Heredity and Variation

159 questions

Question 141Question

Match each non-Mendelian genetic concept or inheritance pattern on the left with its corresponding cellular or phenotypic outcome on the right.

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Items

Incomplete dominance
Codominance
Multiple alleles
Complete dominance (IAI^A over ii)

Matches

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Answer

Incomplete dominance matches blended intermediate phenotypes; Codominance matches simultaneous full expression of both alleles; Multiple alleles match having more than two gene forms in a population; Complete dominance (IAI^A over ii) matches masking of the recessive allele.
Incomplete dominance is defined by intermediate trait blending in heterozygotes. Codominance is defined by simultaneous, unblended expression of both alleles. Multiple alleles refer to gene loci with more than two allele forms in a population pool. Complete dominance describes one allele fully masking a recessive allele in the heterozygous condition.

Step-by-Step Solution

1
Analyze Incomplete Dominance
In incomplete dominance, neither allele is completely dominant, producing an intermediate phenotype in heterozygotes.
Alleles interact such that heterozygotes possess a phenotype between both homozygous conditions.
2
Analyze Codominance
In codominance, both alleles are fully expressed without blending.
Both gene products remain distinct and functional in heterozygotes, as seen in AB blood group.
3
Analyze Multiple Alleles
Multiple alleles refer to three or more allele forms controlling one gene locus across a population.
Individual organisms inherit only two alleles, but population-level variation contains multiple alleles.
4
Analyze Complete Dominance (IAI^A over ii)
The dominant allele IAI^A completely hides the effect of the recessive ii allele.
Heterozygotes with genotype IAiI^A i express blood group A phenotype identically to homozygous IAIAI^A I^A individuals.

Key Concept

Non-Mendelian Inheritance Patterns and Allelic Interactions
Question 142Question

In guinea pigs (*Cavia porcellus*), the allele for rough coat (RR) is completely dominant over the allele for smooth coat (rr). Match each parental cross listed in the left column with its correct expected offspring genotypic or phenotypic ratio in the right column.

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Items

Heterozygous rough coat (RrRr) ×\times Heterozygous rough coat (RrRr)
Heterozygous rough coat (RrRr) ×\times Homozygous smooth coat (rrrr)
Homozygous rough coat (RRRR) ×\times Homozygous smooth coat (rrrr)
Homozygous rough coat (RRRR) ×\times Heterozygous rough coat (RrRr)

Matches

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Answer

Heterozygous cross (Rr×RrRr \times Rr) matches 3 : 1 phenotypic ratio of rough coat to smooth coat; Heterozygous testcross (Rr×rrRr \times rr) matches 1 : 1 phenotypic ratio of rough coat to smooth coat; Pure-breeding cross (RR×rrRR \times rr) matches 100% rough coat phenotype (100% RrRr genotype); Dominant backcross (RR×RrRR \times Rr) matches 100% rough coat phenotype (1 : 1 genotypic ratio of RR:RrRR : Rr).
Each parental genotype combination directly produces specific gamete combinations according to Mendel's Law of Segregation. Crossing two heterozygotes (Rr×RrRr \times Rr) yields a 3 : 1 phenotypic ratio; crossing a heterozygote with a recessive homozygote (Rr×rrRr \times rr) yields a 1 : 1 ratio; crossing pure-breeding opposite homozygotes (RR×rrRR \times rr) produces 100% heterozygous offspring (RrRr); and crossing a homozygous dominant with a heterozygote (RR×RrRR \times Rr) produces 100% dominant phenotypically with equal proportions of RRRR and RrRr genotypes.

Step-by-Step Solution

1
Determine gametes and Punnett square for Rr×RrRr \times Rr.
Gametes are RR and rr from each parent. Offspring genotypes: 1 RR:2 Rr:1 rr1\ RR : 2\ Rr : 1\ rr.
Mendel's Law of Segregation states alleles segregate during gamete formation, resulting in a 3 : 1 phenotypic ratio of dominant (rough) to recessive (smooth).
2
Determine offspring for testcross Rr×rrRr \times rr.
Gametes from heterozygous parent are RR and rr; gametes from homozygous recessive parent are all rr. Offspring genotypes: 1 Rr:1 rr1\ Rr : 1\ rr.
Half the offspring express the dominant phenotype and half express the recessive phenotype, giving a 1 : 1 ratio.
3
Determine offspring for RR×rrRR \times rr.
All gametes from RRRR are RR, all gametes from rrrr are rr. All offspring are RrRr.
100% of the F1 generation is heterozygous and displays the dominant rough coat phenotype.
4
Determine offspring for RR×RrRR \times Rr.
Gametes are RR (from RRRR) and R,rR, r (from RrRr). Offspring genotypes: 1 RR:1 Rr1\ RR : 1\ Rr.
All offspring inherit at least one RR allele, so 100% show the rough coat phenotype, with a 1 : 1 ratio between RRRR and RrRr genotypes.

Key Concept

Mendel's First Law (Law of Segregation) and Monohybrid Cross Ratios
Question 143Question

In *Drosophila melanogaster*, eye color is an X-linked trait where red eye color (XRX^R) is dominant over white eye color (XrX^r). If a heterozygous red-eyed female (XRXrX^R X^r) is crossed with a white-eyed male (XrYX^r Y), what is the probability that a female offspring from this cross will express the white-eye phenotype?

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Answer: 50%

Answer

50%
The female parent donates either XRX^R or XrX^r with equal probability (50% each), while the male parent donates an XrX^r chromosome to all female offspring. As a result, 50% of the female offspring inherit XRXrX^R X^r (red-eyed) and 50% inherit XrXrX^r X^r (white-eyed).

Step-by-Step Solution

1
Determine parental genotypes and gamete contributions.
Female genotype is XRXrX^R X^r (gametes: XRX^R and XrX^r); Male genotype is XrYX^r Y (gametes: XrX^r and YY).
The gene for eye color in Drosophila is located on the X chromosome.
2
Construct the genetic cross to find offspring genotypes.
Offspring genotypes are XRXrX^R X^r (red-eyed female), XrXrX^r X^r (white-eyed female), XRYX^R Y (red-eyed male), and XrYX^r Y (white-eyed male). Each genotype represents 25% of total offspring.
Combining maternal and paternal gametes yields four equally likely genotypic outcomes.
3
Restrict the calculation specifically to female offspring.
Female offspring genotypes are XRXrX^R X^r and XrXrX^r X^r. One out of two female offspring (XrXrX^r X^r) has white eyes, giving a probability of 12=50%\frac{1}{2} = 50\%.
The question specifically asks for the probability among female offspring, not among all offspring.

Key Concept

X-linked recessive inheritance and gender-restricted offspring probability calculations
Question 144Question

A genetics researcher classifies several human phenotypic observations recorded during a demographic study. Match each phenotypic observation on the left with its correct variation category and genetic basis on the right.

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Items

Gradual variation in skin melanin density across a population creating a spectrum of shades
Presence or absence of the chemical tasting perception for phenylthiocarbamide (PTC)
Distinct classification of individuals into dermatoglyphic loop, whorl, or arch categories
Production of specific A, B, or O agglutinogens on erythrocyte membranes

Matches

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Answer

Gradual variation in skin melanin density matches continuous polygenic morphological trait; PTC tasting perception matches discontinuous physiological trait governed by a single gene pair; Fingerprint pattern categories match discontinuous morphological trait uninfluenced by environment; ABO agglutinogen production matches discontinuous physiological trait determined by multiple alleles.
Each phenotype aligns precisely with its biological nature (morphological vs. physiological), distribution pattern (continuous vs. discontinuous), and underlying genetic mode of inheritance.

Step-by-Step Solution

1
Categorize each phenotypic trait as either morphological (structural/physical appearance) or physiological (functional/biochemical process).
Skin melanin density and fingerprint patterns are morphological traits; PTC tasting and erythrocyte agglutinogens are physiological traits.
Morphological traits involve anatomical features, while physiological traits relate to chemical functions and biological processes.
2
Determine whether each trait exhibits continuous variation (spectrum with intermediate phenotypes) or discontinuous variation (discrete non-overlapping categories).
Skin melanin density is continuous; PTC tasting, fingerprint patterns, and ABO blood groups are discontinuous.
Continuous variation displays quantitative grading across a spectrum, whereas discontinuous variation presents distinct qualitative groups.
3
Relate each categorized trait to its correct genetic control mechanism and environmental susceptibility.
Skin color is polygenic and environmental; PTC tasting is monogenic; fingerprints are genetically fixed structural types; ABO blood groups rely on multiple codominant/recessive alleles.
Matching structural, distributional, and genetic parameters confirms the exact paired classifications.

Key Concept

Human Morphological and Physiological Variations (Continuous vs. Discontinuous)
Question 145Question

Match each application of genetics in medicine or agriculture listed on the left with its corresponding biological mechanism or practical outcome on the right.

Click a left item, then click its matching right item

Items

Polyploidy induction in crop breeding
Pre-marital genetic counseling
Recombinant DNA technology in medicine
Inbreeding of livestock lines

Matches

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Answer

Polyploidy induction in crop breeding matches with artificially doubling chromosome sets to produce larger, seedless, or high-yielding crop varieties. Pre-marital genetic counseling matches with evaluating the probability of offspring inheriting autosomal recessive blood disorders such as sickle-cell anemia. Recombinant DNA technology in medicine matches with inserting specific human genes into bacterial plasmids to synthesize therapeutic proteins like insulin. Inbreeding of livestock lines matches with establishing homozygous pure lines for desired traits, which can potentially lead to inbreeding depression.
Each application correctly pairs with its specific biological mechanism or goal: Polyploidy increases chromosome sets for crop improvement; genetic counseling evaluates hereditary disease risks in families; recombinant DNA technology uses bacterial plasmids to produce human therapeutics like insulin; and inbreeding establishes pure homozygous lines in livestock breeding.

Step-by-Step Solution

1
Analyze crop breeding techniques for polyploidy.
Polyploidy refers to having extra sets of chromosomes, used in agriculture to create larger, seedless, or robust varieties.
Polyploid plants often exhibit gigas effects, producing larger fruits and flowers.
2
Analyze medical genetic counseling goals.
Genetic counselors assess carrier statuses (e.g., HbAA vs. HbAS) to inform prospective parents about inheritance probabilities of recessive conditions.
Preventive medicine relies on genetic screening to reduce the incidence of severe hereditary traits.
3
Analyze modern medical biotechnology applications.
Recombinant DNA technology splices targeted human genes into bacterial expression vectors to manufacture proteins such as human insulin or growth hormone.
Bacteria replicate rapidly, allowing large-scale production of human-compatible therapeutic proteins.
4
Analyze animal breeding methods involving close relatives.
Inbreeding concentrates specific alleles to create true-breeding pure lines, but can accumulate harmful recessive alleles.
Homozygosity increases predictability of offspring traits but risks inbreeding depression.

Key Concept

Applications of Genetics in Medicine and Agriculture
Question 146Question

In watermelon plants (*Citrullus lanatus*), solid green rind color (GG) is dominant over striped rind color (gg), and short fruit shape (SS) is dominant over long fruit shape (ss). A pure-breeding plant with solid green, short fruits is crossed with a plant having striped, long fruits. The resulting F1F_1 plants are self-pollinated to produce an F2F_2 generation of 800 offspring. How many of the F2F_2 plants are expected to produce striped, short watermelons?

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Answer: 150

Answer

150
In a classical dihybrid cross involving two heterozygous parents (GgSsGgSs), independent assortment yields an F2 phenotypic ratio of 9:3:3:1. The phenotype with one recessive trait and one dominant trait (striped, short; genotype ggS_ggS\_) represents 3/163/16 of the total population. Multiplying this fraction by 800 gives (3/16)×800=150(3/16) \times 800 = 150 plants.

Step-by-Step Solution

1
Identify parental and F1 genotypes
Parental cross is GGSS×ggssGGSS \times ggss, yielding an F1F_1 genotype of GgSsGgSs.
Pure-breeding parents pass one dominant or recessive allele per gene to the offspring.
2
Determine the F2 phenotypic ratio using Mendel's Law of Independent Assortment
Selfing GgSs×GgSsGgSs \times GgSs yields a 9:3:3:1 phenotypic ratio (9 solid/short : 3 solid/long : 3 striped/short : 1 striped/long).
Alleles for rind color and fruit shape assort independently during gamete formation.
3
Calculate expected offspring count for the target phenotype
Expected count = (3/16)×800=150(3 / 16) \times 800 = 150 plants.
The target phenotype (striped, short) represents 3 out of 16 total phenotypic combinations.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Ratios
Question 147Question

Human physiological traits, such as the ability to roll the tongue and the capacity to taste phenylthiocarbamide (PTC), exhibit distinct phenotypic classes with no intermediate forms, classifying them as examples of discontinuous variation.

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Answer: True

Answer

The statement is TRUE.
Tongue rolling and PTC tasting are classic examples of human physiological variations exhibiting discontinuous variation. They are controlled by major single genes and divide individuals into distinct, qualitative, non-overlapping groups (rollers vs. non-rollers, tasters vs. non-tasters) rather than showing a continuous range of phenotypes.

Step-by-Step Solution

1
Identify the nature of the traits mentioned in the stem.
Tongue rolling ability and PTC tasting are human physiological variations.
Physiological traits relate to the functional capabilities and metabolic responses of an individual.
2
Determine the distribution pattern of these specific physiological traits within human populations.
Individuals are either tongue rollers or non-rollers, and either PTC tasters or non-tasters, with no intermediate phenotype.
Single-gene inheritance produces discrete phenotypic classes unaffected by environmental factors.
3
Evaluate whether discrete phenotypic classification corresponds to discontinuous variation.
Traits divided into clear, non-overlapping categories represent discontinuous variation.
Discontinuous variation is characterized by distinct qualitative phenotypic categories controlled by one or a few genes.

Key Concept

Discontinuous variation in human physiological traits
Question 148Question

In human ABO blood group inheritance, although three main alleles (IAI^A, IBI^B, and ii) exist within the human population, any normal diploid individual can possess at most two of these alleles.

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Answer: True

Answer

The statement is true.
The statement is true because multiple allelism is a population-level concept describing three or more allele variants in a gene pool, whereas every diploid individual is constrained to carrying exactly two alleles for any autosomal gene locus.

Step-by-Step Solution

1
Define multiple allelism in population genetics.
The ABO blood group system exhibits multiple alleles (IAI^A, IBI^B, and ii) across the population.
Multiple alleles refer to three or more alternative forms of a gene occupying the same locus on homologous chromosomes across a species.
2
Analyze individual chromosomal makeup (ploidy).
Humans are diploid organisms with homologous chromosome pairs.
An individual inherits one set of 23 chromosomes from the egg cell and one set from the sperm cell.
3
Determine the maximum number of alleles per individual.
An individual can carry a maximum of two alleles at the ABO gene locus (e.g., IAIAI^A I^A, IAIBI^A I^B, IAiI^A i, IBIBI^B I^B, IBiI^B i, or iiii).
Since a gene locus exists only twice in a diploid cell (once per homologous chromosome), an individual can never possess three alleles.

Key Concept

Multiple Alleles vs Diploid Genotype
Estimated Time:1m 0s
Question 149Question

During meiotic cell division, the failure of homologous chromosomes or sister chromatids to separate properly results in gametes with abnormal numbers of chromosomes. What is the precise biological term for this failure of separation?

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Answer: Nondisjunction

Answer

Nondisjunction is the failure of homologous chromosomes or sister chromatids to separate properly during meiotic cell division.
Nondisjunction specifically describes the failure of homologous chromosomes to separate during Anaphase I or sister chromatids to separate during Anaphase II of meiosis, resulting in daughter cells with missing or extra chromosomes.

Step-by-Step Solution

1
Identify the chromosomal process described in the stem.
The stem describes an error during meiosis where chromosomes or chromatids fail to separate during Anaphase I or Anaphase II.
Accurate biological terminology distinguishes structural chromosome changes from numerical errors caused by separation failures.
2
Differentiate between structural chromosomal aberrations, normal meiotic events, and nuclear division errors.
Translocation is a structural rearrangement, crossing over is normal genetic recombination, and polyploidy involves whole-genome duplication.
Failure of specific chromosome pairs or chromatids to separate during division is termed nondisjunction, which directly causes aneuploidy.

Key Concept

Nondisjunction and Aneuploidy
Estimated Time:1m 0s
Question 150Question

Duchenne muscular dystrophy is an X-linked recessive genetic condition in humans. If a female carrier (XDXdX^D X^d) marries a male affected by the condition (XdYX^d Y), what is the probability that their first child will be an affected female?

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Answer: 25%25\%

Answer

The probability that their first child will be an affected female is 25% (or 1/4).
The correct answer is 25%. A cross between a carrier mother (XDXdX^D X^d) and an affected father (XdYX^d Y) yields four equally likely offspring genotypes: XDXdX^D X^d (carrier female), XdXdX^d X^d (affected female), XDYX^D Y (unaffected male), and XdYX^d Y (affected male). Since only 1 of the 4 possible outcomes represents an affected female (XdXdX^d X^d), the probability is 1 out of 4, or 25%.

Step-by-Step Solution

1
Determine the parental genotypes
Mother = XDXdX^D X^d (carrier female); Father = XdYX^d Y (affected male).
Duchenne muscular dystrophy is an X-linked recessive disorder, so the father must carry the recessive allele on his single X chromosome.
2
Construct a Punnett square for the cross
Gametes: Mother produces XDX^D and XdX^d; Father produces XdX^d and YY.
Offspring Genotypes: XDXdX^D X^d (carrier female), XdXdX^d X^d (affected female), XDYX^D Y (unaffected male), XdYX^d Y (affected male).
Each of the 4 combinations is equally likely (25% chance each).
3
Identify the target phenotype among all possible offspring
The target genotype for an affected female is XdXdX^d X^d, which accounts for 1 out of 4 possible total outcomes.
Probability = 14=25%\frac{1}{4} = 25\%.

Key Concept

X-linked recessive inheritance patterns and probability calculations across total offspring versus sex-specific subsets.
Question 151Question

A poultry breeder recorded two phenotypic traits in a flock of domestic chickens: comb shape (single, pea, rose, or walnut) and body weight at maturity. Which of the following statements correctly categorizes these two traits based on their pattern of variation and underlying genetic control?

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Answer: Comb shape exhibits discontinuous variation controlled by major discrete genes, whereas body weight exhibits continuous variation influenced by polygenes and environmental factors.

Answer

Comb shape exhibits discontinuous variation controlled by major discrete genes, whereas body weight exhibits continuous variation influenced by polygenes and environmental factors.
The option stating that comb shape exhibits discontinuous variation controlled by major discrete genes while body weight exhibits continuous variation influenced by polygenes and environmental factors is correct. Comb shape is a qualitative trait displaying clear-cut, non-overlapping phenotypes (discontinuous variation) under the control of specific gene loci. In contrast, body weight is a quantitative trait exhibiting a smooth gradient of intermediate phenotypes (continuous variation) governed by multiple additive genes (polygenes) interacting with nutrition and environment.

Step-by-Step Solution

1
Analyze the nature of comb shape phenotype distribution.
Comb shape presents clear, distinct, non-overlapping categories (single, pea, rose, walnut) without intermediate gradations.
Discontinuous variation is characterized by qualitative traits controlled by monogenic or oligogenic inheritance with little to no environmental effect.
2
Analyze the nature of body weight phenotype distribution.
Body weight presents a continuous spectrum of quantitative values that can be measured and plotted on a bell-shaped curve.
Continuous variation is characterized by polygenic inheritance where multiple additive genes interact alongside environmental influences.
3
Synthesize the correct classification.
Comb shape is a discontinuous trait, while body weight is a continuous trait.
This correctly pairs the phenotypic distribution pattern with its underlying genetic mechanism.

Key Concept

Continuous vs Discontinuous Variation and Genetic Control
Question 152Question

Match each phenotypic trait or population distribution profile on the left with its corresponding characteristic pattern of variation or graphical feature on the right.

Click a left item, then click its matching right item

Items

Human height and skin color
ABO blood group and tongue rolling ability
Bell-shaped frequency curve
Discrete bar chart with clear gaps

Matches

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Answer

Human height and skin color matches with Continuous variation influenced by polygenes and environmental factors. ABO blood group and tongue rolling ability matches with Discontinuous variation controlled by monogenic inheritance with minimal environmental effect. Bell-shaped frequency curve matches with Graphical representation showing a continuous spectrum of intermediate phenotypes. Discrete bar chart with clear gaps matches with Graphical representation showing non-overlapping, distinct phenotypic categories.
Human height and skin color exhibit continuous variation caused by polygenic inheritance and environmental influences. ABO blood group and tongue rolling ability show discontinuous variation controlled by single genes with minimal environmental impact. Population data for continuous variation forms a smooth bell-shaped curve showing intermediate phenotypes, whereas discontinuous variation forms a discrete bar chart with distinct non-overlapping categories.

Step-by-Step Solution

1
Identify the biological nature and genetic control of phenotypic traits
Human height and skin color exhibit gradual transitions (polygenic continuous variation), while ABO blood group and tongue rolling show clear, separate phenotypic classes (monogenic discontinuous variation).
Continuous traits show a continuum of phenotypes due to multiple genes, whereas discontinuous traits show distinct classes due to single-gene control.
2
Analyze the graphical representations associated with population data for both variation types
Continuous data produces a smooth, bell-shaped normal distribution curve, while discontinuous data forms separate bars representing distinct non-overlapping categories.
Quantitative measurements across a spectrum form a continuous curve, while qualitative discrete categories plot as separate bars.
3
Pair each left item with its corresponding genetic mechanism or graphical characteristic
All four pairs are correctly matched based on variation patterns and graphical profiles.
Ensures complete concept alignment across examples, mechanisms, and graphical representations.

Key Concept

Distinction between continuous and discontinuous variation in inheritance patterns, underlying genetic mechanisms, and graphical representations.
Question 153Question

In domestic cats, coat color is an X-linked trait where the allele for black fur (XBX^B) and the allele for orange fur (XOX^O) are codominant, with heterozygous females (XBXOX^B X^O) exhibiting a tortoiseshell phenotype. If an orange male cat (XOYX^O Y) is mated with a tortoiseshell female cat (XBXOX^B X^O), what percentage of their male offspring is expected to have orange fur?

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Answer: 50

Answer

50%
The female parent (XBXOX^B X^O) transmits her XBX^B allele or her XOX^O allele to male offspring with equal probability (50% each). Because male offspring inherit the Y chromosome from their father, half of the male offspring will inherit XOX^O and have orange fur (XOYX^O Y), giving an expected percentage of 50%.

Step-by-Step Solution

1
Determine the gametes produced by each parent
The male parent (XOYX^O Y) produces XOX^O and YY gametic types in equal proportion (1:1). The female parent (XBXOX^B X^O) produces XBX^B and XOX^O gametic types in equal proportion (1:1).
Segregation of chromosomes during meiosis separates alleles into gametes.
2
Determine the genotypes specifically for male offspring
Male offspring inherit the YY chromosome from the male parent and one XX chromosome from the female parent. The possible male genotypes are XBYX^B Y (black fur) and XOYX^O Y (orange fur).
Sex determination follows the XX-XY system where male sex is determined by inheriting the paternal Y chromosome.
3
Calculate the percentage of orange males among total male offspring
Out of 2 possible male genotypes (XBYX^B Y and XOYX^O Y), 1 is orange (XOYX^O Y). The ratio is 1 out of 2, which equals 50%.
The question asks for the proportion strictly within male offspring.

Key Concept

Sex-linked codominant inheritance and sex determination in XX-XY systems
Question 154Question

In medical practice, understanding the genetic basis of the ABO blood group system is essential for safe blood transfusions. If an individual with blood group AB (IAIBI^A I^B) requires a red blood cell transfusion, which of the following statements correctly explains their biological donor compatibility?

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Answer: They can safely receive red blood cells from any ABO blood group donor because their blood plasma contains neither anti-A nor anti-B antibodies.

Answer

Individuals with blood group AB (IAIBI^A I^B) can safely receive red blood cells from any ABO blood group donor because their plasma contains neither anti-A nor anti-B antibodies.
Individuals with blood group AB (IAIBI^A I^B) express both A and B surface antigens on their erythrocytes due to codominance between the IAI^A and IBI^B alleles. Because their immune system recognizes both antigens as self-antigens, their plasma contains neither anti-A nor anti-B antibodies. Consequently, receiving red blood cells from group A, B, AB, or O donors will not induce antibody-mediated agglutination, making them universal red blood cell recipients.

Step-by-Step Solution

1
Identify the genotype and phenotypic antigen expression of blood group AB.
Genotype IAIBI^A I^B results in the codominant expression of both A and B antigens on the surface of red blood cells.
Both alleles IAI^A and IBI^B are fully expressed simultaneously.
2
Determine the corresponding antibody composition in the blood plasma.
Because both A and B antigens are recognized as self-antigens, the plasma produces neither anti-A nor anti-B antibodies.
Producing antibodies against self-antigens would trigger autoimmune destruction of erythrocytes.
3
Assess transfusion compatibility for red blood cell donation.
The absence of anti-A and anti-B antibodies prevents immune reaction against donated red blood cells from group A, B, AB, or O donors.
Lacking these antibodies makes group AB individuals universal recipients of red blood cells.

Key Concept

ABO blood group codominance, antibody presence, and transfusion compatibility in medical genetics
Estimated Time:1m 0s
Question 155Question

Biological variations in living organisms manifest through distinct phenotypic patterns and underlying genetic mechanisms. Pair each genetic scenario or population trait feature on the left with its correct classification or descriptive feature on the right.

Click a left item, then click its matching right item

Items

Human body mass and adult height
Human Rhesus factor status and ABO blood group
Frequency curve displaying a smooth, continuous spectrum of intermediate phenotypes
Histogram featuring separate, distinct bars without intermediate forms

Matches

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Answer

Human body mass and adult height pairs with continuous variation controlled by polygenes and modified by environmental factors; Human Rhesus factor status and ABO blood group pairs with discontinuous variation controlled by monogenic inheritance independent of environmental influence; Frequency curve displaying a smooth, continuous spectrum pairs with graphical profile typical of quantitative inheritance within a population; Histogram featuring separate, distinct bars pairs with graphical profile typical of qualitative inheritance within a population.
Continuous variation involves quantitative traits such as height and body mass, governed by multiple genes (polygenes) and environmental conditions, producing a smooth distribution curve. Discontinuous variation involves qualitative traits such as blood groups, controlled by major single genes (monogenes), producing discrete, non-overlapping categories represented by separate bars on a histogram.

Step-by-Step Solution

1
Analyze polygenic traits and environmental influence.
Human body mass and height vary continuously across a spectrum and respond to environmental factors like diet.
Multiple additive genes working together with environmental conditions produce a continuous range of phenotypes.
2
Identify traits governed by single genes producing distinct categories.
Blood groups (ABO and Rhesus factor) are strictly categorical without intermediate types.
Monogenic inheritance causes discontinuous phenotypic classes that are immune to environmental modifications.
3
Correlate graph types with quantitative and qualitative variation.
A smooth curve depicts quantitative variation, while isolated bars depict qualitative variation.
Quantitative traits span a continuous spectrum forming a bell curve, whereas qualitative traits yield clear-cut distinct categories.

Key Concept

Distinction between continuous (polygenic, quantitative) and discontinuous (monogenic, qualitative) variation
Estimated Time:1m 30s
Question 156Question

During a pre-marital clinical genetic screening, a medical counselor analyzes the hemoglobin genotypes of a prospective couple. The man is homozygous normal (AAAA), while the woman is a heterozygous carrier (ASAS) for sickle-cell disease. What percentage of their offspring is expected to inherit a genotype that is completely free of the sickle-cell allele (SS)?

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Answer: 50%50\%

Answer

50%50\% of the offspring are expected to be homozygous normal (AAAA) and completely free of the sickle-cell allele.
In genetic inheritance of hemoglobin traits, a homozygous normal individual (AAAA) produces only normal alleles (AA). When crossed with a carrier (ASAS) who produces 50%50\% normal alleles (AA) and 50%50\% mutant alleles (SS), half of the resulting offspring inherit an AA allele from both parents (AAAA). Thus, 50%50\% of the offspring are completely free of the sickle-cell allele.

Step-by-Step Solution

1
Determine the parental genotypes and gametes produced.
Father (AAAA) produces only AA gametic types (100%100\%). Mother (ASAS) produces AA and SS gametes in equal proportions (50%50\% AA, 50%50\% SS).
According to Mendel's Law of Segregation, alleles segregate during gamete formation.
2
Perform a genetic cross using a Punnett square.
Combining gametes yields: 50%50\% AAAA (homozygous normal) and 50%50\% ASAS (heterozygous carrier).
Random fertilization between father's AA gametes and mother's AA or SS gametes yields equal probabilities for AAAA and ASAS genotypes.
3
Identify the proportion of offspring completely free of the sickle-cell allele (SS).
The AAAA genotype has zero copies of the SS allele, which accounts for 50%50\% of the total offspring.
Only individuals with genotype AAAA are completely free of the SS allele; ASAS individuals carry one copy of the allele.

Key Concept

Application of Mendelian genetics in medical counseling for hemoglobinopathy prevention
Estimated Time:1m 30s
Question 157Question

A chromosomal inversion alters the linear sequence of genes on a chromosome without altering the total quantity of genetic material.

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Answer: True

Answer

True. Chromosomal inversion rearranges the linear order of genes without changing the total quantity of genetic material.
Chromosomal inversion is a structural aberration in which a segment of a chromosome breaks off, flips 180180^\circ, and reinserts itself into the chromosome. Since no chromosomal material is deleted or duplicated, the total amount of genetic material is unchanged, even though gene sequence order is rearranged.

Step-by-Step Solution

1
Define the structural alteration that occurs during chromosomal inversion.
A chromosome segment breaks at two points, rotates by 180180^\circ, and reattaches into the same position on the chromosome.
Understanding the mechanism of inversion is necessary to determine if genetic content is modified in quantity.
2
Evaluate if genetic material is added or removed during this process.
Because no DNA segment is excised permanently or duplicated, the total amount of genetic material remains unchanged.
Distinguishing balanced structural changes (inversion, balanced translocation) from unbalanced ones (deletion, duplication) confirms the truth value of the statement.

Key Concept

Chromosomal Inversion and Structural Aberrations
Estimated Time:1m 0s
Question 158Question

Applications of genetics play vital roles in modern agricultural production and medical practice. Match each genetic application listed on the left with its corresponding biological purpose or outcome on the right.

Click a left item, then click its matching right item

Items

Genetic counseling
Selective breeding
DNA profiling
Recombinant DNA technology in insulin production

Matches

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Answer

Genetic counseling matches with evaluating parental genotypes to determine the risk of transmitting hereditary disorders. Selective breeding matches with selecting and mating organisms with desirable traits to enhance agricultural yield. DNA profiling matches with comparing variable regions of DNA to resolve paternity and forensic identity. Recombinant DNA technology in insulin production matches with synthesizing human hormones using genetically engineered bacteria.
Each application correctly maps to its underlying biological mechanism: genetic counseling evaluates inherited disease risks; selective breeding artificially selects desirable agricultural traits over generations; DNA profiling utilizes variable genomic markers for identification; and recombinant DNA technology uses engineered bacterial host cells to produce pharmaceutical proteins like insulin.

Step-by-Step Solution

1
Identify the primary medical application associated with evaluating hereditary risks in prospective parents.
Genetic counseling is matched to evaluating parental genotypes for inheritance risks of genetic disorders.
Prospective parents consult genetic counselors to assess phenotypic and genotypic probabilities before childbirth.
2
Identify the agricultural technique based on artificial selection of beneficial plant or animal traits.
Selective breeding is matched to selecting and mating organisms with desirable traits.
Breeders systematically select parent organisms with high yield or disease resistance to pass favorable alleles to offspring.
3
Identify the molecular technique used for individual identity verification and paternity testing.
DNA profiling is matched to comparing variable regions of DNA sequences for forensic and paternity resolution.
Every individual (except monozygotic twins) has a distinct DNA band pattern derived from variable genomic repeats.
4
Identify the biotechnological method that utilizes transgenic bacteria to synthesize therapeutic proteins.
Recombinant DNA technology in insulin production is matched to synthesizing human hormones using genetically engineered bacteria.
Inserting human insulin coding sequences into bacterial vectors enables industrial-scale synthesis of bio-identical insulin.

Key Concept

Practical utility of Mendelian principles, molecular genetics, and biotechnology in medicine and agriculture
Estimated Time:1m 30s
Question 159Question

An insertion or deletion of a single nucleotide base within the coding region of a gene alters the reading frame, changing all subsequent triplet codons during protein synthesis.

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Answer: True

Answer

The statement is true. Inserting or deleting a single nucleotide base shifts the triplet reading frame during translation, altering every codon following the mutation site.
The statement is correct because adding or removing one nucleotide alters the three-by-three reading frame of codons during protein synthesis, causing a frameshift mutation that alters all downstream amino acids.

Step-by-Step Solution

1
Examine the nature of the genetic code during translation.
Messenger RNA is read by ribosomes in consecutive, non-overlapping triplets called codons.
Each codon specifies a particular amino acid in the synthesizing polypeptide chain.
2
Analyze the impact of inserting or deleting a single nitrogenous base.
Altering the base count by one shifts the triplet alignment down the entire remaining length of the gene.
Because the ribosome continues reading in groups of three, the reading frame boundary changes at the mutation point.
3
Evaluate the outcome on the protein product.
All downstream codons code for different amino acids, or introduce an early stop codon (nonsense mutation).
This frameshift effect changes the entire primary structure of the protein from that point forward.

Key Concept

Frameshift Mutation Mechanism
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