Heredity and Variation

159 questions

Question 61Question

In garden pea plants (*Pisum sativum*), the allele for round seed shape (RR) is completely dominant over the allele for wrinkled seed shape (rr). Match each parental cross in the left column with its corresponding expected offspring phenotypic or genotypic ratio in the right column.

Click a left item, then click its matching right item

Items

Heterozygous round (RrRr) ×\times Heterozygous round (RrRr)
Heterozygous round (RrRr) ×\times Homozygous recessive wrinkled (rrrr)
Homozygous dominant round (RRRR) ×\times Homozygous recessive wrinkled (rrrr)
Homozygous dominant round (RRRR) ×\times Heterozygous round (RrRr)

Matches

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Answer

Heterozygous cross (Rr×RrRr \times Rr) yields a 3:13 : 1 phenotypic ratio (1RR:2Rr:1rr1 RR : 2 Rr : 1 rr). Monohybrid test cross (Rr×rrRr \times rr) yields a 1:11 : 1 phenotypic ratio (1Rr:1rr1 Rr : 1 rr). Pure-breeding dominant cross with recessive (RR×rrRR \times rr) yields 100%100\% heterozygous round (RrRr). Dominant homozygous cross with heterozygote (RR×RrRR \times Rr) yields 100%100\% round phenotype with a 1RR:1Rr1 RR : 1 Rr genotypic ratio.
Each parental monohybrid cross produces a characteristic distribution of alleles according to Mendel's Law of Segregation. Two heterozygous parents (Rr×RrRr \times Rr) yield 3:13 : 1 phenotypic and 1:2:11 : 2 : 1 genotypic ratios. A test cross (Rr×rrRr \times rr) yields a 1:11 : 1 ratio. A cross between pure lines (RR×rrRR \times rr) yields uniform heterozygous dominant offspring (100% Rr100\%\ Rr). A cross between homozygous dominant and heterozygous parents (RR×RrRR \times Rr) yields a 1:11 : 1 genotypic ratio of RR:RrRR : Rr, with 100%100\% exhibiting the dominant phenotype.

Step-by-Step Solution

1
Determine gametes and Punnett square for Heterozygous round (RrRr) ×\times Heterozygous round (RrRr)
Gametes: R,rR, r and R,rR, r. Offspring: 1/4 RR1/4\ RR, 2/4 Rr2/4\ Rr, 1/4 rr1/4\ rr. Phenotypic ratio is 33 round : 11 wrinkled.
Mendel's Law of Segregation states that allele pairs separate during gamete formation.
2
Determine gametes and Punnett square for Heterozygous round (RrRr) ×\times Recessive wrinkled (rrrr)
Gametes: R,rR, r and rr. Offspring: 1/2 Rr1/2\ Rr (round), 1/2 rr1/2\ rr (wrinkled). Ratio is 1:11 : 1.
This is a classic monohybrid test cross used to determine underlying genotypes.
3
Determine gametes and Punnett square for RR×rrRR \times rr
Gametes: RR and rr. All offspring are RrRr (100%100\% heterozygous round).
Homozygous parents pass only one allele type each to offspring.
4
Determine gametes and Punnett square for RR×RrRR \times Rr
Gametes: RR and R,rR, r. Offspring: 1/2 RR1/2\ RR, 1/2 Rr1/2\ Rr. All carry at least one RR allele, so 100%100\% are round.
The dominant allele RR masks the expression of rr in heterozygous conditions.

Key Concept

Mendel's First Law (Law of Segregation) and Monohybrid Inheritance Ratios
Estimated Time:2m 0s
Question 62Question

In tomato plants, the allele for red fruit (RR) is completely dominant over the allele for yellow fruit (rr). If two heterozygous red-fruited plants (RrRr) are crossed, what is the expected genotypic ratio of their offspring?

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Answer: 1:2:11 : 2 : 1

Answer

The expected genotypic ratio among the offspring is 1:2:11 : 2 : 1 (1 RR:2 Rr:1 rr1\ RR : 2\ Rr : 1\ rr).
Crossing two heterozygous parents (Rr×RrRr \times Rr) segregates alleles into gametes such that 25%25\% of offspring receive RRRR (homozygous dominant), 50%50\% receive RrRr (heterozygous), and 25%25\% receive rrrr (homozygous recessive). This results in a 1:2:11 : 2 : 1 genotypic ratio.

Step-by-Step Solution

1
Identify parental genotypes and alleles
Both parents are heterozygous (RrRr), producing gametes containing either the RR allele or the rr allele with equal probability (50%50\% RR, 50%50\% rr).
According to Mendel's Law of Segregation, alleles separate during gamete formation.
2
Construct a Punnett square for Rr×RrRr \times Rr
The possible genetic combinations are 1 RR1\ RR, 2 Rr2\ Rr, and 1 rr1\ rr.
Combining male and female gametes yields 1/4 RR1/4\ RR, 2/4 Rr2/4\ Rr, and 1/4 rr1/4\ rr.
3
Determine the genotypic ratio
The genotypic ratio is 1 RR:2 Rr:1 rr1\ RR : 2\ Rr : 1\ rr, which simplifies to 1:2:11 : 2 : 1.
Genotypic ratio accounts for actual allele combinations regardless of physical appearance.

Key Concept

Mendel's Law of Segregation and Monohybrid Genotypic Ratios
Estimated Time:45s
Question 63Question

In chickens, sex is determined by the ZZ-ZW mechanism, where males are ZZ and females are ZW. Barred plumage is controlled by a Z-linked dominant allele (ZBZ^B), whereas non-barred plumage is controlled by the recessive allele (ZbZ^b). A non-barred rooster (ZbZbZ^b Z^b) is mated with a barred hen (ZBWZ^B W). If an F1F_1 male offspring is subsequently crossed with an F1F_1 female offspring, what is the probability that a female in the F2F_2 generation will have barred feathers?

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Answer: 50% (1/2)

Answer

50% (1/2)
In the F1×F1F_1 \times F_1 cross (ZBZb×ZbWZ^B Z^b \times Z^b W), the male parent produces gametes carrying ZBZ^B and ZbZ^b in equal frequency (50%50\% each). Female offspring inherit the WW chromosome from their mother and one ZZ chromosome from their father. Therefore, 50%50\% of the female offspring receive ZBZ^B and display barred feathers (ZBWZ^B W), while 50%50\% receive ZbZ^b and display non-barred feathers (ZbWZ^b W).

Step-by-Step Solution

1
Determine the genotypes of the P1P_1 parents and F1F_1 offspring
Parental genotypes: Male = ZbZbZ^b Z^b, Female = ZBWZ^B W. F1F_1 male genotype = ZBZbZ^B Z^b (barred male), F1F_1 female genotype = ZbWZ^b W (non-barred female).
The male parent passes a ZbZ^b chromosome to all offspring; female offspring inherit the WW chromosome from the mother.
2
Perform the Punnett square cross for the F1F_1 interbreed (ZBZb×ZbWZ^B Z^b \times Z^b W)
Gametes from F1F_1 male: ZBZ^B, ZbZ^b. Gametes from F1F_1 female: ZbZ^b, WW. F2F_2 genotypes: ZBZbZ^B Z^b (barred male, 25%), ZbZbZ^b Z^b (non-barred male, 25%), ZBWZ^B W (barred female, 25%), ZbWZ^b W (non-barred female, 25%).
Constructing the cross yields all possible F2F_2 genotypic combinations.
3
Calculate the specific probability among female offspring
Female genotypes in F2F_2 are ZBWZ^B W (barred) and ZbWZ^b W (non-barred) in equal proportions. Probability of barred among females = 12=50%\frac{1}{2} = 50\%.
The question asks specifically for the probability within the female subset of offspring, not the total offspring.

Key Concept

ZZ-ZW Sex Determination and Sex-Linked Inheritance Ratios
Estimated Time:1m 30s
Question 64Question

During a cytogenetic investigation of a diploid organism, a researcher observes two distinct nucleotide sequence variations of a single gene located at identical positions on homologous chromosomes. Which genetics term specifically describes these alternative forms of a gene?

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Answer: Alleles

Answer

The term that describes alternative forms of a gene occupying identical positions on homologous chromosomes is alleles.
The correct answer identifies alleles, which are alternative versions of a gene occupying identical loci on homologous chromosomes in diploid organisms.

Step-by-Step Solution

1
Analyze the description given in the question stem
The stem describes variant molecular forms of a single gene at the same locus on homologous chromosomes.
Clear identification of gene variants versus whole structural entities is required.
2
Apply basic genetics terminology definitions
Alleles are defined as different functional or structural versions of a single gene that control contrasting expressions of a character.
This directly satisfies the genetic concept tested.

Key Concept

Alleles are alternative forms of a gene located at the same locus on homologous chromosomes.
Question 65Question

In a genetic study of human physical traits, researchers observed that certain characteristics show distinct, non-overlapping categories with no intermediate forms. Which of the following traits demonstrates this pattern of variation?

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Answer: Ability to roll the tongue

Answer

Ability to roll the tongue
The ability to roll the tongue is an example of discontinuous variation because it presents clear, discrete phenotypic categories with no intermediate phenotypes, controlled primarily by simple monogenic inheritance.

Step-by-Step Solution

1
Define discontinuous variation
Discontinuous variation refers to qualitative traits controlled by one or a few major genes that exhibit clear-cut, non-overlapping phenotypic categories without intermediate forms.
Understanding the definition helps distinguish discontinuous traits from continuous traits.
2
Evaluate the given traits against the definition
Ability to roll the tongue falls into distinct categories (individuals can either roll their tongue or cannot), whereas height, weight, and skin pigmentation exhibit a continuous range of intermediate phenotypes.
Classifying each option based on whether intermediate forms exist identifies the correct discontinuous trait.

Key Concept

Discontinuous Variation Traits
Estimated Time:45s
Question 66Question

In a monohybrid cross between two heterozygous pea plants for flower position, 75%75\% of the F1F_1 offspring are expected to display the dominant phenotype.

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Answer: True

Answer

The statement is True.
The statement is accurate because crossing two heterozygous organisms (Aa×AaAa \times Aa) in a standard monohybrid cross with complete dominance results in a 3:13:1 phenotypic ratio, representing 75%75\% dominant and 25%25\% recessive phenotypes.

Step-by-Step Solution

1
Identify the parental genotypes
Both parent plants are heterozygous (AaAa).
The stem specifies a monohybrid cross between two plants heterozygous for the trait.
2
Determine the offspring genotypic distribution using Mendel's Law of Segregation
The gametes (AA and aa) segregate to produce offspring with genotypes AAAA, AaAa, AaAa, and aaaa in a 1:2:11:2:1 ratio.
Each parent contributes one allele randomly to each offspring during fertilization.
3
Calculate the proportion of offspring expressing the dominant phenotype
Dominant phenotypes (AAAA and AaAa) constitute 33 out of 44 offspring, which equals 75%75\%.
The dominant allele (AA) completely masks the expression of the recessive allele (aa) in heterozygous genotypes.

Key Concept

Mendel's Law of Segregation and Monohybrid Cross Ratios
Question 67Question

In humans, hemophilia is an X-linked recessive disorder. If a carrier woman (XHXhX^H X^h) marries a hemophilic man (XhYX^h Y), what is the probability that any daughter born to this couple will be a carrier of hemophilia?

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Answer: 50%

Answer

50%
For female offspring (XXXX), the father always contributes an XhX^h chromosome. The mother contributes either an XHX^H chromosome (50% chance) or an XhX^h chromosome (50% chance). Thus, 50% of the daughters will have the heterozygous genotype XHXhX^H X^h, making them carriers.

Step-by-Step Solution

1
Determine parental genotypes and gamete types
Mother is XHXhX^H X^h (gametes: XH,XhX^H, X^h). Father is XhYX^h Y (gametes: Xh,YX^h, Y).
Identifying parental gametes is necessary to cross all potential allele combinations.
2
Determine genotypes of female offspring (XXXX)
Female offspring inherit XhX^h from the father and either XHX^H or XhX^h from the mother, resulting in XHXhX^H X^h (carrier female) and XhXhX^h X^h (affected female).
Female children always receive one XX chromosome from each parent.
3
Calculate the probability specific to female offspring
Out of 2 possible female genotypes (XHXhX^H X^h and XhXhX^h X^h), exactly 1 is a carrier (XHXhX^H X^h), giving a probability of 12\frac{1}{2} or 50%.
The question restricts the probability domain specifically to daughters.

Key Concept

Sex-linked inheritance and gender-specific probability calculations
Estimated Time:1m 30s
Question 68Question

In rabbits (*Oryctolagus cuniculus*), black fur color (BB) is dominant over brown fur color (bb), and short hair (SS) is dominant over long hair (ss). If a heterozygous black, short-haired rabbit (BbSsBbSs) is crossed with a homozygous recessive brown, long-haired rabbit (bbssbbss), what proportion of the offspring is expected to possess brown fur and short hair?

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Answer: 14\frac{1}{4}

Answer

The expected proportion of offspring with brown fur and short hair is 14\frac{1}{4} (or 25%25\%).
A dihybrid test cross involves crossing a doubly heterozygous individual (BbSsBbSs) with a doubly homozygous recessive individual (bbssbbss). The heterozygous parent produces four distinct gamete combinations (BSBS, BsBs, bSbS, and bsbs) with equal probability (14\frac{1}{4} each). Combining these with the single gamete type (bsbs) from the recessive parent yields four phenotypic classes in a 1:1:1:11:1:1:1 ratio: Black/Short (BbSsBbSs), Black/Long (BbssBbss), Brown/Short (bbSsbbSs), and Brown/Long (bbssbbss). Therefore, the proportion of offspring exhibiting brown fur and short hair (bbSsbbSs) is 14\frac{1}{4} (or 25%25\%).

Step-by-Step Solution

1
Determine gamete genotypes produced by each parent
The heterozygous parent (BbSsBbSs) produces four types of gametes (BSBS, BsBs, bSbS, bsbs) in equal proportions (14\frac{1}{4} each). The homozygous recessive parent (bbssbbss) produces only one type of gamete (bsbs).
According to Mendel's Law of Independent Assortment, alleles for different traits segregate independently into gametes.
2
Perform the dihybrid test cross (BbSs×bbssBbSs \times bbss)
Offspring genotypes formed are: 14 BbSs\frac{1}{4}\ BbSs, 14 Bbss\frac{1}{4}\ Bbss, 14 bbSs\frac{1}{4}\ bbSs, and 14 bbss\frac{1}{4}\ bbss.
Combine each gamete from the heterozygous parent with the single gamete type (bsbs) from the recessive parent.
3
Identify the target phenotype and calculate its proportion
Brown fur and short hair corresponds to the bbSsbbSs genotype, which has a frequency of 14\frac{1}{4}.
Brown fur requires homozygous recessive alleles (bbbb), and short hair requires at least one dominant allele (SS).

Key Concept

Dihybrid Test Cross Ratio
Question 69Question

In monohybrid crosses obeying Mendel's First Law (Law of Segregation) with complete dominance, match each parental genotype combination on the left with its corresponding expected offspring ratio on the right.

Click a left item, then click its matching right item

Items

Heterozygous ×\times Heterozygous (Aa×AaAa \times Aa)
Heterozygous ×\times Homozygous recessive (Aa×aaAa \times aa)
Homozygous dominant ×\times Homozygous recessive (AA×aaAA \times aa)
Homozygous dominant ×\times Heterozygous (AA×AaAA \times Aa)

Matches

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Answer

The correct pairings are: Aa×AaAa \times Aa matches with a phenotypic ratio of 3:13 : 1; Aa×aaAa \times aa matches with a phenotypic ratio of 1:11 : 1; AA×aaAA \times aa matches with 100%100\% heterozygous offspring (AaAa); and AA×AaAA \times Aa matches with 100%100\% dominant phenotype and a 1:11 : 1 genotypic ratio (AA:AaAA : Aa).
Each monohybrid cross pair follows Mendel's First Law, where allele segregation determines genetic combinations. A heterozygous cross (Aa×AaAa \times Aa) segregates to produce a 3:13 : 1 dominant-to-recessive phenotypic ratio. A test cross (Aa×aaAa \times aa) produces a 1:11 : 1 phenotypic ratio. A pure-line cross (AA×aaAA \times aa) results in 100%100\% heterozygous AaAa offspring. A cross of AA×AaAA \times Aa produces 100%100\% dominant phenotype offspring with a 1:11 : 1 genotypic ratio of AA:AaAA : Aa.

Step-by-Step Solution

1
Determine the gametes and offpsring genotypes for Aa×AaAa \times Aa
Gametes: AA and aa from each parent. Offspring genotypes: 1AA:2Aa:1aa1\,AA : 2\,Aa : 1\,aa. Under complete dominance, 33 express dominant phenotype and 11 expresses recessive phenotype (3:13 : 1 phenotypic ratio).
Mendel's Law of Segregation states that paired alleles separate during gamete formation so each gamete carries only one allele.
2
Determine the outcome of the test cross Aa×aaAa \times aa
Heterozygous parent produces AA and aa gametes; homozygous recessive parent produces only aa gametes. Offspring are 50%Aa50\%\,Aa (dominant) and 50%aa50\%\,aa (recessive), giving a 1:11 : 1 ratio.
Test crosses determine the genotype of an organism displaying the dominant phenotype by crossing it with a homozygous recessive individual.
3
Determine the outcome of crossing true-breeding parents AA×aaAA \times aa
Homozygous dominant parent contributes AA to all gametes, and homozygous recessive parent contributes aa. All F1 offspring are AaAa (100%100\% heterozygous) and show the dominant phenotype.
True-breeding cross produces uniform offspring in the F1 generation.
4
Determine the outcome of crossing AA×AaAA \times Aa
Gametes AA from the first parent combine with AA or aa from the second parent to produce 50%AA50\%\,AA and 50%Aa50\%\,Aa genotypes (1:11 : 1 genotypic ratio). All (100%100\%) present the dominant phenotype.
The presence of the dominant allele AA in all offspring masks the recessive allele aa.

Key Concept

Mendel's First Law of Segregation and Monohybrid Cross Ratios
Question 70Question

A biological study recorded a specific phenotypic trait across a large population of organisms. When the collected measurements were plotted on a frequency graph, the data formed a continuous, bell-shaped normal distribution curve. Which of the following traits was most likely being evaluated in this study?

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Answer: Grain yield per plant in a wheat population

Answer

Grain yield per plant in a wheat population
Traits showing continuous variation, such as grain yield per plant, are polygenic and influenced by environmental conditions. When plotted on a frequency graph across a large population, they generate a smooth, symmetrical bell-shaped curve representing a continuous spectrum of intermediate values.

Step-by-Step Solution

1
Analyze the graphical distribution pattern described in the stem
A smooth, bell-shaped curve indicates a continuous spectrum of phenotypic values across a population.
Continuous variation produces a normal distribution curve because the trait is quantitative, polygenic, and influenced by environmental conditions.
2
Compare the options based on continuous versus discontinuous variation characteristics
Grain yield displays a continuous range of quantitative values, whereas blood groups, horn presence, and Rhesus factor fall into distinct, non-overlapping categories.
Traits under polygenic control show continuous variation, while single-gene traits show discontinuous variation.

Key Concept

Continuous variation involves quantitative traits showing an unbroken spectrum of intermediate phenotypes, producing a bell-shaped normal distribution curve.
Estimated Time:1m 0s
Question 71Question

In humans, Duchenne muscular dystrophy is inherited as an X-linked recessive disorder (XdX^d), while the normal allele is dominant (XDX^D). A phenotypically normal woman seeks genetic counseling. Her maternal grandfather had Duchenne muscular dystrophy, whereas her maternal grandmother was homozygous normal. Her father is phenotypically normal. If this woman marries a phenotypically normal man, what is the probability (expressed as a percentage) that their first male child will be affected by the disorder?

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Answer: 25

Answer

The probability that their first male child will be affected by Duchenne muscular dystrophy is 25%.
The maternal grandfather (XdYX^d Y) passes his XdX^d chromosome to his daughter (the woman's mother), making her an obligate carrier (XDXdX^D X^d). When this carrier mother has a daughter with a normal male (XDYX^D Y), the daughter has a 50%50\% (0.50.5) chance of being a carrier (XDXdX^D X^d). If the woman is a carrier, any male child she has has a 50%50\% (0.50.5) chance of receiving the XdX^d allele and being affected. Multiplying these independent probabilities (0.5×0.50.5 \times 0.5) yields 0.250.25, or 25%25\%.

Step-by-Step Solution

1
Determine the genotype of the woman's mother from her maternal grandparents.
The woman's mother inherited XdX^d from her father (XdYX^d Y) and XDX^D from her mother (XDXDX^D X^D), making her an obligate carrier (XDXdX^D X^d).
Fathers always pass their single X chromosome to their daughters.
2
Calculate the probability that the woman inherited the recessive allele from her mother.
Probability that the woman is a carrier (XDXdX^D X^d) is 0.50.5 (or 50%50\%).
A carrier mother (XDXdX^D X^d) and normal father (XDYX^D Y) have a 50%50\% chance of producing a carrier daughter.
3
Calculate the probability that a male child of a carrier woman receives the recessive X-linked allele.
If the woman is a carrier, the probability of an affected son (XdYX^d Y) is 0.50.5 (or 50%50\%).
A male child receives his only X chromosome from his mother.
4
Multiply the independent probabilities to find the overall risk for the first male child.
P(Affected male child)=0.5 (mother is carrier)×0.5 (son inherits Xd)=0.25=25%P(\text{Affected male child}) = 0.5 \text{ (mother is carrier)} \times 0.5 \text{ (son inherits } X^d) = 0.25 = 25\%.
Both independent events (mother being a carrier and son inheriting the mutated allele) must occur.

Key Concept

Sex-Linked Recessive Inheritance and Pedigree Carrier Probability
Question 72Question

Which of the following characteristics is a defining feature of discontinuous variation in a biological population?

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Answer: The presence of clear-cut, distinct phenotypic categories with no intermediate forms

Answer

The presence of clear-cut, distinct phenotypic categories with no intermediate forms
Discontinuous variation produces non-overlapping phenotypic classes with distinct differences and no intermediate stages between them. Examples include human ABO blood groups, ability to roll the tongue, and presence or absence of horns in cattle.

Step-by-Step Solution

1
Define discontinuous variation in genetic terms.
Discontinuous variation refers to phenotypic traits that are split into distinct, non-overlapping classes.
Understanding the core definition differentiates it from continuous variation.
2
Evaluate the option choices against key features of discontinuous variation.
Discontinuous traits are typically monogenic (controlled by one gene), unaffected by environment, and show discrete phenotypic categories without intermediate states.
Traits like blood groups or sex determination are either present or absent in fixed categories.

Key Concept

Discontinuous Variation Features
Question 73Question

In maize (*Zea mays*), purple endosperm color (PP) is dominant over yellow endosperm color (pp), and starchy kernel texture (SS) is dominant over waxy kernel texture (ss). A plant heterozygous for both traits (PpSsPpSs) is allowed to self-pollinate, producing 480 kernels in the F2F_2 generation. How many of these kernels are expected to display the purple endosperm and waxy kernel texture phenotype?

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Answer: 90

Answer

90 kernels are expected to show the purple endosperm and waxy kernel texture phenotype.
In a dihybrid cross between two heterozygous individuals (PpSs×PpSsPpSs \times PpSs), the offspring phenotypes segregate in a classical 9:3:3:1 ratio. The phenotype representing one dominant trait and one recessive trait (purple endosperm and waxy texture, P_ssP\_ss) occurs with a frequency of 316\frac{3}{16}. Out of 480 kernels, 316×480=90\frac{3}{16} \times 480 = 90 kernels will manifest this specific phenotype.

Step-by-Step Solution

1
Determine parental cross and gametes
The self-pollinated parent is PpSsPpSs. Gametes formed are PSPS, PsPs, pSpS, and psps in equal proportions.
Mendel's Law of Independent Assortment states that alleles of different genes segregate independently during gamete formation.
2
Calculate the expected phenotypic fraction for purple and waxy kernels
Probability of purple (P_P\_) = 34\frac{3}{4}; Probability of waxy (ssss) = 14\frac{1}{4}. Combined probability = 34×14=316\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}.
Because the two genes assort independently, the joint probability is the product of their individual probabilities.
3
Calculate expected count from total offspring
Expected count = 316×480=90\frac{3}{16} \times 480 = 90.
Multiply the expected phenotypic frequency by the total kernel population size.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid F2F_2 Phenotypic Calculations
Question 74Question

In humans, red-green color blindness is an X-linked recessive trait (XcX^c), whereas normal vision is controlled by the dominant allele (XCX^C). Albinism is an autosomal recessive disorder (aa), whereas normal skin pigmentation is controlled by the dominant allele (AA). A woman with normal vision and normal skin pigmentation, whose father was both color-blind and albino, marries a man with normal vision who is a carrier for albinism. If this couple produces a male child (son), what is the percentage probability that the son will be both color-blind and albino?

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Answer: 12.5

Answer

The percentage probability that a son born to this couple will be both color-blind and albino is 12.5%.
The mother's father was albino (aaaa) and color-blind (XcYX^c Y), meaning she inherited aa and XcX^c from him. Given her normal phenotype, her genotype is AaXCXcAa X^C X^c. The father is AaXCYAa X^C Y. When determining traits for a son, the son receives the YY chromosome from the father, so his vision phenotype depends entirely on which XX chromosome he receives from his mother (50% chance of XcX^c). The probability of being albino from two carrier parents (Aa×AaAa \times Aa) is 25% (14\frac{1}{4}). Multiplying these independent probabilities yields 12×14=18\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}, which equals 12.5%.

Step-by-Step Solution

1
Determine parental genotypes from the pedigree information provided.
Mother's genotype: AaXCXcAa X^C X^c; Father's genotype: AaXCYAa X^C Y.
The mother received recessive alleles aa and XcX^c from her affected father (aaXcYaa X^c Y). The father is stated to have normal vision (XCYX^C Y) and to be a carrier for albinism (AaAa).
2
Calculate the probability of the male child inheriting the X-linked color blindness trait.
Probability of color-blind son = 12\frac{1}{2} (50%).
For male offspring, sex is fixed by inheriting the YY chromosome from the father. The mother has a 50% chance of passing her XcX^c allele.
3
Calculate the probability of the child inheriting autosomal albinism.
Probability of albino phenotype (aaaa) = 14\frac{1}{4} (25%).
Crossing two heterozygous carriers (Aa×AaAa \times Aa) yields a 1 in 4 chance of an autosomal recessive aaaa offspring.
4
Apply the product rule for independent genetic events.
Combined probability = 12×14=18=12.5%\frac{1}{2} \times \frac{1}{4} = \frac{1}{8} = 12.5\%.
Autosomal inheritance and X-linked inheritance are independent genetic events, so their probabilities are multiplied.

Key Concept

Independent assortment of an autosomal recessive trait and an X-linked recessive trait in human pedigree analysis.
Question 75Question

A study conducted on a human population recorded two phenotypic traits: resting systolic blood pressure (Trait I) and the presence or absence of the Rhesus D antigen on red blood cells (Trait II). Trait I produced a smooth, bell-shaped frequency distribution curve across a continuous gradient of values, whereas Trait II resulted in two distinct, non-overlapping categories with no intermediate forms. Which of the following correctly explains the genetic mechanism and environmental susceptibility responsible for these observed patterns of variation?

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Answer: Trait I is polygenic and modified by environmental factors, whereas Trait II is controlled by a single gene locus and unaffected by environmental conditions.

Answer

Trait I represents continuous variation, which is polygenic (controlled by multiple genes) and influenced by environmental factors, whereas Trait II represents discontinuous variation, which is monogenic (controlled by one gene pair or single locus) and independent of environmental changes.
The option stating that Trait I is polygenic and modified by environmental factors, whereas Trait II is controlled by a single gene locus and unaffected by environment is correct. Continuous traits like blood pressure are polygenic (controlled by multiple genes whose effects combine additively) and show significant variation due to external environmental factors like diet and stress, resulting in a continuous bell-shaped curve. In contrast, discontinuous traits like the Rhesus factor are monogenic (determined by alleles at a single locus), resulting in distinct, clear-cut phenotypic classes that environment cannot alter.

Step-by-Step Solution

1
Analyze the phenotypic distribution of Trait I (systolic blood pressure).
The smooth, bell-shaped normal distribution curve indicates continuous variation with quantitative grading between extremes.
Continuous variation occurs when traits are polygenic (governed by multiple additive genes) and sensitive to environmental influences.
2
Analyze the phenotypic distribution of Trait II (Rhesus antigen presence/absence).
Two distinct, non-overlapping categories with no intermediate forms indicate discontinuous variation.
Discontinuous variation is driven by monogenic inheritance (a single gene pair or major locus) and is largely unaffected by environmental conditions.
3
Synthesize the genetic mechanisms and environmental impacts for both traits.
Trait I is polygenic and environmentally modified, while Trait II is monogenic and environmentally stable.
Matching phenotypic distribution patterns (bell-shaped vs discrete categories) directly to their genetic architecture resolves the question.

Key Concept

Polygenic inheritance causing continuous variation versus monogenic inheritance causing discontinuous variation
Question 76Question

In classic Mendelian genetics, a testcross is carried out by mating an organism exhibiting a dominant phenotype with an individual that is homozygous dominant for the trait in question.

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Answer: False

Answer

The statement is False because a testcross requires mating with a homozygous recessive individual, not a homozygous dominant one.
The statement is false because a testcross relies on a homozygous recessive tester (aaaa) so that the offspring phenotypes directly reveal the gametic contribution and genotype of the dominant parent.

Step-by-Step Solution

1
Identify the biological purpose of a testcross.
A testcross determines whether an organism expressing a dominant trait is homozygous dominant (AAAA) or heterozygous (AaAa).
Organisms with genotypes AAAA and AaAa are phenotypically identical.
2
Determine the required genotype of the tester parent.
The tester individual must be homozygous recessive (aaaa).
A homozygous recessive tester produces only recessive gametes (aa), allowing hidden recessive alleles from the tested parent to be expressed in the offspring's phenotype.
3
Evaluate the effect of using a homozygous dominant tester (AAAA).
Crossing either AAAA or AaAa with AAAA yields 100% dominant phenotype offspring.
The dominant allele from the AAAA tester masks any recessive allele contributed by a heterozygous parent.

Key Concept

Definition and methodology of a genetic testcross
Estimated Time:1m 0s
Question 77Question

A plant possesses two different alleles (TT and tt) for the gene controlling stem height. Which of the following statements correctly identifies the genetic condition, genotype, and phenotype of this plant?

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Answer: The plant is heterozygous with a genotype of TtTt and displays a tall phenotype.

Answer

The plant is heterozygous with a genotype of TtTt and displays a tall phenotype.
Having two different alleles (TT and tt) defines the heterozygous condition. The exact combination of alleles (TtTt) represents the genotype, while the expressed physical characteristic (tall) represents the phenotype under complete dominance.

Step-by-Step Solution

1
Identify the allele composition of the plant.
The plant possesses two non-identical alleles (TT for tallness and tt for dwarfness).
An individual carrying two different alleles at a specific gene locus is defined as heterozygous.
2
Determine the genotype.
The genotype is expressed as TtTt.
Genotype refers to the specific genetic makeup or combination of alleles of an organism.
3
Determine the phenotype under complete dominance.
The phenotype is tall.
Phenotype refers to the observable physical trait. Because the dominant allele (TT) completely masks the expression of the recessive allele (tt), the expressed physical appearance is tall.

Key Concept

Heterozygous genotype versus physical phenotype in complete dominance
Estimated Time:1m 0s
Question 78Question

During a biology practical session, students recorded various inherited characteristics in their peer group. Which of the following traits demonstrates discontinuous variation?

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Answer: ABO blood group

Answer

ABO blood group
ABO blood group exhibits discontinuous variation because individuals fall into clear-cut, distinct phenotypic categories (A, B, AB, and O) without intermediate phenotypes, and the trait is strictly inherited without environmental modification.

Step-by-Step Solution

1
Identify the distinguishing characteristic of discontinuous variation.
Discontinuous variation produces distinct, non-overlapping phenotypic categories without intermediate forms, usually under monogenic control.
Qualitative traits are determined by major alleles at one or very few gene loci.
2
Evaluate the phenotypic distribution of each listed trait.
Height, body weight, and skin color display continuous gradations across a population spectrum. In contrast, ABO blood group classifies individuals strictly into discrete groups (A, B, AB, or O).
Blood groups show clear-cut phenotypic distinction unaffected by environmental factors.

Key Concept

Discontinuous phenotypic variation in human traits
Estimated Time:45s
Question 79Question

Match each sex determination mechanism or sex-linked inheritance phenomenon on the left with its corresponding biological characteristic or inheritance pattern on the right.

Click a left item, then click its matching right item

Items

XX-XO sex-determination system in grasshoppers (*Melanoplus* species)
ZZ-ZW sex-determination system in birds (*Gallus gallus*)
X-linked recessive phenotypic expression in Turner syndrome females (45,X45, X)
Holandric (Y-linked) trait inheritance in humans

Matches

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Answer

XX-XO in grasshoppers matches males being heterogametic (X0X0) and females homogametic (XXXX); ZZ-ZW in birds matches females being heterogametic (ZWZW) and males homogametic (ZZZZ); X-linked recessive expression in Turner syndrome females matches hemizygosity due to monosomy X; Holandric inheritance matches exclusive father-to-son transmission without female carriers.
Each mechanism accurately corresponds to its defining chromosomal configuration or inheritance pattern: XX-XO grasshoppers have X0X0 heterogametic males; ZZ-ZW birds have ZWZW heterogametic females; Turner syndrome females are hemizygous (45,X45, X) expressing X-linked recessives directly; holandric Y-linked traits transmit exclusively from fathers to sons.

Step-by-Step Solution

1
Analyze the sex determination system in grasshoppers (XX-XO).
Identify that grasshopper females are XXXX (homogametic) and males are X0X0 (heterogametic).
The absence of a Y chromosome means males have 23 chromosomes (22+X022 + X0) while females have 24 (22+XX22 + XX).
2
Analyze the sex determination system in birds (ZZ-ZW).
Identify that female birds are ZWZW (heterogametic) and male birds are ZZZZ (homogametic).
This reverses the male heterogametic pattern seen in mammals.
3
Evaluate the genetic condition of Turner syndrome females (45,X45, X) regarding X-linked traits.
Determine that monosomy X creates a hemizygous state in females.
Without a second X chromosome to mask a recessive allele, a single X-linked recessive allele is expressed phenotypically.
4
Evaluate holandric (Y-linked) inheritance in humans.
Determine that Y-linked genes pass strictly from male parent to male offspring.
Females do not inherit a Y chromosome and therefore cannot carry or pass on holandric traits.

Key Concept

Chromosomal mechanisms of sex determination (XX-XY, XX-XO, ZZ-ZW) and hemizygous expression of sex-linked genes.
Estimated Time:2m 0s
Question 80Question

In human ABO blood group inheritance, an individual carrying both the IAI^A and IBI^B alleles expresses both A and B antigens on the surface of their red blood cells, resulting in blood type AB. Which genetic phenomenon is directly demonstrated by the simultaneous, full expression of both alleles in the heterozygous state?

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Answer: Codominance

Answer

Codominance
The correct answer is codominance because both the IAI^A and IBI^B alleles contribute fully and independently to the phenotype. Heterozygous individuals (IAIBI^A I^B) produce both A and B functional agglutinogens (antigens) on their red blood cell surfaces without blending or masking.

Step-by-Step Solution

1
Analyze the expression of alleles IAI^A and IBI^B in a heterozygous individual (IAIBI^A I^B).
Both antigen A and antigen B are independently produced and present on the erythrocyte membrane.
Neither allele masks the other, nor do they blend to form an intermediate antigen structure.
2
Relate this joint phenotypic expression to standard non-Mendelian genetic definitions.
The simultaneous full expression of two different alleles at a locus is defined as codominance.
This contrasts with complete dominance (where one allele masks another) and incomplete dominance (where a intermediate phenotype is formed).

Key Concept

Codominance in human blood groups
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