Heredity and Variation

159 questions

Question 41Question

Which of the following human traits represents a physiological variation rather than a morphological variation?

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Answer: Ability to taste phenylthiocarbamide (PTC)

Answer

Ability to taste phenylthiocarbamide (PTC)
The ability to taste phenylthiocarbamide (PTC) depends on functional chemical receptors on the tongue, which makes it a physiological variation.

Step-by-Step Solution

1
Distinguish between morphological and physiological variations in humans.
Morphological variations involve external appearance, anatomical structures, and physical form. Physiological variations involve body functions, biochemical processes, and cellular mechanisms.
Separating physical structure from biological function is necessary to classify human traits accurately.
2
Classify the given options based on structure versus function.
Fingerprint patterns, nose shape, and earlobe structure are physical body features (morphological). The ability to taste PTC is a sensory functional response based on receptor chemistry (physiological).
Identifying the trait driven by functional biochemistry pinpoints the physiological variation.

Key Concept

Classification of human variations into morphological (structural/physical form) and physiological (functional/biochemical process) categories.
Question 42Question

An allele whose phenotype is expressed in an organism only when two copies of it are present, and whose effect is masked in the presence of a contrasting allele, is best described as which of the following?

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Answer: A recessive allele

Answer

A recessive allele
The correct option is 'A recessive allele' because recessive traits require two identical recessive alleles (homozygous state) to be observed phenotypically. In a heterozygous individual, the dominant allele overrides and masks the recessive trait.

Step-by-Step Solution

1
Analyze the definition given in the stem.
The stem describes an allele that requires two copies (homozygous condition) to manifest phenotypically and is masked when paired with a different allele.
By definition in classical Mendelian genetics, an allele that is concealed in the heterozygous condition is recessive.
2
Evaluate the choices against basic genetics terminology.
A recessive allele fits this exact criterion, whereas a dominant allele masks others and a codominant allele expresses alongside others.
Distinguishing between dominant, recessive, and codominant expression is fundamental to genetic analysis.

Key Concept

Recessive Allele Expression
Estimated Time:45s
Question 43Question

In a diploid eukaryotic organism, alternative forms of a single gene (alleles) controlling a specific trait occupy different loci on non-homologous chromosomes.

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Answer: False

Answer

The statement is False. Alleles of a given gene occupy identical loci on homologous chromosomes, rather than different loci on non-homologous chromosomes.
The statement is false because homologous chromosomes carry alleles controlling the exact same character at identical loci. Non-homologous chromosomes contain non-allelic genes governing unrelated traits.

Step-by-Step Solution

1
Define the terms allele, locus, and homologous chromosomes.
Alleles are variant forms of a gene. A locus is the specific physical location of a gene on a chromosome. Homologous chromosomes are matching pairs that carry the same genes in the same sequence.
Clear definitions are needed to determine chromosome and locus arrangements.
2
Determine where alleles of the same gene reside in diploid cells.
In diploid cells, one allele is inherited from each parent, placing them at the exact same relative locus on each member of a homologous chromosome pair.
Non-homologous chromosomes carry completely different sets of genes, not alternative alleles of the same gene.
3
Evaluate the validity of the statement.
Since the statement asserts that alleles occupy different loci on non-homologous chromosomes, it contradicts fundamental principles of chromosome structure.
Concluding that the statement is false.

Key Concept

Gene Locus and Homologous Chromosomes
Question 44Question

In sorghum plants, tall stem height (TT) is completely dominant over dwarf stem height (tt). In a monohybrid cross between two heterozygous tall plants (Tt×TtTt \times Tt), the conditional probability that a plant displaying the tall phenotype is homozygous dominant (TTTT) is 1/31/3.

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Answer: True

Answer

The statement is true because restricting the sample space to offspring displaying the dominant tall phenotype (TTTT and TtTt) yields 3 out of 4 total genotypic combinations, of which exactly 1 is homozygous dominant (TTTT), giving a conditional probability of 1/31/3.
The statement is correct because out of the 3 possible genotype combinations that produce the dominant tall phenotype (1 TT1\ TT and 2 Tt2\ Tt), exactly 1 is homozygous dominant (TTTT), making the conditional probability 1/31/3.

Step-by-Step Solution

1
Determine the genotypic ratio resulting from a monohybrid cross of two heterozygotes (Tt×TtTt \times Tt).
The genotypic ratio is 1 TT:2 Tt:1 tt1\ TT : 2\ Tt : 1\ tt.
According to Mendel's Law of Segregation, alleles segregate independently during gamete formation, resulting in a 1:2:11:2:1 genotypic ratio.
2
Identify the genotypes that express the dominant tall phenotype.
Tall plants possess either the homozygous dominant genotype (TTTT) or the heterozygous genotype (TtTt), totaling 33 out of 44 genotypic units (1 TT+2 Tt1\ TT + 2\ Tt).
Dominance dictates that both TTTT and TtTt produce the tall phenotype.
3
Calculate the conditional probability that a tall plant is homozygous dominant (TTTT).
The probability is Number of TT combinationsTotal number of tall combinations=11+2=13\frac{\text{Number of } TT \text{ combinations}}{\text{Total number of tall combinations}} = \frac{1}{1 + 2} = \frac{1}{3}.
The question restricts the sample space to only tall offspring.

Key Concept

Conditional probability in monohybrid crosses and Mendel's Law of Segregation
Question 45Question

In pea plants, the allele for green pod color (GG) is completely dominant over the allele for yellow pod color (gg). What is the expected genotypic ratio of the offspring when two heterozygous green-podded plants (GgGg) are crossed?

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Answer: 1:2:11 : 2 : 1

Answer

The expected genotypic ratio of the offspring is 1:2:11 : 2 : 1 (1 GG:2 Gg:1 gg1\ GG : 2\ Gg : 1\ gg).
In a monohybrid cross between two heterozygous individuals (Gg×GgGg \times Gg), the Law of Segregation dictates that gametes carry either allele GG or gg with equal probability. Combining these gametes produces offspring with genotypes 1/4 GG1/4\ GG, 2/4 Gg2/4\ Gg, and 1/4 gg1/4\ gg, resulting in a genotypic ratio of 1:2:11 : 2 : 1.

Step-by-Step Solution

1
Identify parental genotypes
Both parent plants are heterozygous green-podded (Gg×GgGg \times Gg).
The stem specifies crossing two heterozygous green-podded plants.
2
Determine gamete formation according to Mendel's Law of Segregation
Each parent produces two types of gametes: GG and gg in equal proportions (50%50\% GG, 50%50\% gg).
Alleles separate during gamete formation so each gamete carries only one allele for the trait.
3
Perform the monohybrid cross using a Punnett square
Genotype combinations are 1/4 GG1/4\ GG, 2/4 Gg2/4\ Gg, and 1/4 gg1/4\ gg.
Random fertilization yields 1 GG1\ GG, 2 Gg2\ Gg, and 1 gg1\ gg among four equal outcomes.
4
State the genotypic ratio
The genotypic ratio is 1:2:11 : 2 : 1.
The ratio compares homozygous dominant (GGGG), heterozygous (GgGg), and homozygous recessive (gggg) combinations.

Key Concept

Monohybrid Cross Genotypic Ratio
Estimated Time:45s
Question 46Question

In a monohybrid cross between two heterozygous green-pod pea plants (GgGg), Mendel's Law of Segregation predicts that 75%75\% of the F2F_2 offspring will possess the heterozygous (GgGg) genotype.

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Answer: False

Answer

False
The statement is false because a monohybrid cross of two heterozygous individuals (Gg×GgGg \times Gg) yields a genotypic ratio of 1 GG:2 Gg:1 gg1\ GG : 2\ Gg : 1\ gg. Exactly 50%50\% (22 out of 44) of the F2F_2 offspring inherit the heterozygous genotype (GgGg). The claimed 75%75\% proportion applies to the dominant phenotype, improperly interchanging genotypic composition with phenotypic expression.

Step-by-Step Solution

1
Identify parental genotypes and apply Mendel's First Law
Both parents are heterozygous (Gg×GgGg \times Gg). During gamete formation, alleles segregate so each gamete receives either GG or gg with equal probability (50%50\%).
Mendel's Law of Segregation states that allele pairs separate during gametogenesis.
2
Determine expected offspring genotypic proportions
Random combination of gametes yields 1/4 GG1/4\ GG, 2/4 Gg2/4\ Gg, and 1/4 gg1/4\ gg. The genotypic ratio is 1:2:11 : 2 : 1.
A standard Punnett square for a monohybrid cross combines GG and gg alleles from both parents.
3
Evaluate the statement's genotypic percentage claim
The heterozygous (GgGg) fraction is 2/42/4, which equals 50%50\%. The 75%75\% value corresponds to the dominant phenotype (GG+Gg=25%+50%=75%GG + Gg = 25\% + 50\% = 75\%).
The statement incorrectly equates the dominant phenotypic percentage (75%75\%) with the heterozygous genotypic percentage (50%50\%).

Key Concept

Mendel's Law of Segregation and Monohybrid Genotypic vs Phenotypic Ratios
Estimated Time:1m 30s
Question 47Question

Pair each basic genetic term listed on the left with its correct biological description on the right.

Click a left item, then click its matching right item

Items

Heterozygous
Genotype
Dominant allele
Phenotype

Matches

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Answer

Heterozygous matches with possessing two non-identical alleles; Genotype matches with the specific allele composition; Dominant allele matches with an allele that masks its contrasting form; Phenotype matches with the observable physical traits.
Each genetic term matches its corresponding definition precisely: Heterozygous indicates possessing two non-identical alleles at a locus; Genotype represents the organism's specific allele composition; Dominant allele describes an allele that masks the expression of its contrasting alternative form; and Phenotype refers to the observable physical or physiological traits.

Step-by-Step Solution

1
Define Heterozygous
Heterozygous refers to having different alleles at a given gene locus.
The prefix 'hetero-' means different, indicating non-identical alleles.
2
Define Genotype
Genotype represents the internal genetic code or allele constitution.
Genotype specifies the genetic information rather than the outward appearance.
3
Define Dominant allele
A dominant allele masks the phenotypic effect of its alternative allele.
Dominance means the trait is expressed whenever at least one copy of the allele is present.
4
Define Phenotype
Phenotype represents the observable attributes and physical characteristics.
Phenotype is the external expression resulting from genotypic and environmental factors.

Key Concept

Basic Genetics Terminology and Concepts
Estimated Time:1m 0s
Question 48Question

In genetics, an organism's traits are determined by its underlying genetic composition as well as its interaction with the environment. Which term specifically describes the observable physical, physiological, or behavioral characteristics of an organism?

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Answer: Phenotype

Answer

The observable physical and physiological characteristics of an organism are referred to as its phenotype.
The term phenotype describes all observable characteristics of an organism, such as morphology, development, biochemical properties, and behavior, which are produced by the interaction of its genotype with the environment.

Step-by-Step Solution

1
Identify the core biological concept requested by the question stem
The stem asks for the term that denotes the expressed physical or functional traits of an organism.
Genetics distinguishes between the underlying genetic sequence/alleles and the outward physical manifestation.
2
Differentiate between genotype and phenotype
Genotype represents the internal hereditary information (alleles), whereas phenotype represents the observable external attributes (e.g., height, flower color).
Selecting the correct term requires distinguishing the genetic code from its expression.

Key Concept

Phenotype vs Genotype
Estimated Time:45s
Question 49Question

In fruit flies (*Drosophila melanogaster*), the allele for normal wings (VV) is completely dominant over the allele for vestigial wings (vv). Match each parental cross listed on the left with its corresponding expected offspring phenotypic or genotypic ratio on the right.

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Items

Cross between two heterozygous normal-winged flies (Vv×VvVv \times Vv)
Test cross of a heterozygous normal-winged fly (Vv×vvVv \times vv)
Cross between a homozygous normal-winged fly and a vestigial-winged fly (VV×vvVV \times vv)
Cross between a homozygous normal-winged fly and a heterozygous fly (VV×VvVV \times Vv)

Matches

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Answer

Cross Vv×VvVv \times Vv matches 3:1 phenotypic ratio (normal:vestigial). Test cross Vv×vvVv \times vv matches 1:1 phenotypic ratio. Cross VV×vvVV \times vv matches 100% normal wings (all VvVv). Cross VV×VvVV \times Vv matches 100% normal wings (1:1 genotypic ratio VV:VvVV:Vv).
Each monohybrid cross yields specific offspring ratios according to Mendel's Law of Segregation. Crossing two heterozygotes (Vv×VvVv \times Vv) yields a 3:1 phenotypic ratio. A test cross (Vv×vvVv \times vv) yields a 1:1 phenotypic ratio. Crossing homozygous dominant with homozygous recessive (VV×vvVV \times vv) produces 100% heterozygous offspring (VvVv). Crossing homozygous dominant with a heterozygote (VV×VvVV \times Vv) yields 100% dominant phenotype with a 1:1 genotypic ratio of VVVV to VvVv.

Step-by-Step Solution

1
Analyze the cross Vv×VvVv \times Vv
Produces genotypes 1 VV:2 Vv:1 vv1\ VV : 2\ Vv : 1\ vv. Since VV is dominant, 3 parts show normal wings and 1 part shows vestigial wings (3:1 phenotypic ratio).
Mendel's Law of Segregation states that two alleles of a gene separate during gamete formation.
2
Analyze the test cross Vv×vvVv \times vv
Gametes from VvVv are VV and vv; gametes from vvvv are vv. Offspring are 50% VvVv and 50% vvvv (1:1 phenotypic ratio).
A monohybrid test cross pairs a heterozygous individual with a homozygous recessive individual.
3
Analyze the cross VV×vvVV \times vv
All offspring inherit VV from the dominant parent and vv from the recessive parent, resulting in 100% VvVv (100% normal wings).
Homozygous dominant crossed with homozygous recessive produces uniformly heterozygous F1 offspring.
4
Analyze the cross VV×VvVV \times Vv
Offspring genotypes are 50% VVVV and 50% VvVv. Because all possess at least one dominant allele VV, 100% display normal wings.
The dominant allele masks the recessive allele in heterozygous individuals.

Key Concept

Mendel's First Law and Monohybrid Inheritance Ratios
Question 50Question

In rabbits, the allele for short hair (HH) is completely dominant over the allele for long hair (hh). A pure-breeding long-haired female rabbit is crossed with a heterozygous short-haired male rabbit to produce an F1F_1 generation. If two short-haired offspring from this F1F_1 generation are subsequently mated to produce an F2F_2 generation, what is the expected ratio of homozygous dominant to heterozygous genotypes strictly among the short-haired F2F_2 individuals?

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Answer: 1:21 : 2

Answer

The expected ratio of homozygous dominant to heterozygous genotypes among the short-haired F2F_2 individuals is 1:21 : 2.
Crossing the pure-breeding long-haired female (hhhh) with the heterozygous short-haired male (HhHh) produces short-haired offspring that are all heterozygous (HhHh). Intercrossing these short-haired F1F_1 individuals (Hh×HhHh \times Hh) generates F2F_2 genotypes in the proportion 1 HH:2 Hh:1 hh1\ HH : 2\ Hh : 1\ hh. Among only the short-haired individuals (HHHH and HhHh), the proportion of homozygous dominant (HHHH) to heterozygous (HhHh) is 1:21 : 2.

Step-by-Step Solution

1
Determine the genotypes of the parental generation and the F1F_1 short-haired offspring.
Parental cross is hh×Hhhh \times Hh. Offspring genotypes are 50% Hh50\%\ Hh (short-haired) and 50% hh50\%\ hh (long-haired). Therefore, all short-haired F1F_1 individuals must be heterozygous (HhHh).
Mendel's Law of Segregation dictates that the homozygous recessive parent contributes only hh gametes while the heterozygous parent contributes HH or hh gametes.
2
Determine the genotypic distribution of the F2F_2 generation from crossing two short-haired F1F_1 rabbits (Hh×HhHh \times Hh).
The F2F_2 genotypic distribution is 1/4 HH1/4\ HH, 1/2 Hh1/2\ Hh, and 1/4 hh1/4\ hh (1 HH:2 Hh:1 hh1\ HH : 2\ Hh : 1\ hh).
Random fertilization between gametes carrying HH and hh alleles yields the standard monohybrid F2F_2 genotypic ratio.
3
Filter for the target phenotype (short-haired offspring) and calculate the ratio between HHHH and HhHh genotypes.
Short-haired F2F_2 individuals comprise 1/4 HH1/4\ HH and 2/4 Hh2/4\ Hh. Comparing these two genotypes gives 1 HH:2 Hh1\ HH : 2\ Hh, or a ratio of 1:21 : 2.
The question specifies restricting the ratio calculation strictly to the short-haired individuals, excluding the hhhh (long-haired) individuals from the denominator.

Key Concept

Conditional probability and genotypic ratio determination in monohybrid inheritance
Estimated Time:1m 30s
Question 51Question

In human genetics, the ABO blood group system is controlled by multiple alleles (IAI^A, IBI^B, and ii). Which of the following blood group phenotypes directly demonstrates codominance between alleles?

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Answer: Blood group AB

Answer

Blood group AB
Blood group AB is determined by the heterozygous genotype IAIBI^A I^B. In this condition, both alleles IAI^A and IBI^B are fully and simultaneously expressed on red blood cell membranes without masking each other, making it a classic example of codominance.

Step-by-Step Solution

1
Define codominance in contrast to complete dominance and recessiveness.
Codominance occurs when two different alleles at a locus are both fully expressed in the heterozygous condition, resulting in a phenotype that displays both traits simultaneously.
Identifying the phenotypic expression of heterozygous genotypes determines the inheritance pattern.
2
Analyze the allele interactions in the ABO blood group system.
Alleles IAI^A and IBI^B are both completely dominant over allele ii, but are codominant to each other.
Evaluating the relationship between alleles clarifies which genotype produces a codominant phenotype.
3
Determine which phenotype expresses both functional alleles simultaneously.
An individual with blood group AB has the genotype IAIBI^A I^B and produces both A and B antigens on red blood cells.
The presence of both distinct antigens demonstrates codominance.

Key Concept

Codominant Allele Expression in ABO Blood Group System
Estimated Time:45s
Question 52Question

In a dihybrid organism with the genotype AaBbAaBb, the two gene pairs assort independently during meiosis according to Mendel's Second Law. What is the expected phenotypic or genotypic ratio of the gametes produced by this individual?

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Answer: 1 AB:1 Ab:1 aB:1 ab1\ AB : 1\ Ab : 1\ aB : 1\ ab

Answer

The expected ratio of gametes produced is 1 AB : 1 Ab : 1 aB : 1 ab.
The correct answer states a 1:1:1:1 ratio of ABAB, AbAb, aBaB, and abab. A dihybrid organism heterozygous for two unlinked genes (AaBbAaBb) undergoes independent assortment during meiosis. Allele AA or aa has a 50% chance of combining with allele BB or bb, generating four distinct haploid gametes in equal proportions (25% each).

Step-by-Step Solution

1
Determine the possible allele combinations for gamete formation from genotype AaBbAaBb.
Gene AA produces alleles AA and aa; Gene BB produces alleles BB and bb.
Meiosis separates homologous chromosomes so each gamete receives one allele per gene.
2
Apply Mendel's Law of Independent Assortment to combine alleles from both genes.
The four possible gamete combinations are ABAB, AbAb, aBaB, and abab.
Alleles of unlinked genes segregate independently into gametes.
3
Calculate the probability of each gamete combination.
Each combination has a probability of 12×12=14\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}, yielding a ratio of 1:1:1:11 : 1 : 1 : 1.
Since segregation of each allele pair is equally likely, all four gamete types occur in equal frequencies.

Key Concept

Mendel's Law of Independent Assortment states that alleles of different genes segregate independently of one another during gamete formation, resulting in equal proportions of all possible haploid allele combinations.
Estimated Time:45s
Question 53Question

In garden pea plants (*Pisum sativum*), tall stem height (TT) is dominant over dwarf stem height (tt), and yellow seed color (YY) is dominant over green seed color (yy). According to Mendel's Law of Independent Assortment, how many genetically distinct types of gametes can be produced by an F1F_1 plant with the genotype TtYyTtYy?

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Answer: 4 types

Answer

4 types of gametes (TYTY, TyTy, tYtY, and tyty)
According to Mendel's Second Law (Law of Independent Assortment), alleles for different traits segregate independently during meiosis. For a dihybrid individual with genotype TtYyTtYy, the two alleles of the stem height gene (TT and tt) combine randomly with the two alleles of the seed color gene (YY and yy). Using the formula 2n2^n (where nn is the number of heterozygous gene pairs, here n=2n = 2), the number of distinct gamete types produced is 22=42^2 = 4, which are TYTY, TyTy, tYtY, and tyty.

Step-by-Step Solution

1
Identify the number of heterozygous gene pairs in the given genotype.
The genotype TtYyTtYy has two heterozygous gene pairs (n=2n = 2).
Mendel's Law of Independent Assortment states that allele pairs separate independently during gamete formation.
2
Calculate the number of possible unique gamete types using the formula 2n2^n.
22=42^2 = 4 distinct gamete types.
Where nn represents the number of heterozygous loci, 222^2 yields the total combinations.
3
List the distinct allele combinations to verify.
The four possible gametes are TYTY, TyTy, tYtY, and tyty.
Each gamete receives one allele from each gene pair independently.

Key Concept

Mendel's Law of Independent Assortment and Gamete Formation in Dihybrids
Question 54Question

In watermelon plants (*Citrullus lanatus*), solid green rind color (GG) is dominant over striped rind color (gg), and short fruit shape (RR) is dominant over long fruit shape (rr). If a watermelon plant heterozygous for both traits (GgRrGgRr) is self-pollinated and yields 800800 offspring, how many of these offspring are expected to possess solid green rind and long fruit shape?

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Answer: 150150

Answer

150 offspring exhibit the solid green rind and long fruit shape phenotype.
In a dihybrid cross between two heterozygous parents (GgRr×GgRrGgRr \times GgRr), the genes assort independently, yielding an F2F_2 phenotypic ratio of 9:3:3:19 : 3 : 3 : 1 (9 solid green short : 3 solid green long : 3 striped short : 1 striped long). The proportion of offspring showing solid green rind (G_G\_) and long fruit (rrrr) is 316\frac{3}{16}. Among 800800 offspring, the expected number is 316×800=150\frac{3}{16} \times 800 = 150.

Step-by-Step Solution

1
Determine the parental genotypes and set up the cross.
Parental cross is GgRr×GgRrGgRr \times GgRr.
Self-pollination of a dihybrid plant involves crossing two organisms heterozygous for both genes.
2
Calculate the expected phenotypic ratio for the F2F_2 generation according to Mendel's Second Law.
The standard dihybrid phenotypic ratio is 9:3:3:19 : 3 : 3 : 1.
Genes for rind color and fruit shape assort independently during gamete formation.
3
Identify the probability of the target phenotype (solid green rind, long fruit).
Target phenotype genotype is G_rrG\_rr, which corresponds to a probability of 316\frac{3}{16}.
Dominant for rind color (G_G\_) has probability 34\frac{3}{4}, and recessive for fruit shape (rrrr) has probability 14\frac{1}{4}. Thus, 34×14=316\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}.
4
Multiply the phenotype probability by the total offspring count.
316×800=150\frac{3}{16} \times 800 = 150.
Multiplying total progeny count by the expected probability yields the expected phenotype frequency.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Calculations
Question 55Question

In cattle, coat color is determined by a single gene exhibiting complete dominance, where the allele for black coat (BB) is dominant over the allele for red coat (bb). Match each monohybrid cross scenario with its correct genotypic or phenotypic outcome outcome/deduction.

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Items

Cross between a heterozygous black bull (BbBb) and a red cow (bbbb)
Cross between two heterozygous black cattle (Bb×BbBb \times Bb)
Cross between a homozygous black bull (BBBB) and a heterozygous black cow (BbBb)
Test cross of an individual with dominant phenotype producing 100%100\% black offspring

Matches

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Answer

The monohybrid cross scenarios match their expected outcomes as follows: The cross between a heterozygous black bull and a red cow yields a phenotypic ratio of 1 black : 1 red; the cross between two heterozygous black cattle yields a genotypic ratio of 1 BB : 2 Bb : 1 bb; the cross between a homozygous black bull and a heterozygous black cow produces a genotypic ratio of 1 BB : 1 Bb (100% black); and a test cross producing 100% black offspring leads to the deduction that the tested parent is homozygous dominant (BB).
Each monohybrid inheritance scenario follows Mendel's First Law (Law of Segregation). Gametes segregate cleanly during meiosis, recombining in predictable frequencies: a heterozygous backcross (Bb×bbBb \times bb) yields equal proportions of black and red progeny (1:11:1); two heterozygotes (Bb×BbBb \times Bb) yield the classic 1 BB:2 Bb:1 bb1\ BB : 2\ Bb : 1\ bb genotypic ratio (3:13:1 phenotype); a dominant homozygote crossed with a heterozygote (BB×BbBB \times Bb) yields equal proportions of BBBB and BbBb (100%100\% black); and a test cross yielding zero recessive phenotypes confirms homozygosity of the dominant parent.

Step-by-Step Solution

1
Analyze the cross Bb×bbBb \times bb
Gametes from BbBb are BB (50%50\%) and bb (50%50\%). Gametes from bbbb are all bb (100%100\%). Offspring genotypes are 50% Bb50\%\ Bb (black coat) and 50% bb50\%\ bb (red coat).
Demonstrates a standard testcross ratio of 1:11 : 1 for phenotypes.
2
Analyze the cross Bb×BbBb \times Bb
Punnett square analysis gives 25% BB25\%\ BB, 50% Bb50\%\ Bb, and 25% bb25\%\ bb.
This confirms Mendel's classic monohybrid F2 genotypic ratio of 1 BB:2 Bb:1 bb1\ BB : 2\ Bb : 1\ bb and phenotypic ratio of 3:13 : 1.
3
Analyze the cross BB×BbBB \times Bb
The homozygous dominant parent provides only BB alleles. Offspring genotypes are 50% BB50\%\ BB and 50% Bb50\%\ Bb. All offspring display the black coat phenotype.
This matches the genotypic ratio 1 BB:1 Bb1\ BB : 1\ Bb with 100%100\% dominant phenotype.
4
Analyze the test cross principle for an unknown dominant phenotype
Crossing B_B\_ with bbbb gives bbbb offspring only if the parent carries a hidden bb allele. Receiving 100%100\% dominant offspring confirms the parent is BBBB.
Confirms the diagnostic utility of test crosses for establishing zygosity.

Key Concept

Mendel's Law of Segregation and Monohybrid Genotypic/Phenotypic Ratios
Estimated Time:2m 0s
Question 56Question

In guinea pigs, the allele for a rough coat (RR) is completely dominant over the allele for a smooth coat (rr). A monohybrid cross is performed between two heterozygous rough-coated parents (Rr×RrRr \times Rr). What is the expected ratio of homozygous dominant individuals to heterozygous individuals among the offspring possessing the rough coat phenotype?

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Answer: 1:21 : 2

Answer

The expected ratio of homozygous dominant individuals to heterozygous individuals among the rough-coated offspring is 1 : 2.
In a cross between two heterozygous parents (Rr×RrRr \times Rr), the expected offspring genotypic breakdown is 1RR:2Rr:1rr1\,RR : 2\,Rr : 1\,rr. The rough coat phenotype comprises both the 1RR1\,RR and 2Rr2\,Rr individuals. Comparing homozygous dominant (RRRR) to heterozygous (RrRr) within this rough phenotype class gives a ratio of 1:21 : 2.

Step-by-Step Solution

1
Determine the parental genotypes and set up the monohybrid cross.
The cross between two heterozygous parents is represented as Rr×RrRr \times Rr.
Both parents carry one dominant allele (RR) and one recessive allele (rr).
2
Determine the genotypic ratio of the F1 generation using Mendel's Law of Segregation.
The resulting offspring genotypes are 1RR:2Rr:1rr1\,RR : 2\,Rr : 1\,rr.
Gametes segregate independently during meiosis, giving a 25%25\% chance for RRRR, 50%50\% chance for RrRr, and 25%25\% chance for rrrr.
3
Identify the subset of offspring that display the rough coat phenotype.
Rough-coated offspring include genotypes RRRR (homozygous dominant) and RrRr (heterozygous), totaling 33 out of 44 expected outcomes.
Because allele RR is completely dominant, both RRRR and RrRr express the rough coat phenotype.
4
Calculate the ratio of homozygous dominant (RRRR) to heterozygous (RrRr) within the rough-coated phenotype group.
Ratio of RR:Rr=1:2RR : Rr = 1 : 2.
For every 11 homozygous dominant (RRRR) rough-coated individual, there are 22 heterozygous (RrRr) rough-coated individuals.

Key Concept

Mendel's Law of Segregation and Monohybrid Genotypic Ratios
Estimated Time:1m 0s
Question 57Question

A man with blood group AB (IAIBI^A I^B) marries a woman with blood group A who is heterozygous (IAiI^A i). What is the probability that their first child will have blood group A?

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Answer: 50%50\%

Answer

The probability that their child will have blood group A is 50%50\% (or 1/21/2).
The option stating 50% is correct because crossing parental genotypes IAIBI^A I^B and IAiI^A i yields four equally likely offspring genotypes: IAIAI^A I^A (blood group A), IAiI^A i (blood group A), IAIBI^A I^B (blood group AB), and IBiI^B i (blood group B). Combining the two genotypes that result in blood group A (25%+25%25\% + 25\%) gives a total probability of 50%50\%.

Step-by-Step Solution

1
Identify the parental genotypes and gametes produced.
Father's genotype is IAIBI^A I^B (gametes: IAI^A, IBI^B). Mother's genotype is IAiI^A i (gametes: IAI^A, ii).
Heterozygous blood group A carries the recessive allele ii, while blood group AB expresses both IAI^A and IBI^B codominantly.
2
Construct a Punnett square for the cross IAIB×IAiI^A I^B \times I^A i.
The four possible offspring genotypes are IAIAI^A I^A (25%25\%), IAiI^A i (25%25\%), IAIBI^A I^B (25%25\%), and IBiI^B i (25%25\%).
Combining male and female gametes gives all expected genetic ratios.
3
Determine the phenotypic expression for each genotype.
IAIAI^A I^A and IAiI^A i both express blood group A (25%+25%=50%25\% + 25\% = 50\%). IAIBI^A I^B expresses blood group AB (25%25\%). IBiI^B i expresses blood group B (25%25\%).
Alleles IAI^A and IBI^B are codominant with each other, and both are completely dominant over the recessive allele ii.

Key Concept

Codominance and Multiple Alleles in Human ABO Blood Groups
Question 58Question

In sweet pea plants (*Lathyrus odoratus*), purple flower color (PP) is dominant over red flower color (pp), and long pollen grain (LL) is dominant over round pollen grain (ll). If a heterozygous dihybrid plant with the genotype PpLlPpLl is self-pollinated and yields a total of 160160 F2 seeds, how many of these seeds are expected to produce plants with the double recessive phenotype of red flowers and round pollen grains?

Show answer & explanation

Answer: 10

Answer

The expected number of seeds producing plants with red flowers and round pollen grains is 10.
In a dihybrid cross involving two heterozygous parents (PpLl×PpLlPpLl \times PpLl), independent assortment results in a 9:3:3:1 phenotypic ratio in the F2 generation. The double recessive phenotype (red flowers and round pollen grains, genotype ppllppll) accounts for 1 out of 16 total offspring. Multiplying this fraction (1/16) by the total yield of 160 seeds produces an expected value of 10 seeds.

Step-by-Step Solution

1
Determine the dihybrid F2 phenotypic ratio
The phenotypic ratio for a cross between two heterozygous dihybrid parents (PpLl×PpLlPpLl \times PpLl) is 9:3:3:1.
According to Mendel's Law of Independent Assortment, the alleles for flower color and pollen shape segregate independently during gamete formation.
2
Calculate the proportion of double recessive offspring
The fraction of offspring displaying both recessive traits (red flowers and round pollen grains, genotype ppllppll) is 1/16.
Out of 16 equal Punnett square combinations, exactly 1 combination represents the homozygous double recessive phenotype.
3
Compute the expected number of double recessive seeds
(1 / 16) * 160 = 10 seeds.
Multiplying the phenotypic probability (1/16) by the total seed population (160) yields the expected quantity.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Ratios
Question 59Question

In domestic fowl (*Gallus gallus*), rose comb (RR) is dominant over single comb (rr), and feathered legs (FF) are dominant over clean legs (ff). A rooster heterozygous for both traits (RrFfRrFf) is test-crossed with a hen possessing a single comb and clean legs (rrffrrff). If a total of 400400 chicks are hatched from this cross, how many are expected to exhibit a rose comb and clean legs?

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Answer: 100100

Answer

The expected number of offspring displaying rose comb and clean legs is 100100.
A cross between a double heterozygote (RrFfRrFf) and a double recessive organism (rrffrrff) is a dihybrid test cross. According to Mendel's Law of Independent Assortment, the heterozygous parent forms four types of gametes (RFRF, RfRf, rFrF, rfrf) in equal 1:1:1:11:1:1:1 frequencies, while the recessive parent yields only rfrf gametes. This produces four phenotypic classes in equal proportions (14\frac{1}{4} each). Out of 400400 total offspring, the expected count for rose comb and clean legs (RrffRrff) is 14×400=100\frac{1}{4} \times 400 = 100.

Step-by-Step Solution

1
Determine the parental genotypes and gamete types.
The heterozygous rooster (RrFfRrFf) produces four gamete types in equal proportions: RFRF, RfRf, rFrF, and rfrf. The homozygous recessive hen (rrffrrff) produces only one gamete type: rfrf.
Mendel's Law of Independent Assortment states that alleles of different genes segregate independently into gametes during meiosis.
2
Derive the offspring genotypes and phenotypic proportions.
Combining gametes yields four distinct offspring genotypes: RrFfRrFf (rose comb, feathered legs), RrffRrff (rose comb, clean legs), rrFfrrFf (single comb, feathered legs), and rrffrrff (single comb, clean legs) in a 1:1:1:11:1:1:1 ratio.
A dihybrid test cross always generates a 1:1:1:1 phenotypic frequency among offspring.
3
Calculate the expected count for the target phenotype.
Target phenotype proportion (RrffRrff) = 14=0.25\frac{1}{4} = 0.25. Expected count = 0.25×400=1000.25 \times 400 = 100.
Multiplying the phenotypic probability by the total sample size gives the expected frequency.

Key Concept

Dihybrid Test Cross Ratio
Estimated Time:2m 0s
Question 60Question

Match each dihybrid parental cross genotype involving independently assorting genes with its corresponding expected phenotypic ratio in the offspring.

Click a left item, then click its matching right item

Items

AaBb×AaBbAaBb \times AaBb
AaBb×aabbAaBb \times aabb
AaBB×AaBBAaBB \times AaBB
AABB×aabbAABB \times aabb

Matches

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Answer

The correct pairings match AaBb×AaBbAaBb \times AaBb to 9:3:3:19:3:3:1, AaBb×aabbAaBb \times aabb to 1:1:1:11:1:1:1, AaBB×AaBBAaBB \times AaBB to 3:13:1, and AABB×aabbAABB \times aabb to 100% dominant for both traits.
Each cross results from independent assortment during meiosis. Selfing a double heterozygote (AaBb×AaBbAaBb \times AaBb) yields 9:3:3:1; testcrossing a double heterozygote (AaBb×aabbAaBb \times aabb) yields 1:1:1:1; selfing a single heterozygote with one homozygous locus (AaBB×AaBBAaBB \times AaBB) yields a 3:1 ratio; and crossing true-breeding dominant and recessive lines (AABB×aabbAABB \times aabb) produces uniform offspring exhibiting both dominant traits.

Step-by-Step Solution

1
Determine gamete types produced by each parent genotype
AaBbAaBb produces 4 gamete types (AB,Ab,aB,abAB, Ab, aB, ab), aabbaabb produces 1 type (abab), AaBBAaBB produces 2 types (AB,aBAB, aB), and AABBAABB produces 1 type (ABAB).
Mendel's Law of Independent Assortment states that gene pairs segregate independently during gamete formation.
2
Combine gametes to calculate phenotypic probabilities for each cross
Combining gametes yields the classic dihybrid phenotypic ratios for each cross type.
The phenotypic ratio depends on allele interactions and dominance relationships across both gene loci.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid Cross Ratios
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