Heredity and Variation Practice Questions — JAMB UTME — Page 2

Question 21Question →

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

True

False

Show answer & explanation

Answer: True

Answer

The statement is True.

Reciprocal translocation represents a structural chromosomal mutation where segments of non-homologous chromosomes are exchanged without altering the total chromosome count (

2n

). Non-disjunction, on the other hand, is a nuclear division error where chromosomes fail to segregate properly during Anaphase I or II of meiosis, producing gametes with missing or extra whole chromosomes, which leads to numerical aneuploidy (such as Down syndrome,

47, +21

).

Step-by-Step Solution

1

Analyze the nature of reciprocal translocation

Reciprocal translocation involves non-homologous chromosomes swapping segments. This rearranges gene loci (a structural alteration) but leaves the total count of chromosomes intact.

Structural chromosomal aberrations modify chromosome architecture rather than chromosome complement number.

2

Analyze the mechanism and outcome of meiotic non-disjunction

Non-disjunction occurs when homologous chromosomes fail to segregate in Anaphase I or sister chromatids fail to separate in Anaphase II, yielding abnormal gametes (

n+1

or

n-1

).

Failure of proper spindle separation directly alters gametic chromosome counts, causing numerical aberrations (aneuploidy) upon fertilization.

3

Evaluate the complete comparative statement

The distinction drawn between structural alteration (translocation preserving number) and numerical alteration (non-disjunction causing aneuploidy) is scientifically accurate.

Both biological concepts are correctly defined and contrasted.

Key Concept

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

Question 22Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

Substitution mutation matches replacement of a single nucleotide base; Frameshift mutation matches addition or deletion altering the reading frame; Aneuploidy matches gain or loss of specific individual chromosomes (

2n+1

or

2n-1

); Polyploidy matches possession of extra complete chromosome sets (

3n

or

4n

).

Substitution mutation corresponds to swapping a single nucleotide base for another. Frameshift mutation occurs when additions or deletions of nucleotides shift the triplet codon reading sequence. Aneuploidy describes the loss or gain of individual chromosomes, while polyploidy describes organisms having additional full sets of chromosomes.

Step-by-Step Solution

1

Distinguish between gene mutations and numerical chromosomal aberrations.

Substitution and frameshift are point gene mutations, while aneuploidy and polyploidy involve changes in chromosome number.

Gene mutations affect nucleotide sequences within a gene, whereas numerical aberrations alter chromosome counts.

2

Identify the mechanisms of point mutations.

Substitution replaces a single base, whereas frameshift adds or deletes bases shifting downstream codons.

Codons are read in triplets, so non-multiple-of-three insertions or deletions disrupt all subsequent amino acid translation.

3

Distinguish between aneuploidy and polyploidy.

Aneuploidy affects individual chromosome counts (

2n \pm 1

), whereas polyploidy involves entire extra sets of chromosomes (

3n, 4n

).

Non-disjunction of a single chromosome pair causes aneuploidy, whereas failure of spindle formation during total cell division leads to polyploidy.

Key Concept

Gene Mutations vs Chromosomal Numerical Aberrations

Question 23Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match each monohybrid inheritance scenario to its precise Mendelian ratio or outcome: (1) Heterozygous black crossed with white matches the 1:1 genotypic and phenotypic ratio; (2) Two heterozygous black parents producing 160 offspring matches 120 black and 40 white offspring (3:1 ratio); (3) A test cross producing 100% dominant offspring confirms a homozygous dominant parent (BB); and (4) Crossing pure-breeding parents yields 100% uniform heterozygous F1 offspring.

Each cross directly demonstrates a fundamental aspect of Mendel's First Law. Heterozygote

\times

homozygous recessive gives a

1:1

ratio; two heterozygotes produce a

3:1

phenotypic ratio (120:40 out of 160); a test cross yielding zero recessive offspring confirms homozygous dominance; and contrasting pure lines produce uniform

F_1

heterozygotes.

Step-by-Step Solution

1

Analyze the cross between

Bb

and

bb

Gametes: Parent 1 produces

B

and

b

in equal proportion (

1:1

), Parent 2 produces only

b

. Offspring genotypes:

\frac{1}{2} Bb

(black fur) and

\frac{1}{2} bb

(white fur).

Mendel's Law of Segregation dictates that alleles segregate during gamete formation so that each gamete carries only one allele for each gene.

2

Calculate expected offspring numbers for

Bb \times Bb

with total

N = 160

Monohybrid phenotypic ratio is

3

dominant :

1

recessive. Dominant count

= \frac{3}{4} \times 160 = 120

; Recessive count

= \frac{1}{4} \times 160 = 40

.

A monohybrid cross of two heterozygotes generates a genotypic ratio of

1 BB : 2 Bb : 1 bb

, which simplifies to a

3:1

phenotypic ratio under complete dominance.

3

Evaluate the test cross of an unknown black parent with a recessive

bb

parent

If the parent were

Bb

, white offspring would appear in approximately

50\%

of cases. Since

100\%

of offspring are black, the unknown parent must be homozygous dominant (

BB

).

A test cross uses a known homozygous recessive individual to reveal whether an organism expressing a dominant trait is homozygous or heterozygous.

4

Determine the outcome of crossing homozygous contrasting parents (

BB \times bb

)

All

F_1

offspring receive

B

from the black parent and

b

from the white parent, making

100\%

of the

F_1

generation heterozygous (

Bb

).

This illustrates Mendel's principle of uniformity in the

F_1

generation when crossing pure lines.

Key Concept

Mendel's Law of Segregation and Monohybrid Inheritance Ratios

Question 24Question →

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

100%

75%

50%

25%

Show answer & explanation

Answer: 50%

Answer

50% of the resulting zygotes will exhibit trisomy 21.

When non-disjunction of a chromosome pair takes place during Meiosis I, the pair fails to segregate. Consequently, one daughter cell receives both homologous chromosomes while the other receives none. Following normal chromatid separation in Meiosis II, half of the gametes carry an extra chromosome (

n+1

) and half are deficient by one chromosome (

n-1

). Fertilization of the

n+1

gametes with normal haploid gametes (

n

) produces trisomic (

2n+1

) zygotes. Therefore, 50% of the zygotes will have trisomy 21.

Step-by-Step Solution

1

Analyze the meiotic stage where non-disjunction occurs

In Meiosis I, non-disjunction means homologous chromosomes fail to separate into distinct daughter cells.

Understanding the exact stage of meiotic failure determines the chromosome distribution in the daughter cells after Meiosis I.

2

Determine the chromosomal content of the resulting four gametes after Meiosis II

Meiosis I yields one cell with an extra chromosome (

n+1

) and one cell missing that chromosome (

n-1

). Normal chromatid separation in Meiosis II creates two

(n+1)

gametes and two

(n-1)

gametes.

All four gametes produced are genetically abnormal after a Meiosis I non-disjunction event.

3

Calculate the proportion of trisomic zygotes post-fertilization

Fertilization of the two

(n+1)

gametes by normal haploid (

n

) gametes gives

(2n+1)

, which is trisomy 21. Two out of four total zygotes (50%) will be trisomic (and 50% will be monosomic).

Trisomy requires the combination of an

(n+1)

gamete with a normal

n

gamete.

Key Concept

Chromosomal Non-disjunction and Aneuploidy

Question 25Question →

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

60

180

20

240

Show answer & explanation

Answer: 60

Answer

60 offspring are expected to have a black body with normal wings.

In a dihybrid cross of two heterozygous individuals (

BbVv \times BbVv

), the independent assortment of alleles yields an F2 phenotypic ratio of

9 : 3 : 3 : 1

. The proportion of offspring displaying the recessive trait for the first gene (black body,

bb

) and the dominant trait for the second gene (normal wings,

V\_

) is

\frac{3}{16}

. Multiplying

\frac{3}{16}

by the total offspring count of

320

gives exactly

60

.

Step-by-Step Solution

1

Determine parental genotypes and set up the dihybrid cross.

Parental cross is

BbVv \times BbVv

, where both parents are heterozygous for both traits.

Mendel's Second Law states that alleles of different genes assort independently during gamete formation.

2

Identify the expected phenotypic ratio for the offspring.

The phenotypic ratio for a dihybrid cross between two double heterozygotes (

BbVv \times BbVv

) is

9 : 3 : 3 : 1

.
- Grey body, normal wings (

B\_V\_

) =

\frac{9}{16}

- Grey body, vestigial wings (

B\_vv

) =

\frac{3}{16}

- Black body, normal wings (

bbV\_

) =

\frac{3}{16}

- Black body, vestigial wings (

bbvv

) =

\frac{1}{16}

The probability of recessive body color (

bb = \frac{1}{4}

) combined with dominant wing length (

V\_ = \frac{3}{4}

) is calculated as

\frac{1}{4} \times \frac{3}{4} = \frac{3}{16}

.

3

Calculate the expected number of offspring with black body and normal wings.

\frac{3}{16} \times 320 = 60

.

Multiplying the expected phenotypic fraction by the total offspring count yields the specific expected count.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Ratios

Estimated Time:1m 30s

Question 26Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

Human skin pigmentation matches continuous variation governed by polygenic inheritance with environmental modification. ABO blood group classification matches discontinuous variation controlled by multiple alleles at a single locus. Widow's peak hairline matches discontinuous variation governed by simple monogenic Mendelian inheritance. Total dermal ridge count matches continuous variation determined predominantly by additive polygenes with minimal environmental influence.

Continuous traits (such as skin pigmentation and fingerprint ridge counts) form a continuous spectrum of phenotypic values controlled by polygenes, differing in their environmental sensitivity. Discontinuous traits (such as ABO blood groups and widow's peak) form distinct qualitative classes without intermediate phenotypes, controlled by single loci with either multiple alleles or simple Mendelian dominance.

Step-by-Step Solution

1

Analyze the phenotypic distribution pattern for each feature to classify it as either continuous (quantitative spectrum) or discontinuous (discrete qualitative categories).

Skin pigmentation and dermal ridge count are continuous traits; ABO blood groups and widow's peak are discontinuous traits.

Continuous traits show intermediate gradation across a population, whereas discontinuous traits exhibit distinct non-overlapping phenotypes.

2

Examine the underlying genetic architecture (monogenic vs. polygenic vs. multiple alleles) for each trait.

Skin pigmentation and ridge count involve polygenes; ABO blood group involves multiple alleles at one locus (

I^A, I^B, i

); widow's peak involves a single monogenic locus.

Polygenic inheritance produces continuous variation through additive effects of multiple genes, whereas single loci yield distinct phenotypic classes.

3

Evaluate the impact of environmental factors on the expression of each continuous trait.

Skin tone varies significantly with solar radiation exposure (high environmental effect), while finger ridge counts are genetically fixed early in development with minimal environmental susceptibility.

Environmental modification alters phenotypic expression of certain polygenic traits without altering underlying genotype.

Key Concept

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

Question 27Question →

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

True

False

Show answer & explanation

Answer: True

Answer

True

The statement accurately distinguishes between population-level multiple alleles and individual diploid genetic constitution, while correctly identifying the codominant relationship between

I^A

and

I^B

alongside their complete dominance over the recessive

i

allele.

Step-by-Step Solution

1

Analyze allele distribution in populations versus diploid individuals.

Multiple alleles refer to the presence of more than two allele options in a population gene pool (

I^A

,

I^B

,

i

), but diploid organisms possess only two locus copies (one maternal, one paternal).

Meiotic segregation and diploid chromosome organization restrict an individual to possessing a maximum of two alleles for any autosomal gene.

2

Examine the phenotypic expression pattern of heterozygous genotype

I^A I^B

.

Both antigen A and antigen B are independently and fully expressed on red blood cell surfaces, defining codominance.

Codominance occurs when neither allele masks the other, leading to simultaneous, distinct expression of both allele products.

3

Evaluate the interaction between alleles

I^A

/

I^B

and allele

i

.

Genotypes

I^A i

and

I^B i

result in blood group A and blood group B phenotypes respectively, as allele

i

produces a non-functional enzyme.

Allele

i

is recessive, so

I^A

and

I^B

exhibit complete dominance over

i

.

Key Concept

Multiple Alleles and Codominance in ABO Blood Groups

Estimated Time:1m 30s

Question 28Question →

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

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. A monohybrid cross between two heterozygous parents yields a genotypic ratio of

1:2:1

(

1\ SS : 2\ Ss : 1\ ss

) and a phenotypic ratio of

3:1

(3 smooth : 1 wrinkled).

Evaluating the statement as False is correct because Mendel's Law of Segregation dictates that crossing two monohybrids (

Ss \times Ss

) yields three distinct genotypes (

SS

,

Ss

, and

ss

) in a

1:2:1

ratio, while the

3:1

ratio applies only to the physical phenotypes.

Step-by-Step Solution

1

Determine the parental genotypes and gametes produced during meiosis according to Mendel's First Law.

Both parents are heterozygous (

Ss

). Each parent produces gametes with either the

S

allele (

50\%

) or the

s

allele (

50\%

).

Mendel's Law of Segregation states that allele pairs separate during gamete formation so that each gamete carries only one allele for each gene.

2

Construct a Punnett square for the monohybrid cross (

Ss \times Ss

).

Offspring genotypes are

1\ SS

,

2\ Ss

, and

1\ ss

.

Combining the gametes gives

25\%\ SS

,

50\%\ Ss

, and

25\%\ ss

.

3

Compare the resulting genotypic ratio with the ratio stated in the prompt.

The genotypic ratio is

1:2:1

, whereas the phenotypic ratio (smooth to wrinkled) is

3:1

.

The prompt incorrectly asserts that

3:1

is the genotypic ratio, confusing genotypic outcome with phenotypic outcome.

Key Concept

Mendel's Law of Segregation and Monohybrid Cross Ratios

Estimated Time:1m 0s

Question 29Question →

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

Alleles

Loci

Phenotypes

Sister chromatids

Show answer & explanation

Answer: Alleles

Answer

Alleles are the alternative forms of a gene occupying identical positions on homologous chromosomes.

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

Step-by-Step Solution

1

Identify the biological entity described in the stem.

The stem describes alternative molecular forms of a gene (axial

A

vs terminal

a

) controlling the same trait.

Genes often exist in multiple structural variations that lead to contrasting characteristics.

2

Differentiate between gene location and gene variant.

The physical position on a chromosome is a locus, whereas the contrasting gene versions located at that locus are alleles.

Distinguishing between structural variants (alleles) and positional coordinates (loci) is essential in basic genetics.

Key Concept

Alleles and Gene Loci

Estimated Time:1m 0s

Question 30Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

Locus matches the specific physical position of a gene on a chromosome; Allele matches an alternative functional form of a gene; Genotype matches the specific allelic composition of an organism; Phenotype matches the observable physical or physiological characteristics of an organism.

Each genetic term matches its corresponding definition: Locus is the physical gene location on a chromosome, Allele is a gene variant, Genotype is the allelic composition, and Phenotype is the observable structural or functional trait.

Step-by-Step Solution

1

Identify the chromosomal position term.

Locus corresponds to the fixed physical site of a gene on a chromosome.

Chromosomal positioning terminology distinguishes the locus from the gene itself.

2

Distinguish gene variants.

Allele corresponds to alternative sequence versions of the same gene.

Genes exist in variant molecular forms called alleles.

3

Separate internal genetic makeup from outward appearance.

Genotype is the inherited genetic constitution, while Phenotype is the observable physical feature.

Genotype provides the instructions; phenotype is the expressed physical manifestation.

Key Concept

Basic Genetics Terminology

Estimated Time:1m 30s

Question 31Question →

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

1 : 2 : 1

3 : 1

1 : 1

9 : 3 : 3 : 1

Show answer & explanation

Answer: $1 : 2 : 1$

Answer

The expected genotypic ratio among the offspring is

1 : 2 : 1

(

1\,AA : 2\,Aa : 1\,aa

).

In a monohybrid cross between two heterozygous individuals (

Aa \times Aa

), Mendel's Law of Segregation dictates that alleles segregate into gametes independently. Punnett square analysis gives

1/4\,AA

,

2/4\,Aa

, and

1/4\,aa

, producing an exact genotypic ratio of

1 : 2 : 1

.

Step-by-Step Solution

1

Identify parental genotypes and alleles produced

Both parents have the genotype

Aa

. Gametes produced by each parent are

A

and

a

in equal proportions (

1/2\,A

and

1/2\,a

).

Mendel's Law of Segregation states that allele pairs separate during gamete formation so that each gamete carries only one allele for each gene.

2

Construct a Punnett square for the cross

Aa \times Aa

Offspring combinations are:

AA

(

1/4

),

Aa

(

2/4

), and

aa

(

1/4

).

Random fertilization combines gametes from both parents to form offspring genotypes.

3

Determine the genotypic ratio

The ratio of genotypes is

1\,AA : 2\,Aa : 1\,aa

, which simplifies to

1 : 2 : 1

.

The question specifically asks for the ratio of genotypes (

AA : Aa : aa

), not physical appearances (phenotypes).

Key Concept

Mendel's Law of Segregation and Monohybrid Cross Genotypic Ratio

Estimated Time:1m 0s

Question 32Question →

In tomato plants (*Solanum lycopersicum*), purple stem ( $P$ ) is dominant over green stem ( $p$ ), and cut leaf ( $C$ ) is dominant over potato leaf ( $c$ ). A plant heterozygous for both traits ( $PpCc$ ) is crossed with a homozygous recessive plant ( $ppcc$ ). What proportion of the offspring is expected to have purple stems and potato leaves?

1/4 (25%)

3/16 (18.75%)

9/16 (56.25%)

1/16 (6.25%)

Show answer & explanation

Answer: 1/4 (25%)

Answer

1/4 (25%)

In a dihybrid test cross between a double heterozygote (

PpCc

) and a double homozygous recessive individual (

ppcc

), the heterozygous parent produces four gamete types (

PC

,

Pc

,

pC

,

pc

) in equal proportions (

1/4

each). When combined with the recessive parent's

pc

gamete, four phenotypic classes result with an expected frequency of

1/4

(

25\%

) each. Therefore, the proportion of offspring showing purple stems and potato leaves (

Ppcc

) is

1/4

.

Step-by-Step Solution

1

Identify parental genotypes and determine gametes produced

The heterozygous parent (

PpCc

) produces four types of gametes in equal proportions:

PC

,

Pc

,

pC

, and

pc

(

1/4

each). The homozygous recessive parent (

ppcc

) produces only one type of gamete:

pc

(

100\%

).

Mendel's Law of Independent Assortment states that alleles for different traits segregate independently during gamete formation.

2

Determine offspring genotypes and phenotypes

Combining gametes yields four offspring genotypes:

PpCc

(purple stem, cut leaf),

Ppcc

(purple stem, potato leaf),

ppCc

(green stem, cut leaf), and

ppcc

(green stem, potato leaf), each occurring at a frequency of 1/4.

A test cross matches each distinct gamete from the heterozygous parent with a recessive allele pair from the tester parent.

3

Calculate the expected proportion for the targeted phenotype

The proportion of offspring displaying purple stems (

P_

) and potato leaves (

cc

), represented by genotype

Ppcc

, is 1/4 or 25%.

Exactly one of the four equally probable phenotypic outcomes exhibits the dominant stem color and recessive leaf form.

Key Concept

Dihybrid Test Cross Ratio

Estimated Time:1m 30s

Question 33Question →

In tomato plants, the allele for red fruit ( $R$ ) is completely dominant over the allele for yellow fruit ( $r$ ). If a heterozygous red-fruited plant is self-pollinated and produces $400$ offspring in the $F_2$ generation, how many of these plants are expected to be heterozygous red-fruited?

200

300

100

400

Show answer & explanation

Answer: $200$

Answer

The expected number of heterozygous red-fruited plants is 200.

Crossing two heterozygous individuals (

Rr \times Rr

) produces genotypes in a

1\ RR : 2\ Rr : 1\ rr

ratio according to Mendel's First Law. Therefore, half (

50\%

) of the

400

total offspring are expected to have the heterozygous genotype (

Rr

), which equals

200

plants.

Step-by-Step Solution

1

Determine the parental genotypes and set up the monohybrid cross.

The self-pollinated plant is heterozygous (

Rr

), so the cross is

Rr \times Rr

.

Mendel's First Law (Law of Segregation) states that paired alleles segregate during gamete formation so each gamete carries only one allele.

2

Calculate the genotypic ratio of the F2 generation.

The offspring genotypic ratio is

1\ RR : 2\ Rr : 1\ rr

, meaning

1/4

homozygous dominant,

2/4

(or

1/2

) heterozygous, and

1/4

homozygous recessive.

Random combination of parental gametes (

R

and

r

) yields probabilities of

25\%\ RR

,

50\%\ Rr

, and

25\%\ rr

.

3

Compute the expected count of heterozygous offspring out of 400 total plants.

Heterozygous count =

\frac{2}{4} \times 400 = 200

plants.

Multiplying the probability of the heterozygous genotype (

1/2

) by the total number of offspring gives the expected number.

Key Concept

Mendel's Law of Segregation and Monohybrid Genotypic Ratios

Question 34Question →

In guinea pigs (*Cavia porcellus*), black coat color ( $B$ ) is dominant over white coat color ( $b$ ), and short hair ( $S$ ) is dominant over long hair ( $s$ ). If two heterozygous guinea pigs ( $BbSs$ ) are mated and produce a total of $400$ offspring, how many offspring are expected to display the black coat and long hair phenotype?

Show answer & explanation

Answer: 75

Answer

75 offspring are expected to display the black coat and long hair phenotype.

In a dihybrid cross between two individuals heterozygous for both traits (

BbSs \times BbSs

), independent assortment produces an

F_2

phenotypic ratio of

9:3:3:1

. The phenotype with one dominant and one recessive trait (black coat and long hair,

B\_ss

) represents

\frac{3}{16}

of the total offspring population. Out of

400

offspring, the expected count is calculated as

\frac{3}{16} \times 400 = 75

.

Step-by-Step Solution

1

Determine the phenotypic ratio from the dihybrid cross

BbSs \times BbSs

.

The expected ratio is

9 : 3 : 3 : 1

for (black coat, short hair) : (black coat, long hair) : (white coat, short hair) : (white coat, long hair).

Mendel's Law of Independent Assortment dictates that the two gene pairs segregate independently during gamete formation.

2

Determine the proportion of offspring showing the black coat and long hair phenotype (

B\_ss

).

The proportion is

\frac{3}{16}

.

The probability of inheriting the dominant coat trait (

B\_

) is

\frac{3}{4}

, and the probability of inheriting the recessive hair length trait (

ss

) is

\frac{1}{4}

. Combined probability =

\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}

.

3

Multiply the proportion by the total population size to find the expected number of offspring.

\frac{3}{16} \times 400 = 75

.

Applying the theoretical phenotypic probability to the sample size of 400 yields the expected count.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Ratios

Estimated Time:1m 30s

Question 35Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair

AaBb \times AaBb

with

9:3:3:1

,

AaBb \times aabb

with

1:1:1:1

,

AaBb \times Aabb

with

3:3:1:1

, and

AABb \times aaBb

with

3:1

.

Each parent cross is accurately matched to its offspring phenotypic distribution by applying the product rule of probability to independently assorting gene pairs:

AaBb \times AaBb

produces

9:3:3:1

,

AaBb \times aabb

produces

1:1:1:1

,

AaBb \times Aabb

produces

3:3:1:1

, and

AABb \times aaBb

produces

3:1

.

Step-by-Step Solution

1

Analyze the dihybrid self-cross

AaBb \times AaBb

Combining independent monohybrid crosses

(3 \text{ dominant} : 1 \text{ recessive}) \times (3 \text{ dominant} : 1 \text{ recessive})

produces the phenotypic ratio

9 : 3 : 3 : 1

.

According to Mendel's Law of Independent Assortment, the segregation of alleles for one gene occurs independently of the segregation of alleles for another gene.

2

Analyze the dihybrid test cross

AaBb \times aabb

The heterozygous parent produces four distinct gamete types (

AB, Ab, aB, ab

) in equal frequencies of

25\%

each, while the homozygous recessive parent produces only

ab

gametes, yielding a

1:1:1:1

phenotypic ratio.

Test cross phenotypic ratios directly mirror the gametic frequencies produced by the heterozygous parent.

3

Analyze the cross

AaBb \times Aabb

For gene A (

Aa \times Aa

), expected probabilities are

\frac{3}{4}

dominant (

A\_

) and

\frac{1}{4}

recessive (

aa

). For gene B (

Bb \times bb

), expected probabilities are

\frac{1}{2}

dominant (

B\_

) and

\frac{1}{2}

recessive (

bb

). Multiplying probabilities gives

\frac{3}{8} A\_B\_ : \frac{3}{8} A\_bb : \frac{1}{8} aaB\_ : \frac{1}{8} aabb

, simplified as

3:3:1:1

.

Applying the product rule of probability for independent genetic events.

4

Analyze the cross

AABb \times aaBb

For gene A (

AA \times aa

),

100\%

of offspring express the dominant phenotype (

Aa

). For gene B (

Bb \times Bb

),

\frac{3}{4}

express the dominant phenotype (

B\_

) and

\frac{1}{4}

express the recessive phenotype (

bb

). The total offspring ratio simplifies to

3:1

.

Since trait A displays zero phenotypic variation in the progeny, the overall phenotypic ratio is determined solely by the segregation of trait B.

Key Concept

Mendel's Law of Independent Assortment and Dihybrid Phenotypic Ratios

Question 36Question →

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

25%

0%

50%

75%

Show answer & explanation

Answer: 25%

Answer

The probability that their first child will have blood group O is 25%.

Both parents inherited a recessive allele

i

from their respective blood group O parents, giving them heterozygous genotypes (

I^A i

and

I^B i

). Crossing these genotypes produces four offspring genotypes (

I^A I^B

,

I^A i

,

I^B i

,

ii

) in equal proportions, resulting in a 25% (1 in 4) chance for a child with blood group O (

ii

).

Step-by-Step Solution

1

Determine parental genotypes from their phenotypes and family histories.

The man has blood group A (

I^A I^A

or

I^A i

). Because his mother was blood group O (

ii

), he inherited an

i

allele, giving him genotype

I^A i

. The woman has blood group B (

I^B I^B

or

I^B i

) and her father was blood group O (

ii

), giving her genotype

I^B i

.

An individual with blood group O has genotype

ii

and must pass a recessive

i

allele to all offspring.

2

Perform a genetic cross between the two parents (

I^A i \times I^B i

).

The offspring genotypes formed are

I^A I^B

(blood group AB),

I^A i

(blood group A),

I^B i

(blood group B), and

ii

(blood group O) in a 1:1:1:1 ratio.

Punnett square analysis of two heterozygous individuals yields four equally likely genotype combinations.

3

Calculate the probability of obtaining genotype

ii

(blood group O).

1 out of 4 possible outcomes is genotype

ii

, which equals 25% or

\frac{1}{4}

.

Blood group O is expressed only when the homozygous recessive genotype

ii

is present.

Key Concept

Multiple Alleles and Codominance in ABO Blood Group Inheritance

Estimated Time:1m 30s

Question 37Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

Incomplete Dominance matches with pink flower color in snapdragons; Codominance matches with simultaneous expression of

I^A

and

I^B

antigens in AB blood group; Multiple Alleles matches with the existence of

I^A

,

I^B

, and

i

gene variants in a population.

Matching Incomplete Dominance to pink snapdragons correctly identifies intermediate phenotypic blending. Matching Codominance to blood type AB correctly identifies joint expression of two functional alleles. Matching Multiple Alleles to the ABO gene locus (

I^A

,

I^B

,

i

) correctly identifies gene loci that have more than two alternative allele variants within a population.

Step-by-Step Solution

1

Analyze Incomplete Dominance

Incomplete dominance involves a blending of phenotypes in heterozygous offspring.

Crossing red and white homozygous snapdragons produces pink heterozygous offspring because the single functional dominant allele produces insufficient pigment for full red expression.

2

Analyze Codominance

Codominance involves full, distinct, non-blended phenotypic expression of both inherited alleles.

Heterozygotes carrying both

I^A

and

I^B

alleles produce both A and B glycoprotein antigens on red blood cell surfaces equally.

3

Analyze Multiple Alleles

Multiple alleles describes a population-level genetic phenomenon where a single locus has more than two allele options.

While any diploid individual inherits only two alleles, the human population possesses three distinct alleles (

I^A

,

I^B

, and

i

) governing the ABO blood group locus.

Key Concept

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

Question 38Question →

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

25\%

50\%

0\%

100\%

Show answer & explanation

Answer: $50\%$

Answer

The probability that any male child born to this couple will be color-blind is

50\%

.

The mother is a carrier (

X^C X^c

) because she inherited the color-blindness allele (

X^c

) from her affected father. When crossed with a normal male (

X^C Y

), sons inherit their Y chromosome from the father and one of the mother's two X chromosomes. Therefore, there is a

50\%

chance that a son receives the

X^c

allele and manifests color blindness.

Step-by-Step Solution

1

Determine the parental genotypes.

The woman's father was color-blind (

X^c Y

), so she inherited the

X^c

allele. Being phenotypically normal, her genotype is

X^C X^c

(heterozygous carrier). The man has normal vision, so his genotype is

X^C Y

.

Red-green color blindness is an X-linked recessive trait.

2

Construct a Punnett square for the cross

X^C X^c \times X^C Y

.

The possible offspring genotypes are

X^C X^C

(normal female),

X^C X^c

(carrier female),

X^C Y

(normal male), and

X^c Y

(color-blind male).

Each parent contributes one sex chromosome to offspring.

3

Calculate the probability specific to male offspring.

The male offspring genotypes are

X^C Y

and

X^c Y

. Out of these

2

possible male outcomes,

1

is color-blind (

X^c Y

). The probability among male children is

\frac{1}{2} = 50\%

.

The question asks specifically for the probability among sons, requiring evaluation within the male subset only.

Key Concept

Sex-Linked Inheritance and X-Linked Recessive Traits

Estimated Time:1m 30s

Question 39Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: XX-XO System matches grasshoppers with

XO

males; ZZ-ZW System matches birds with heterogametic

ZW

females; X-linked Recessive Inheritance matches traits passed from carrier mothers to sons; Y-linked Inheritance matches traits passed exclusively from an affected father to all sons.

Each mechanism accurately corresponds to its characteristic biological model: XX-XO occurs in grasshoppers (

XO

males); ZZ-ZW occurs in birds where females are heterogametic (

ZW

); X-linked recessive traits pass from carrier mothers to sons; and Y-linked traits pass exclusively from fathers to sons.

Step-by-Step Solution

1

Identify the chromosomal structure of the XX-XO system.

Determine that males lack a Y chromosome (

XO

) while females have two X chromosomes (

XX

), which is characteristic of grasshoppers.

Distinguishing insect sex determination from mammalian systems.

2

Identify the heterogametic sex in the ZZ-ZW system.

Recognize that in birds, females are heterogametic (

ZW

) and males are homogametic (

ZZ

).

Reversing the standard male-heterogametic concept seen in humans.

3

Compare transmission pathways of X-linked versus Y-linked traits.

X-linked recessive conditions are passed from carrier females to sons, whereas Y-linked traits are passed strictly along the male line (father to son).

Applying rules of sex-linked pedigree analysis.

Key Concept

Sex Determination Systems and Sex-Linked Inheritance Patterns

Question 40Question →

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

They express the same phenotype but possess different genotypes.

They possess the same genotype but express different phenotypes.

Plant 1 is heterozygous dominant, whereas Plant 2 is homozygous recessive.

They share identical genotypes because environmental factors mask the recessive allele.

Show answer & explanation

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

Answer

They express the same phenotype but possess different genotypes.

Phenotype refers to the observable physical characteristics of an organism, while genotype refers to its specific genetic composition. Because allele

R

is dominant over allele

r

, both

RR

and

Rr

produce the disease-resistant phenotype. However, their genotypes differ because one plant is homozygous dominant (

RR

) while the other is heterozygous (

Rr

).

Step-by-Step Solution

1

Identify the phenotypic expression of both plants.

Both Plant 1 (

RR

) and Plant 2 (

Rr

) exhibit resistance to the fungal pathogen, meaning they have identical phenotypes.

Phenotype is the observable physical or physiological trait of an organism produced by the interaction of its genotype with the environment.

2

Analyze the genotypic constitution of each plant.

Plant 1 is homozygous dominant (

RR

) and Plant 2 is heterozygous (

Rr

).

Genotype refers to the actual allele combination present at a specific gene locus.

3

Compare the phenotype and genotype between the two plants.

The two plants share the same observable phenotype (disease resistance) despite having different genotypes (

RR

versus

Rr

).

The dominant allele (

R

) completely masks the presence of the recessive allele (

r

) in the heterozygous state.

Key Concept

Phenotype vs. Genotype and Allelic Dominance

Heredity and Variation

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept