AaAa). A back/test cross (Aa×aaAa \times aa) yields a 1:1 ratio of normal to albino. A cross between homozygous dominant and heterozygous (AA×AaAA \times Aa) produces 100% normal offspring, split equally between non-carriers (AAAA) and carriers (AaAa).

Step-by-Step Solution

1
Analyze the cross Aa×AaAa \times Aa using Mendel's Law of Segregation.
Offspring genotypes are 1AA:2Aa:1aa1 AA : 2 Aa : 1 aa, giving a 3:13:1 ratio of normal skin pigmentation to albinism.
Both parents contribute alleles AA and aa in equal proportions (50% each).
2
Analyze the cross AA×aaAA \times aa.
All offspring receive AA from the first parent and aa from the second parent, yielding 100%Aa100\% Aa.
The dominant allele AA masks the recessive allele aa, making all offspring phenotypically normal carriers.
3
Analyze the cross Aa×aaAa \times aa.
Offspring genotypes are 1Aa:1aa1 Aa : 1 aa, resulting in a 1:11:1 phenotypic ratio of normal to albino offspring.
This is a monohybrid test cross where the phenotypic ratio directly reflects gamete segregation in the heterozygous parent.
4
Analyze the cross AA×AaAA \times Aa.
Offspring genotypes are 1AA:1Aa1 AA : 1 Aa, resulting in 100%100\% normal phenotypes.
The homozygous dominant parent guarantees that every offspring receives at least one AA allele.

Key Concept

Mendel's First Law and Monohybrid Cross Inheritance Ratios
Estimated Time:1m 30s
Question 85Question

A geneticist analyzed the phenotypic distribution of two traits in a population of fruit flies (*Drosophila melanogaster*): wing length measured in millimeters and eye color (red versus white). When cultures were reared across a range of ambient temperatures, the wing lengths across the population formed a smooth spectrum whose mean value shifted significantly with temperature changes. In contrast, eye color consistently segregated into two discrete, non-overlapping categories in predictable Mendelian ratios regardless of rearing temperature. What does this observation indicate regarding the genetic control and environmental sensitivity of these two traits?

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Answer: Wing length is polygenic and influenced by environmental factors, whereas eye color is monogenic and unaffected by temperature variations.

Answer

Wing length is polygenic and influenced by environmental factors, whereas eye color is monogenic and unaffected by temperature variations.
Continuous variation (such as wing length) is quantitative, controlled by multiple additive genes (polygenic inheritance), and readily modified by environmental conditions such as temperature. Discontinuous variation (such as eye color) is qualitative, controlled by one or a few major genes (monogenic inheritance), and produces distinct, non-overlapping categories that remain stable despite environmental fluctuations.

Step-by-Step Solution

1
Analyze the phenotypic distribution pattern of wing length.
Wing length exhibits a smooth, quantitative spectrum of variation whose population mean shifts with ambient temperature.
Traits that exhibit a unbroken range of phenotypes and show environmental sensitivity are characteristic of continuous variation governed by polygenic inheritance.
2
Analyze the phenotypic distribution pattern of eye color.
Eye color falls into two distinct, non-overlapping qualitative classes in discrete ratios regardless of environmental temperature.
Traits with clear-cut, qualitative categories independent of environmental conditions represent discontinuous variation governed by monogenic inheritance.
3
Synthesize the findings to match genetic mechanisms with variation types.
Wing length is polygenic and environmentally modified, while eye color is monogenic and environmental resistant.
This correctly aligns the observed phenotypic patterns with their underlying genetic and environmental determinants.

Key Concept

Polygenic vs Monogenic Control in Continuous and Discontinuous Variation
Estimated Time:1m 30s
Question 86Question

The human ABO blood group system is classified as a morphological trait that displays continuous variation across a population.

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Answer: False

Answer

The statement is false because the ABO blood group system is a physiological trait exhibiting discontinuous variation.
The statement is incorrect because the ABO blood group system represents a physiological trait governed by specific alleles, resulting in discrete phenotypic classes (discontinuous variation) rather than physical structural features along a continuous spectrum.

Step-by-Step Solution

1
Classify the nature of the trait as morphological or physiological.
ABO blood group involves biochemical antigens present on red blood cells, which makes it a physiological function rather than an anatomical structural trait.
Physiological variations relate to internal body chemistry, metabolism, and organ functioning.
2
Identify whether the distribution pattern is continuous or discontinuous.
Human blood groups fall into four discrete, non-overlapping phenotypes (A, B, AB, and O) without intermediate types.
Discontinuous variations consist of clear-cut qualitative categories determined by genetic inheritance, whereas continuous variations show a smooth spectrum.

Key Concept

Classification of human traits into physiological vs. morphological and continuous vs. discontinuous variations
Question 87Question

Phenotypic traits exhibiting discontinuous variation, such as human ABO blood groups and tongue-rolling ability, are controlled by polygenic inheritance and show a continuous spectrum of intermediates modified by environmental factors.

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Answer: False

Answer

The statement is False. Discontinuous variation features discrete, non-overlapping phenotypic classes governed by monogenic inheritance and is not modified by environmental factors.
The statement is false because discontinuous variation is characterized by clear-cut, non-overlapping phenotypic categories (such as blood types AA, BB, ABAB, and OO) controlled by single genes or single loci with few alleles (monogenic inheritance), operating independently of environmental influence.

Step-by-Step Solution

1
Identify the biological traits mentioned in the stem (ABO blood groups and tongue-rolling ability).
These phenotypic features are classic examples of discontinuous variation in humans.
Classifying the traits establishes the correct genetic framework.
2
Evaluate the genetic control and environmental influence described in the statement.
The statement attributes polygenic control and environmental modification to discontinuous traits.
Polygenic inheritance and environmental factors produce continuous variation (such as height or skin color), whereas discontinuous traits are monogenic and environment-independent.
3
Determine the validity of the statement.
Because the statement incorrectly pairs discontinuous traits with the characteristics of continuous variation, it is false.
A proposition containing contradicted biological definitions is false.

Key Concept

Genetic mechanisms of continuous vs. discontinuous variation
Question 88Question

Match each sex-linked inheritance pattern or sex determination concept on the left with its correct biological description on the right.

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Items

Y-linked gene transmission
X-linked recessive carrier mother
Homogametic male sex determination
Heterogametic female sex determination

Matches

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Answer

Y-linked gene transmission matches with direct father-to-son inheritance; X-linked recessive carrier mother matches with 50% allele transmission probability; Homogametic male sex determination matches with ZZ males in birds; Heterogametic female sex determination matches with ZW females in birds and reptiles.
Each genetic concept correctly aligns with its characteristic mode of chromosome transmission or gametic constitution: Y-linked traits are holandric (father-to-son), X-linked recessive carrier mothers have a 50% chance per child of passing the mutated allele, ZZ represents homogametic males in birds, and ZW represents heterogametic females in birds and reptiles.

Step-by-Step Solution

1
Identify the inheritance mode of Y-linked genes.
Since only males possess the Y chromosome, Y-linked traits pass exclusively from fathers to sons.
Y chromosomes are inherited solely through the paternal line.
2
Determine the transmission probability for a heterozygous carrier mother.
A carrier mother (XHXhX^H X^h) passes the XhX^h allele to 50% of her offspring.
Segregation during meiosis distributes each X chromosome into half of the gametes.
3
Differentiate between homogametic and heterogametic sex determination systems.
In birds (ZZ-ZW system), males have two identical sex chromosomes (ZZ, homogametic) while females have two distinct sex chromosomes (ZW, heterogametic).
Homogametic individuals produce only one type of sex gamete, whereas heterogametic individuals produce two different types.

Key Concept

Sex Determination Mechanisms and Sex-Linked Inheritance
Question 89Question

In maize plants (*Zea mays*), the allele for purple kernels (PP) is completely dominant to the allele for yellow kernels (pp). A true-breeding purple-kernel plant is crossed with a true-breeding yellow-kernel plant. If the F1F_1 plants are self-pollinated, what is the expected genotypic ratio in the F2F_2 generation?

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Answer: 1:2:11 : 2 : 1

Answer

The expected genotypic ratio in the F2F_2 generation is 1:2:11 : 2 : 1 (1 PP:2 Pp:1 pp1\ PP : 2\ Pp : 1\ pp).
Crossing two heterozygous F1F_1 plants (Pp×PpPp \times Pp) yields genotypes PPPP, PpPp, and pppp in the proportions 1/4 PP1/4\ PP, 1/2 Pp1/2\ Pp, and 1/4 pp1/4\ pp. This corresponds to a genotypic ratio of 1:2:11 : 2 : 1.

Step-by-Step Solution

1
Determine parental genotypes and F1F_1 genotype
True-breeding parents are PPPP and pppp; all F1F_1 offspring are PpPp.
Homozygous dominant (PPPP) crossed with homozygous recessive (pppp) produces offspring that all inherit one PP and one pp allele.
2
Perform the F1F_1 self-cross (Pp×PpPp \times Pp)
Gametes PP and pp combine to produce PPPP, PpPp, pPpP, and pppp.
According to Mendel's Law of Segregation, alleles segregate into gametes with equal probability.
3
Calculate the genotypic ratio of the F2F_2 offspring
Genotypic ratio is 1 PP:2 Pp:1 pp1\ PP : 2\ Pp : 1\ pp, or 1:2:11 : 2 : 1.
Grouping offspring by genetic constitution yields 1 homozygous dominant (PPPP), 2 heterozygous (PpPp), and 1 homozygous recessive (pppp).

Key Concept

Monohybrid Cross Genotypic Ratio under Mendel's First Law
Estimated Time:1m 0s
Question 90Question

In diploid eukaryotic organisms, genetic information is organized into chromosomes containing specific functional sequences. Which of the following statements accurately describes the biological distinction between alleles and gene loci during meiotic division?

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Answer: A gene locus defines the fixed physical position of a gene on a chromosome, whereas alleles are alternative sequence variants of that gene located at identical loci on homologous chromosomes.

Answer

A gene locus defines the fixed physical position of a gene on a chromosome, whereas alleles are alternative sequence variants of that gene located at identical loci on homologous chromosomes.
The correct response accurately highlights that a gene locus is the precise structural coordinate on a chromosome, whereas alleles represent the specific molecular sequence variations of that gene present at corresponding loci on homologous pairs.

Step-by-Step Solution

1
Define the term 'gene locus'.
A gene locus (plural: loci) refers to the specific, fixed location of a particular gene along the length of a chromosome.
Establishing physical chromosomal position is essential before analyzing variant forms.
2
Define the term 'allele' in relation to homologous chromosomes.
Alleles are different functional variants of the same gene that occupy identical corresponding loci on homologous chromosome pairs.
Diploid organisms carry two alleles for each autosomal gene, one inherited from each parent.
3
Compare definitions to evaluate statement validity.
The option stating that a locus is the fixed position while alleles are sequence variants at identical loci on homologous chromosomes is scientifically correct.
This maintains precise distinction between physical locus coordinates and allele variants.

Key Concept

Gene Locus vs. Allele in Homologous Chromosomes
Estimated Time:1m 30s
Question 91Question

In maize (*Zea mays*), the allele for starchy endosperm (SuSu) is completely dominant over the allele for sugary endosperm (susu). Based on Mendel's First Law (Law of Segregation), which expected progeny outcome on the right correctly matches each monohybrid parental cross scenario on the left?

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Items

Cross between two heterozygous starchy plants (Susu×SusuSu\,su \times Su\,su), evaluated for the genotypic ratio among starchy progeny only
Test cross of a heterozygous starchy plant (Susu×susuSu\,su \times su\,su), evaluated for the ratio of dominant allele (SuSu) carriers to non-carriers
Self-pollination of an F1F_1 plant (SusuSu\,su), evaluated for the total percentage of pure-breeding (homozygous) progeny
Cross between a heterozygous starchy plant (SusuSu\,su) and a homozygous starchy plant (SuSuSu\,Su), evaluated for the percentage of sugary (sususu\,su) progeny

Matches

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Answer

Cross between two heterozygous starchy plants matches 1 SuSu:2 Susu1\ Su\,Su : 2\ Su\,su; Test cross of a heterozygous starchy plant matches 1:11 : 1 ratio (50%50\% carriers to 50%50\% non-carriers); Self-pollination of an F1F_1 plant matches 50%50\% of total progeny; Cross between a heterozygous starchy plant and homozygous starchy plant matches 0%0\% of total progeny.
Each monohybrid parental cross is correctly matched by calculating the specific genotypic or phenotypic probabilities based on allele segregation during meiosis according to Mendel's First Law.

Step-by-Step Solution

1
Analyze the cross between two heterozygous starchy plants (Susu×SusuSu\,su \times Su\,su).
Genotypes produced are 1/4 SuSu1/4\ Su\,Su, 1/2 Susu1/2\ Su\,su, and 1/4 susu1/4\ su\,su.
Mendel's Law of Segregation states alleles segregate into gametes with equal probability (1/2 Su1/2\ Su, 1/2 su1/2\ su). Starchy plants comprise SuSuSu\,Su and SusuSu\,su. Among starchy progeny, the genotypic ratio is 1 SuSu:2 Susu1\ Su\,Su : 2\ Su\,su.
2
Analyze the test cross of a heterozygous starchy plant (Susu×susuSu\,su \times su\,su).
Genotypes produced are 1/2 Susu1/2\ Su\,su and 1/2 susu1/2\ su\,su.
Carriers of the dominant allele SuSu are SusuSu\,su (50%50\%), while non-carriers are sususu\,su (50%50\%), yielding a ratio of 1:11 : 1.
3
Analyze self-pollination of an F1F_1 plant (Susu×SusuSu\,su \times Su\,su) for pure-breeding individuals.
Pure-breeding individuals (SuSuSu\,Su and sususu\,su) constitute 50%50\% of offspring.
Pure-breeding offspring are homozygous. 25% SuSu+25% susu=50%25\%\ Su\,Su + 25\%\ su\,su = 50\% of total progeny.
4
Analyze the cross Susu×SuSuSu\,su \times Su\,Su for sugary (sususu\,su) progeny.
Proportion of sususu\,su offspring is 0%0\%.
The homozygous dominant parent contributes a dominant SuSu allele to all offspring, preventing the expression of the recessive sususu\,su genotype.

Key Concept

Mendel's First Law and Monohybrid Cross Proportions
Estimated Time:2m 30s
Question 92Question

Human physiological traits such as blood pressure and resting pulse rate exhibit discontinuous variation because individuals strictly belong to distinct, non-overlapping phenotypic categories controlled solely by single-gene inheritance.

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Answer: False

Answer

False
The statement is false because blood pressure and resting pulse rate exhibit continuous physiological variation, characterized by a smooth quantitative gradient across a population under polygenic control and environmental modification.

Step-by-Step Solution

1
Analyze the pattern of variation exhibited by physiological traits like blood pressure and resting pulse rate.
Blood pressure and pulse rate display a continuous spectrum of phenotypic values across a population spectrum without distinct breaks.
Continuous variation presents a gradual quantitative range of phenotypic expressions rather than distinct categories.
2
Examine the underlying genetic basis and environmental interaction for these physiological traits.
These traits are polygenic (controlled by multiple genes working additively) and are substantially modified by environmental factors.
Polygenic inheritance combined with environmental modifications produces a smooth, bell-shaped normal distribution curve.
3
Contrast continuous physiological traits with discontinuous variations.
Discontinuous variations (such as ABO blood groups or tongue rolling) consist of clear-cut non-overlapping phenotypic categories controlled by single genes and are generally unaffected by environmental conditions.
Mislabelling continuous physiological traits as discontinuous fails to account for polygenic control and environmental susceptibility.

Key Concept

Continuous vs. Discontinuous Variation in Human Physiological Traits
Question 93Question

Human skin color exhibits continuous variation because it is governed by polygenic inheritance with additive gene action, allowing environmental exposure to modify phenotypic expression along a continuous spectrum, whereas discontinuous traits like the ABO blood group system are controlled monogenically and are exempt from environmental modification.

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Answer: True

Answer

True. Continuous variation is characterized by polygenic inheritance and environmental influence, whereas discontinuous variation is governed by monogenic inheritance without environmental modification.
The statement accurately presents biological facts: continuous traits like skin color are polygenic and influenced by environmental factors such as sunlight, whereas discontinuous traits like ABO blood groups are controlled by a single gene locus and remain completely unaffected by the environment.

Step-by-Step Solution

1
Analyze the genetic basis of continuous variation in phenotypic traits.
Traits showing continuous variation (such as skin color, height, and body mass) are controlled by multiple independent gene pairs acting additively (polygenic inheritance).
Polygenic inheritance creates a continuous distribution of phenotypes rather than discrete categories.
2
Assess the role of environmental factors on continuous versus discontinuous traits.
Environmental exposure (such as sunlight altering melanin production for skin color or nutrition altering height) shifts phenotypes along a smooth gradient. Discontinuous traits (such as ABO blood groups or tongue rolling) are genetically fixed and immune to environmental modification.
Differentiating environmental influence helps distinguish continuous variation from discontinuous variation.
3
Evaluate the genetic control mechanism of discontinuous traits.
Discontinuous traits are inherited monogenically (or via a single gene locus with major alleles), leading to distinct, non-overlapping phenotypic classes.
Single-gene inheritance prevents intermediate phenotypic gradations from forming in populations.
4
Determine the validity of the complete statement.
Both assertions regarding genetic mechanism (polygenic vs monogenic) and environmental sensitivity (modifiable vs exempt) are scientifically accurate, making the statement True.
The contrast drawn between skin color and ABO blood group inheritance correctly reflects the mechanisms of biological variation.

Key Concept

Polygenic vs Monogenic Control of Continuous and Discontinuous Variation
Question 94Question

In humans, hypertrichosis of the ear pinna is a Y-linked (holandric) trait, whereas red-green color blindness is an X-linked recessive disorder (XcX^c). A man with hypertrichosis and normal vision marries a phenotypically normal woman whose father was color-blind. What is the probability that their first child will be a male displaying both hypertrichosis and red-green color blindness?

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Answer: 25%

Answer

25%
The correct answer is 25% because the mother is a carrier (XCXcX^C X^c) and the father carries hypertrichosis on his Y chromosome (XCYHX^C Y^H). For a child to be a male with both conditions, he must receive the YHY^H chromosome from his father (probability 0.5) and the XcX^c allele from his mother (probability 0.5). Multiplying these independent events gives 0.5×0.5=0.250.5 \times 0.5 = 0.25 or 25%.

Step-by-Step Solution

1
Determine parental genotypes
Father = XCYHX^C Y^H; Mother = XCXcX^C X^c
The father has normal vision (XCX^C) and hypertrichosis (YHY^H). The mother is phenotypically normal but inherited XcX^c from her color-blind father (XcYX^c Y).
2
Determine offspring gamete combinations via a Punnett square
Female offspring: XCXCX^C X^C (25%), XCXcX^C X^c (25%); Male offspring: XCYHX^C Y^H (25%), XcYHX^c Y^H (25%)
Sons receive the YHY^H chromosome from the father and either XCX^C or XcX^c from the mother.
3
Calculate the total probability for a male child with both traits
P(Male with both traits)=P(Inheriting YH)×P(Inheriting Xc)=0.50×0.50=0.25=25%P(\text{Male with both traits}) = P(\text{Inheriting } Y^H) \times P(\text{Inheriting } X^c) = 0.50 \times 0.50 = 0.25 = 25\%
The question asks for the probability among all potential offspring, requiring the product of receiving the YHY^H chromosome (50%) and the XcX^c allele (50%).

Key Concept

Simultaneous X-linked and Y-linked trait inheritance
Estimated Time:2m 0s
Question 95Question

Match each sex determination mechanism or sex-linked trait concept on the left with its corresponding biological description on the right.

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Items

XX-XY determination mechanism
ZZ-ZW determination mechanism
Haemophilia inheritance
Holandric traits

Matches

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Answer

XX-XY determination mechanism corresponds to the system with heterogametic males (XY); ZZ-ZW determination mechanism corresponds to the system with heterogametic females (ZW); Haemophilia inheritance corresponds to the X-linked recessive blood-clotting disorder; Holandric traits correspond to Y-linked traits passed exclusively from father to son.
Each concept is correctly paired with its biological definition: XX-XY features heterogametic males; ZZ-ZW features heterogametic females; Haemophilia is an X-linked recessive disorder impairing blood clotting; and Holandric traits are Y-linked traits inherited strictly from father to son.

Step-by-Step Solution

1
Identify the chromosomal composition of heterogametic and homogametic sexes in different organisms.
In XX-XY systems (mammals), males are heterogametic (XY). In ZZ-ZW systems (birds), females are heterogametic (ZW).
The heterogametic sex produces two different types of gametes that determine the offspring's sex.
2
Examine the inheritance pattern of Haemophilia.
Haemophilia is an X-linked recessive trait that affects blood coagulation and is primarily expressed in males.
Males receive their single X chromosome from their mother, so a recessive allele on the X chromosome will always be expressed.
3
Examine the inheritance pattern of Holandric traits.
Holandric traits are Y-linked traits passed directly from fathers to all male offspring.
Only males inherit the Y chromosome, ensuring strict paternal transmission.

Key Concept

Sex Determination Mechanisms and Sex-Linked Inheritance Patterns
Question 96Question

A mother with blood group B gives birth to a child with blood group O. An alleged father involved in a paternity case has blood group AB. Based on the genetic principles of codominance and multiple alleles governing the ABO blood group system, which of the following conclusions regarding paternity is correct?

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Answer: The alleged father is excluded because he lacks the recessive ii allele required to produce a blood group O child.

Answer

The alleged father is excluded because he lacks the recessive ii allele required to produce a child with blood group O.
Blood group O is a recessive phenotype resulting from the homozygous genotype iiii. For a child to have blood group O, both biological parents must pass on a recessive ii allele. A person with blood group AB has the genotype IAIBI^A I^B due to codominance between the IAI^A and IBI^B alleles. Since this individual carries no ii allele, he can only pass either IAI^A or IBI^B to his offspring. Therefore, he is genetically excluded from fathering a child with blood group O.

Step-by-Step Solution

1
Determine the genotype of the child with blood group O.
The child has the genotype iiii, requiring one recessive ii allele from each biological parent.
Blood group O is an autosomal recessive phenotype in the ABO system.
2
Determine the genotype and possible gametes produced by the alleged father.
The alleged father has genotype IAIBI^A I^B and produces gametes containing either the IAI^A allele or the IBI^B allele.
The IAI^A and IBI^B alleles are codominant, resulting in blood group AB.
3
Evaluate whether the alleged father can contribute to the child's genotype.
The alleged father cannot contribute an ii allele to the child.
An individual with genotype IAIBI^A I^B completely lacks the ii allele.
4
Formulate the final conclusion regarding paternity.
The alleged father is definitively excluded as the biological father.
A child cannot inherit an allele from a biological parent who does not possess that allele.

Key Concept

Codominance and Multiple Alleles in ABO Blood Group Inheritance
Question 97Question

Agricultural plant breeders often cross two genetically distinct inbred lines of maize to produce offspring that display superior growth rate, pest resistance, and crop yield compared to either parent. Which of the following biological terms best describes this application of genetics?

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Answer: Hybrid vigor

Answer

Hybrid vigor (heterosis) is the increased vigor, growth, and fertility of offspring resulting from crossing two genetically distinct inbred lines.
The term hybrid vigor (or heterosis) describes the phenomenon where the crossbred offspring of two distinct, homozygous inbred parental strains exhibit enhanced traits such as greater size, higher yield, and better disease resistance due to increased heterozygosity.

Step-by-Step Solution

1
Identify the genetic process described in the scenario
Two distinct inbred lines are crossed to create superior offspring.
Plant breeders deliberately combine diverse gene pools to mask harmful recessive alleles.
2
Match the process with its correct genetic term
The enhancement of performance in heterozygous offspring is termed hybrid vigor (heterosis).
Heterozygosity at multiple loci often leads to improved physiological performance over homozygous inbred parents.

Key Concept

Hybrid vigor (Heterosis) in crop and animal breeding
Question 98Question

A geneticist crosses two heterozygous tall pea plants (TtTt). In the resulting offspring, some plants display the dwarf trait even though both parent plants were tall. Which genetic principle best explains why the dwarf trait was masked in the parent generation?

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Answer: The allele for dwarfness is recessive and masked by the dominant allele for tallness

Answer

The allele for dwarfness is recessive and masked by the dominant allele for tallness in the heterozygous parent generation.
The option stating that the allele for dwarfness is recessive and masked by the dominant allele for tallness is correct because in heterozygous organisms (TtTt), the dominant allele determines the phenotype, keeping the recessive allele hidden until inherited in a homozygous recessive state (tttt).

Step-by-Step Solution

1
Identify the parental genotypes and phenotypes
Both parent plants have the heterozygous genotype TtTt and display the tall phenotype.
Since both parents are tall but carry the dwarf allele (tt), the tall allele (TT) must be dominant.
2
Analyze the expression of alleles in the heterozygous state
In TtTt individuals, the recessive allele (tt) is masked phenotypically by the dominant allele (TT).
By definition of dominance and recessiveness in basic genetics, a recessive allele is only expressed when two copies are present (tttt).
3
Relate to the appearance of dwarf offspring in the F1 generation
Segregation of alleles produces tttt offspring with a 25% probability, revealing the masked recessive trait.
The reappearance of dwarf plants proves that the dwarf allele was present but masked in the heterozygous parents.

Key Concept

Dominant and Recessive Alleles in Heterozygous Organisms
Estimated Time:1m 0s
Question 99Question

A woman who is a carrier for red-green colour blindness (XCXcX^C X^c) marries a man with normal colour vision (XCYX^C Y). What is the probability that any son born to this couple will be colour-blind?

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Answer: 50%50\%

Answer

The probability that any son born to this couple will be colour-blind is 50%50\%.
A carrier mother has the genotype XCXcX^C X^c, meaning half of her eggs carry the normal allele (XCX^C) and half carry the recessive colour-blindness allele (XcX^c). All sons inherit a Y chromosome from their father and an X chromosome from their mother. Therefore, each son has a 50%50\% chance of inheriting the XcX^c chromosome and being colour-blind (XcYX^c Y).

Step-by-Step Solution

1
Determine parental genotypes and gametes
Mother (XCXcX^C X^c) produces gametes XCX^C and XcX^c in equal proportions (50%50\% each). Father (XCYX^C Y) produces gametes XCX^C and YY.
Red-green colour blindness is an X-linked recessive trait.
2
Determine genotypes of male offspring
Sons inherit the Y chromosome from their father and an X chromosome from their mother. Possible male genotypes are XCYX^C Y (normal vision) and XcYX^c Y (colour-blind).
Male offspring inherit their sex-defining Y chromosome strictly from the father.
3
Calculate the probability among sons
Out of 2 possible male genotypes (XCYX^C Y and XcYX^c Y), 1 represents a colour-blind son, giving a probability of 12=50%\frac{1}{2} = 50\%.
The question specifically asks for the probability among sons, not total offspring.

Key Concept

X-linked recessive inheritance and gender-restricted offspring probabilities
Question 100Question

During a genetic survey of a fruit fly (*Drosophila melanogaster*) population, scientists observed variations in eye color caused by different molecular versions of the gene controlling pigment production located at the same locus on homologous chromosomes. Which term correctly identifies these alternative functional forms of a single gene?

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Answer: Alleles

Answer

Alleles are the alternative forms of a gene occupying the same gene locus on homologous chromosomes.
The term allele refers directly to one of two or more alternative versions of a gene that arise by mutation and are found at the same place (locus) on a chromosome.

Step-by-Step Solution

1
Identify the biological concept described in the stem.
The stem describes different structural/molecular versions of a single gene that reside at the same chromosomal position (locus).
Genes often exist in more than one form within a population, giving rise to variations in specific traits.
2
Differentiate between gene structure terms and chromosomal/phenotypic terms.
The term 'allele' specifically denotes these alternative versions of a gene (e.g., red vs. white eye color genes in *Drosophila*).
Understanding the distinction between a gene locus, an allele, a chromatid, and a phenotypic trait is essential in basic genetics.

Key Concept

Alleles as alternative forms of a gene at a specific locus
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