Pre AP Biology -GEN 5.1 Inheritance Patterns- MCQ Exam Style Questions -New Syllabus 2025-2026
Pre AP Biology -GEN 5.1 Inheritance Patterns- MCQ Exam Style Questions – New Syllabus 2025-2026
Pre AP Biology -GEN 5.1 Inheritance Patterns- MCQ Exam Style Questions – Pre AP Biology – per latest Pre AP Biology Syllabus.
▶️ Answer/Explanation
The correct option is c. temperature, because it determines their sex.
In many reptile species, such as crocodiles and turtles, the sex of the offspring is not determined by chromosomes.
Instead, it is determined by the incubation temperature of the eggs during a critical period of development.
This biological phenomenon is known as Temperature-dependent Sex Determination (TSD).
Higher or lower temperatures can result in a clutch being entirely male or entirely female.
Environmental temperature changes can therefore significantly impact the population sex ratios.
Other options are incorrect as humidity does not regulate internal body temperature in reptiles.
▶️ Answer/Explanation
Let $P$ be the dominant perky allele and $p$ be the recessive droopy allele.
The female parent is droopy-eared, so her genotype must be $pp$.
The male parent is homozygous perky-eared, so his genotype is $PP$.
All offspring (sons and daughters) will inherit the genotype $Pp$ ($P$ from father, $p$ from mother).
In daughters, genomic imprinting is not mentioned as sex-specific, so $P$ is dominant over $p$, resulting in perky ears.
In sons, the paternal allele ($P$) is silenced by methylation, meaning only the maternal allele ($p$) is expressed.
Since sons only express the maternal $p$ allele, they will have droopy ears.
Therefore, the correct result is option a.
▶️ Answer/Explanation
To get 100% white-eyed offspring, every child must receive the recessive allele (\(X^r\)) and no dominant alleles (\(X^R\)).
The father must be white-eyed (\(X^r Y\)) to pass the allele to daughters (and Y to sons).
The mother must be homozygous recessive (\(X^r X^r\)) to pass the white allele to all offspring.
Therefore, the cross is \(X^r X^r \times X^r Y\).
Males have one X chromosome, so a male with a white-eye allele has the genotype \(X^r Y\) (white eyes).
Experiment 1 produced 100% red-eyed males (\(X^R Y\)).
Experiment 2 produced 50% white-eyed males (\(X^r Y\)).
Experiment 3 produced 50% white-eyed males (\(X^r Y\)).
Experiment 4 produced 100% white-eyed males (\(X^r Y\)).
Thus, the correct experiments are 2, 3, and 4.
Experiment 2 is a cross between offspring of Exp 1: \(X^R X^r\) (female) \(\times\) \(X^R Y\) (male).
The possible genotypes are: \(X^R X^R\) (Red Female), \(X^R X^r\) (Red Female), \(X^R Y\) (Red Male), and \(X^r Y\) (White Male).
Counting the phenotypes: 3 offspring are Red-eyed and 1 offspring is White-eyed.
Therefore, the ratio of Red:White is 3:1.
▶️ Answer/Explanation
A: A (\(3\))
A: B (\(7.81\))
A3: D
Explanation:
For Q: A standard dihybrid cross results in \(4\) possible phenotypic classes (Tall/Violet, Tall/White, Dwarf/Violet, Dwarf/White). The degrees of freedom are calculated as \(df = n – 1 = 4 – 1 = 3\).
For Q: Referring to the Chi-square table provided, find the intersection of the column for \(df = 3\) and the row for \(p = 0.05\). The value listed is \(7.81\).
For Q: The conclusion relies on comparing the calculated \(\chi^2\) to the critical value (\(7.81\)). Only option D correctly identifies the critical value for \(3\) degrees of freedom (\(7.81\)).
For Q: Since option D in Q184 is the only logically consistent conclusion (citing the correct critical value), the calculated \(\chi^2\) value must be \(719.21\). This extremely high value suggests the genes are linked rather than independent.
▶️ Answer/Explanation
Let $B$ represent the dominant allele for black fur and $b$ represent the recessive allele for white fur.
The homozygous black parent has the genotype $BB$.
The heterozygous black parent has the genotype $Bb$.
The cross is $BB \times Bb$.
The resulting offspring genotypes are $50\%$ $BB$ and $50\%$ $Bb$.
Since both $BB$ and $Bb$ genotypes result in the black fur phenotype, $100\%$ of the offspring will be black.
Therefore, the predicted phenotype ratio is $100\%$ black.
The correct option is C.
▶️ Answer/Explanation
The parental phenotypes (Gray/Long and Black/Apterous) appear in much higher frequencies ($42$ and $41$) than the non-parental types.
If independent assortment occurred, we would expect a phenotypic ratio of approximately $1:1:1:1$ (roughly $25$ each).
The high frequency of parental combinations indicates the genes are linked on the same chromosome.
The small number of Gray/Apterous ($9$) and Black/Long ($8$) offspring are recombinants.
These recombinants resulted from crossing over during meiosis in the parent.
Therefore, the data supports that these genes are linked with a recombination frequency of approximately $17\%$.
▶️ Answer/Explanation
The correct option is D. X-linked inheritance.
In the $P$ generation, the cross is $X^W X^W$ (red female) $\times$ $X^w Y$ (white male).
All $F_1$ offspring ($X^W X^w$ and $X^W Y$) have red eyes, showing red is dominant.
In the $F_2$ cross ($X^W X^w \times X^W Y$), all females receive a $X^W$ allele from the father.
This results in $100\%$ red-eyed females ($X^W X^W$ and $X^W X^w$).
Males receive either $X^W$ or $X^w$ from the mother, creating a $1:1$ red to white ratio.
The unequal distribution of the trait between sexes confirms the gene is on the $X$ chromosome.
▶️ Answer/Explanation
The man has blood type A, so his genotype is either $I^A I^A$ or $I^A i$.
The woman has blood type O, meaning her genotype must be $ii$.
If the father is $I^A I^A$, all children will be $I^A i$ (Type A).
If the father is $I^A i$, the children could be $I^A i$ (Type A) or $ii$ (Type O).
In either scenario, the only possible phenotypes are blood type A or blood type O.
Therefore, the correct option is A.
▶️ Answer/Explanation
The cross $GgPp \times GgPp$ involves two organisms heterozygous for two distinct traits.
According to Mendel’s Law of Independent Assortment, alleles for different traits segregate independently.
The four possible phenotypes resulting from this cross are expressed in a specific distribution.
Dominant for both traits ($G\_P\_$) occurs in $9/16$ of the offspring.
Dominant for one and recessive for the other ($G\_pp$ and $ggP\_$) each occur in $3/16$ of the offspring.
Recessive for both traits ($ggpp$) occurs in $1/16$ of the offspring.
Therefore, the classic Mendelian dihybrid ratio is $9:3:3:1$.
The correct option is C.
▶️ Answer/Explanation
The correct answer is C.
Mendel’s Principle of Dominance states that in a heterozygote, one allele can mask the presence of another.
Organisms possess two alleles for a single trait, often inherited as one from each parent.
An allele is defined as dominant if it determines the phenotype even when only one copy is present.
An allele is recessive if its expression is masked by the dominant allele in a hybrid ($F_1$) generation.
Therefore, this principle establishes that some alleles are dominant and others are recessive.
▶️ Answer/Explanation
Let the genotype for red be $RR$ and white be $WW$, resulting in pink $RW$.
The cross requested is between a red ($RR$) and a pink ($RW$) offspring.
The gametes from the red parent are all $R$.
The gametes from the pink parent are $50\% R$ and $50\% W$.
Offspring genotypes will be $50\% RR$ (red) and $50\% RW$ (pink).
Therefore, the phenotype ratio is $50\%$ red and $50\%$ pink.
The correct option is D.
▶️ Answer/Explanation
The correct answer is D. Polygenic inheritance.
Polygenic inheritance occurs when a single characteristic is controlled by two or more genes.
Unlike Mendelian traits, these genes exert an additive effect on the final phenotype.
This results in continuous variation, creating a wide spectrum of possibilities (\(>60\) in this case).
Human skin color, height, and eye color are classic examples of this complex pattern.
Options A, B, and C typically involve a single gene locus and fewer distinct phenotypes.
▶️ Answer/Explanation
A: B
The definitive proof of a recessive trait is the appearance of the condition in the offspring of unaffected parents.
If a trait were dominant, every affected child must have at least one affected parent.
In this pedigree, Individual \(14\) is affected, yet both parents (Individuals \(3\) and \(4\)) are unaffected.
This “skipping of generations” confirms that individuals \(3\) and \(4\) are heterozygous carriers for a recessive allele.
A: B
We test for sex-linkage (specifically X-linked recessive) by looking for violations of X-linked inheritance rules.
Rule 1: An affected daughter (\(X^r X^r\)) must have an affected father (\(X^r Y\)). Individual \(19\) is affected, but her father (\(7\)) is not.
Rule 2: An affected mother (\(X^r X^r\)) must pass the trait to all her sons (\(X^r Y\)). Individual \(8\) is affected, but her sons (\(18, 20\)) are not.
Therefore, the existence of affected females \(8\) and \(19\) (given their family context) provides the best evidence that the trait is autosomal, not sex-linked.
▶️ Answer/Explanation
A. Correct Answer: B
The cross shown is between two heterozygotes (\(Tt \times Tt\)).
This produces genotypes: \(TT\) (homozygous dominant), \(2 \times Tt\) (heterozygous), and \(tt\) (homozygous recessive).
Both \(TT\) and \(Tt\) genotypes result in the dominant “Tall” phenotype, totaling 3 offspring.
The \(tt\) genotype results in the recessive “Short” phenotype, totaling 1 offspring.
Therefore, the phenotypic ratio is 3 tall and 1 short.
A. Correct Answer: C
Individual 1 inherits a \(T\) from both parents, making it homozygous dominant (\(TT\)).
Individuals 2 and 3 inherit one \(T\) and one \(t\), making them heterozygous (\(Tt\)).
Individual 4 inherits a \(t\) allele from both parents (\(t \times t\)).
This results in the genotype \(tt\), which is classified as homozygous recessive.
A. Correct Answer: D
The \(Tt\) genotype represents a heterozygous individual.
In the Punnett square, Box 2 is formed by \(t\) (top) and \(T\) (side), resulting in \(Tt\).
Box 3 is formed by \(T\) (top) and \(t\) (side), resulting in \(Tt\).
Boxes 1 and 4 are homozygous (\(TT\) and \(tt\)).
Thus, only Individual 2 and 3 have the \(Tt\) genotype.
A. Correct Answer: D
A heterozygous pea plant has the genotype \(Yy\).
Organisms inherit one allele for a specific trait from each parent.
To possess both a \(Y\) and a \(y\) allele, the offspring must receive \(Y\) from one parent and \(y\) from the other.
It is not required for parents to be homozygous; they just need to contribute those specific alleles.
Therefore, one parent must have contributed a \(Y\) allele while the other contributed a \(y\) allele.
▶️ Answer/Explanation
A. Correct Answer: A
The dependent variable (DV) is the variable being measured and is typically plotted on the Y-axis. The Y-axis label in Figure 1 is “Relative Concentration of Product”. Therefore, the rate is being measured by tracking the relative product produced over time.
A. Correct Answer: B
The rate of an enzymatic reaction can be determined by measuring the appearance of product or the disappearance of substrate. Since the conversion of reactants to products is stoichiometric, measuring the rate at which substrates/reactants are consumed is a valid alternative method.
A. Correct Answer: C
The slope of the curve (representing the rate) flattens over time, indicating the rate is decreasing. This happens because the substrate concentration drops as it is converted to product. Option D is incorrect because enzymes are not consumed in the reaction. Option C correctly describes the kinetic shift: as product accumulates and substrate depletes (increasing the product:substrate ratio), the reaction slows.
A. Correct Answer: C
At \(t=1\) minute, the product concentration is approximately \(4.2\). Under normal conditions, at \(t=2\) minutes, it reaches approximately \(7.8\). A competitive inhibitor would slow the reaction rate but not stop it immediately. Therefore, the product concentration would increase from \(4.2\), but to a level lower than the uninhibited \(7.8\). The range of \(5\) to \(7\) fits a reaction that is proceeding at a reduced rate.
▶️ Answer/Explanation
The $\text{HTT}$ gene is located on chromosome 4, which is a non-sex chromosome (autosome).
Therefore, the inheritance pattern of the condition is autosomal.
The text states that an individual with one faulty copy will have the disease.
This indicates a dominant inheritance, as only one allele is needed for the phenotype.
Combining these facts, Huntington’s disease is a dominant, autosomal genetic disease.
This matches option (D).
▶️ Answer/Explanation
Achondroplasia is an autosomal dominant disorder, meaning affected individuals have at least one dominant allele ($A$).
Individual $\text{I-1}$ is unaffected (white square), so he must carry two recessive alleles ($aa$).
Even though his mother was affected, he did not inherit the trait, making him homozygous recessive.
Individual $\text{II-2}$ is also unaffected (white square), meaning his genotype must be ($aa$).
An unaffected person in a dominant pedigree cannot be a carrier; they are always homozygous recessive.
Therefore, both $\text{I-1}$ and $\text{II-2}$ are homozygous recessive.
The most accurate selection based on the visual evidence is that both are unaffected.
Correct Option: (None of the above strictly match if searching for homozygous recessive for both, but usually implies (C) or (B) if the question contains a typo; however, based strictly on the pedigree, both are homozygous recessive.)