Health Education Systems Inc (HESI) A2 Exam Quiz: Genes Chromosomes And Inheritance
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Genes Chromosomes And InheritanceQuestion 1 of 20

A segment of a DNA template strand has the sequence: 3'-T A C G G A-5'. What is the corresponding sequence of the mRNA molecule that will be transcribed from this segment?

5'-A U G C C U-3'
5'-A T G C C T-3'
3'-A U G C C U-5'
5'-U A C G G A-3'
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Health Education Systems Inc (HESI) A2 Exam Quiz

Health Education Systems Inc (HESI) A2 Exam Quiz: Genes Chromosomes And Inheritance

Practice Genes Chromosomes And Inheritance in Health Education Systems Inc (HESI) A2 Exam with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Genes Chromosomes And Inheritance, giving you a quick way to practice the rules, question types, and explanations that matter most for Health Education Systems Inc (HESI) A2 Exam.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A segment of a DNA template strand has the sequence: 3'-T A C G G A-5'. What is the corresponding sequence of the mRNA molecule that will be transcribed from this segment?

  1. 5'-A U G C C U-3' (correct answer)
  2. 5'-A T G C C T-3'
  3. 3'-A U G C C U-5'
  4. 5'-U A C G G A-3'
Explanation: When you encounter DNA transcription questions, remember that you're translating the genetic code from DNA's language to RNA's language using specific base-pairing rules. During transcription, RNA polymerase reads the DNA template strand in the 3' to 5' direction and synthesizes mRNA in the 5' to 3' direction. The key base-pairing rules are: A pairs with U (not T in RNA), T pairs with A, G pairs with C, and C pairs with G. Starting with the template strand 3'-T A C G G A-5', you read it from 3' to 5' and create the complementary mRNA strand from 5' to 3'. Each base pairs as follows: T→A, A→U, C→G, G→C, G→C, A→U. This gives you 5'-A U G C C U-3', which is answer A. Let's examine why the other options are incorrect. Answer B (5'-A T G C C T-3') uses thymine instead of uracil, but RNA contains uracil, not thymine. Answer C (3'-A U G C C U-5') has the correct bases and uracil, but the directionality is backwards—mRNA is synthesized 5' to 3', not 3' to 5'. Answer D (5'-U A C G G A-3') simply copies the DNA template sequence while substituting U for T, but this ignores the complementary base-pairing requirement entirely. For HESI success, always remember the acronym "AUGC" for RNA bases, and that transcription creates a complementary strand, not a copy. Pay close attention to the 5' and 3' directionality—it's frequently tested and often trips up students.

Question 2

A couple has three children, all with Type A blood. The father has Type O blood. If the mother is heterozygous for the ABO blood type, what is the probability that their next child will have Type O blood?

  1. 0% because all previous children had Type A blood
  2. 25% because the mother must be homozygous dominant
  3. 50% because the mother contributes either A or O allele (correct answer)
  4. 75% because Type O is recessive and more likely
Explanation: The father has Type O blood (genotype OO) and can only contribute O alleles. The mother is heterozygous (genotype AO) and can contribute either an A allele (50% chance) or an O allele (50% chance). For a child to have Type O blood, they must receive O from both parents. Since the father always contributes O, the probability depends only on the mother contributing O, which is 50%. The previous children's blood types don't affect future offspring probabilities.

Question 3

In pea plants, tall (T) is dominant to short (t), and purple flowers (P) are dominant to white flowers (p). If a plant that is heterozygous for both traits (TtPp) is crossed with a plant that is short and has white flowers (ttpp), what is the probability of producing an offspring that is short with purple flowers?

  1. 12.5%
  2. 25% (correct answer)
  3. 50%
  4. 75%
Explanation: When you encounter a genetics cross problem, you're dealing with probability and the principle of independent assortment. This is a testcross (heterozygote × homozygous recessive), which is perfect for determining the probability of specific trait combinations. Set up the cross systematically. The heterozygous parent (TtPp) can produce four types of gametes: TP, Tp, tP, and tp, each with equal probability (25%). The homozygous recessive parent (ttpp) can only produce one type of gamete: tp. When you cross these gametes, you get four equally likely offspring genotypes:
  • TP × tp = TtPp (tall, purple)
  • Tp × tp = Ttpp (tall, white)
  • tP × tp = ttPp (short, purple) ← This is what we want
  • tp × tp = ttpp (short, white)
Since each outcome has equal probability, the chance of getting short with purple flowers (ttPp) is 14=25%\frac{1}{4} = 25\%. Looking at the wrong answers: A) 12.5% would be the probability if you needed a specific combination of three traits, but we only have two. C) 50% incorrectly assumes you're looking at just one trait independently. D) 75% represents the probability of getting the dominant phenotype for a single trait, not the specific combination asked for. Remember this testcross pattern: when crossing a heterozygote with a homozygous recessive for two traits, each of the four possible phenotype combinations occurs 25% of the time. This makes testcrosses straightforward probability problems.

Question 4

A patient's chart notes that they are heterozygous for the gene associated with Tay-Sachs disease, an autosomal recessive disorder, but they do not exhibit symptoms. Which statement correctly interprets this information?

  1. The patient's phenotype is heterozygous, while their genotype is unaffected.
  2. The patient's genotype is heterozygous, while their phenotype is unaffected. (correct answer)
  3. The patient's genotype is homozygous recessive, and their phenotype is affected.
  4. The patient's phenotype and genotype are both described as heterozygous.
Explanation: When you encounter genetics questions involving carriers and disease expression, you need to distinguish between genotype (the actual genetic makeup) and phenotype (the observable characteristics or traits). This distinction is crucial for understanding autosomal recessive inheritance patterns. The patient is heterozygous for Tay-Sachs disease, meaning they have one normal allele and one disease allele. Since Tay-Sachs is autosomal recessive, you need two copies of the disease allele to actually express the condition. With only one disease allele, this patient is a carrier who doesn't show symptoms. Therefore, their genotype is heterozygous (having two different alleles), while their phenotype is unaffected (no observable disease symptoms). Option A reverses the definitions by calling the phenotype "heterozygous" - but phenotypes describe observable traits, not genetic combinations. You can't observe whether someone is heterozygous just by looking at them. Option C incorrectly states the genotype is homozygous recessive, which would mean two disease alleles and would result in an affected phenotype. Option D misuses terminology by describing the phenotype as heterozygous, which is impossible since phenotypes describe observable characteristics, not allele combinations. Remember this pattern: for autosomal recessive disorders, heterozygous individuals are carriers with normal phenotypes, while only homozygous recessive individuals express the disease. Always match genotype with genetic makeup (homozygous/heterozygous) and phenotype with observable traits (affected/unaffected). This concept frequently appears on the HESI when testing genetic inheritance patterns.

Question 5

In the ABO blood group system, the I^A and I^B alleles are codominant, and both are dominant to the i allele. A person with the genotype I^Ai has which blood type, and what term describes their genetic state?

  1. Type A, homozygous dominant
  2. Type B, heterozygous
  3. Type A, heterozygous (correct answer)
  4. Type AB, codominant
Explanation: When you encounter ABO blood group genetics problems, focus on understanding how codominance and dominance relationships work together. The key is recognizing that IAI^A and IBI^B are codominant with each other, while both are completely dominant over the recessive ii allele. A person with genotype IAiI^A i will express type A blood because the IAI^A allele is dominant over the recessive ii allele. Since this person carries two different alleles (IAI^A and ii), they are heterozygous. The ii allele doesn't produce any functional enzyme, so only the A antigen appears on red blood cells, resulting in type A blood. Looking at the wrong answers: Choice A incorrectly calls this "homozygous dominant" - homozygous means having two identical alleles, but IAiI^A i has two different alleles. Choice B gives the wrong blood type entirely; IAiI^A i cannot produce type B blood since there's no IBI^B allele present. Choice D suggests type AB blood, but that only occurs with genotype IAIBI^A I^B where both codominant alleles are present and expressed equally. Remember this pattern: for ABO genetics questions, first identify which alleles are present, then apply the dominance hierarchy (IAI^A and IBI^B > ii), and finally determine if the genotype is homozygous (two identical alleles) or heterozygous (two different alleles). The term "codominant" describes the relationship between alleles, not the genetic state of an individual.

Question 6

A mutation occurs in a skin cell of an adult due to excessive sun exposure. Why is this specific mutation not typically passed on to the person's offspring?

  1. The mutation will be completely repaired by the body's DNA repair mechanisms before reproduction can occur.
  2. Mutations caused by radiation are not heritable, unlike mutations caused by chemical mutagens.
  3. The mutation is in a somatic cell, and only mutations in germline cells can be inherited. (correct answer)
  4. The offspring's immune system will recognize and destroy any gametes carrying the parental mutation.
Explanation: When you encounter questions about inheritance and mutations, you need to distinguish between the two main types of cells in your body: somatic cells and germline cells. This distinction determines whether genetic changes can be passed to offspring. The skin cell mutation from sun exposure occurs in a somatic cell - one of the regular body cells that make up tissues and organs. Somatic cells include skin, muscle, liver, and brain cells. While these mutations can cause problems like cancer in the individual, they cannot be inherited because somatic cells don't contribute genetic material to offspring. Only germline cells (sperm and egg cells) pass their DNA to the next generation, so only mutations in these reproductive cells can be inherited. Let's examine why the other options are incorrect: Option A suggests DNA repair mechanisms will completely fix the mutation before reproduction. While cells do have repair systems, they're not 100% effective, and many somatic mutations persist throughout a person's lifetime. Option B incorrectly claims that radiation-induced mutations aren't heritable while chemical mutations are. The source of the mutation doesn't determine inheritability - the cell type does. Both radiation and chemical mutagens can cause heritable mutations if they affect germline cells. Option D describes offspring immune systems destroying mutated gametes, which doesn't happen. The immune system doesn't routinely screen gametes for mutations during fertilization. Remember this key distinction for the HESI: somatic cell mutations affect only the individual, while germline cell mutations can be passed to offspring. The location of the mutation, not its cause, determines inheritance.

Question 7

A single nucleotide substitution in a gene's coding sequence is generally less disruptive than a single nucleotide insertion near the beginning of the sequence. What is the most likely reason for this difference?

  1. A substitution always results in a silent mutation, which does not change the amino acid sequence.
  2. An insertion shifts the reading frame for all subsequent codons, altering the entire downstream amino acid sequence. (correct answer)
  3. Insertions are repaired by cellular machinery much less effectively than substitutions.
  4. A substitution changes the DNA but has no effect on the transcribed mRNA molecule's sequence.
Explanation: When you encounter questions about DNA mutations, focus on how different types of changes affect the genetic code's reading frame - the way nucleotides are grouped into three-base codons during translation. A single nucleotide substitution replaces one base with another but maintains the original sequence length. This affects only one codon, potentially changing one amino acid (missense mutation) or having no effect if it creates a synonymous codon (silent mutation). The rest of the protein sequence remains unchanged because all downstream codons stay in their correct reading frame. In contrast, a single nucleotide insertion near the beginning of a coding sequence creates a frameshift mutation. Since codons are read in groups of three, adding one nucleotide shifts the reading frame for every subsequent codon. This dramatically alters the amino acid sequence from the insertion point onward, often creating a completely different and usually nonfunctional protein. Choice A is incorrect because substitutions don't always result in silent mutations - they can create missense mutations that change amino acids. Choice C misrepresents the issue, as this isn't primarily about DNA repair efficiency but about the structural consequences of each mutation type. Choice D is false because substitutions absolutely affect the mRNA sequence - that's how they influence protein synthesis. For HESI questions on molecular biology, remember that frameshift mutations (insertions and deletions) are generally more disruptive than point mutations (substitutions) because they affect multiple codons rather than just one.

Question 8

In a certain species of flower, a cross between a true-breeding red-flowered plant and a true-breeding white-flowered plant produces F1 offspring that all have pink flowers. This pattern of inheritance is best described as which of the following?

  1. Simple dominance
  2. Codominance
  3. Incomplete dominance (correct answer)
  4. Polygenic inheritance
Explanation: When you encounter genetics problems involving flower color crosses, pay attention to what happens in the F1 generation - this tells you the inheritance pattern. Here, two true-breeding parents (red and white) produce offspring with an intermediate phenotype (pink), which is the hallmark of incomplete dominance. In incomplete dominance, neither allele is completely dominant over the other. Instead, the heterozygous offspring show a blended phenotype that's intermediate between the two homozygous parents. The red allele and white allele each contribute partially to the phenotype, resulting in pink flowers. If you crossed these F1 pink flowers, you'd expect a 1:2:1 ratio of red:pink:white in the F2 generation. Choice A (simple dominance) is incorrect because in simple dominance, one allele masks the other completely - you'd see only red OR only white flowers in the F1, not pink. Choice B (codominance) is wrong because codominance means both alleles are fully expressed simultaneously, like AB blood type or roan coat color in cattle where you see both colors distinctly, not blended. Choice D (polygenic inheritance) involves multiple genes affecting one trait, typically showing continuous variation across a range - not the simple three-phenotype pattern seen here. For HESI genetics questions, remember this key distinction: incomplete dominance produces a blend (pink from red + white), while codominance produces distinct expression of both traits (red AND white patches). The intermediate phenotype in F1 is your clue for incomplete dominance.

Question 9

Chromosomal nondisjunction during meiosis I results in gametes with an abnormal number of chromosomes. If this event occurs during spermatogenesis, what is the composition of the four resulting gametes at the end of meiosis II?

  1. One gamete with n+1, one with n-1, and two normal gametes (n).
  2. Two gametes with an extra chromosome (n+1) and two gametes missing a chromosome (n-1). (correct answer)
  3. Four gametes with an extra chromosome (n+1).
  4. Two normal gametes (n) and two gametes missing a chromosome (n-1).
Explanation: When you encounter questions about chromosomal nondisjunction during meiosis, focus on tracking what happens to chromosome pairs through both divisions. Nondisjunction means chromosomes fail to separate properly, creating an unequal distribution. During meiosis I nondisjunction, homologous chromosomes fail to separate and both go to the same daughter cell. This creates two abnormal secondary spermatocytes: one receives both homologs (giving it an extra chromosome) while the other receives neither (missing that chromosome). Meiosis II then proceeds normally, with sister chromatids separating properly in both cells. The cell with the extra chromosome undergoes normal meiosis II, producing two gametes that each have n+1 chromosomes. The cell missing the chromosome also completes meiosis II normally, yielding two gametes with n-1 chromosomes each. This gives you two n+1 gametes and two n-1 gametes total. Choice A incorrectly suggests you get normal gametes, but nondisjunction in meiosis I affects all four final products since it occurs before the second division. Choice C assumes all gametes gain chromosomes, ignoring that one secondary spermatocyte loses chromosomes. Choice D incorrectly includes normal gametes and fails to account for the cell that gained extra chromosomes. For HESI questions on meiotic errors, remember that the timing matters crucially. Nondisjunction in meiosis I affects all four gametes because it disrupts the initial separation of homologs, while nondisjunction in meiosis II would only affect two of the four final gametes.

Question 10

Human traits such as height and skin color show a continuous range of variation rather than a few distinct forms. This is because these traits are determined by the cumulative effect of multiple genes. What is this pattern of inheritance called?

  1. Pleiotropy
  2. Polygenic inheritance (correct answer)
  3. Incomplete dominance
  4. Epistasis
Explanation: When you encounter questions about traits that show continuous variation rather than distinct categories, you're dealing with quantitative genetics. These questions test your understanding of how multiple genes can work together to produce observable characteristics. Polygenic inheritance (B) is the correct answer because it describes exactly what the question presents: multiple genes contributing small, additive effects to produce a continuous range of phenotypes. Height is a classic example - you don't see just "tall" and "short" people, but every gradation in between because dozens of genes each contribute a small amount to final height. The same applies to skin color, where multiple genes control melanin production and distribution. Choice A, pleiotropy, is the opposite concept - it describes one gene affecting multiple different traits, not multiple genes affecting one trait. Choice C, incomplete dominance, occurs when heterozygotes show a blended phenotype between two parental forms, but this still produces distinct categories rather than continuous variation. Choice D, epistasis, involves one gene masking or modifying the expression of another gene, which can create complex inheritance patterns but doesn't typically produce the smooth continuous variation described in the question. For HESI success, remember this key distinction: if a question mentions "continuous variation," "range of phenotypes," or gives examples like height, weight, or skin color, think polygenic inheritance. These traits contrast sharply with simple Mendelian traits that show clear dominant/recessive patterns with distinct categories.

Question 11

In a population study, researchers found that individuals with genotype AA have a 90% survival rate, genotype Aa have an 85% survival rate, and genotype aa have a 60% survival rate. If the current allele frequencies are A = 0.7 and a = 0.3, what evolutionary process is most likely occurring?

  1. Genetic drift because survival rates vary randomly among genotypes
  2. Gene flow because different populations have different survival patterns
  3. Natural selection favoring the A allele due to higher survival fitness (correct answer)
  4. Mutation pressure increasing the frequency of beneficial alleles over time
Explanation: The data shows a clear fitness advantage for genotypes containing the A allele (AA and Aa have higher survival rates than aa). This differential survival based on genotype is the definition of natural selection. The A allele confers higher fitness, so natural selection will favor it and increase its frequency over time. Genetic drift involves random changes, gene flow involves migration, and mutation pressure involves new allele creation.

Question 12

A woman carries a recessive X-linked allele for color blindness. Her husband has normal color vision. If they have four sons, what is the most likely outcome regarding color blindness in their male offspring?

  1. All four sons will have normal vision because the father determines male traits
  2. Approximately two sons will be color blind and two will have normal vision (correct answer)
  3. All four sons will be color blind because they inherit X chromosomes from mother
  4. One son will be color blind because X-linked traits affect only one child per family
Explanation: The woman is heterozygous (XCX^C XcX^c) and the man has normal vision (XCX^C Y). Each son receives his X chromosome from the mother and Y chromosome from the father. The mother has a 50% chance of passing either the normal vision allele (XCX^C) or color blind allele (XcX^c) to each son. With four sons, the most likely outcome is approximately 2 color blind and 2 normal, though exact ratios may vary due to chance.

Question 13

During meiosis I, a pair of homologous chromosomes fails to separate properly. If this nondisjunction event involves chromosome 21 in a human female, and the resulting egg is fertilized by a normal sperm, which condition would most likely result?

  1. Turner syndrome because the offspring lacks a complete chromosome set
  2. Klinefelter syndrome because of the extra genetic material present
  3. Down syndrome because of trisomy 21 in the developing embryo (correct answer)
  4. Monosomy 21 because one chromosome was lost during meiosis
Explanation: Nondisjunction of chromosome 21 during meiosis I in the female would produce eggs with either two copies of chromosome 21 or no copies. If an egg with two copies is fertilized by a normal sperm (which carries one copy of chromosome 21), the resulting zygote would have three copies of chromosome 21, causing Down syndrome (trisomy 21). Turner and Klinefelter syndromes involve sex chromosomes, not chromosome 21.

Question 14

A genetics counselor is working with a family where the mother has phenylketonuria (PKU), a recessive genetic disorder. The father does not have PKU but his brother does have the condition. The couple is concerned about the risk of having children with PKU.

Based on the family history provided, what is the probability that this couple's first child will have PKU?

  1. 33% because the father has a 2/3 probability of being a carrier (correct answer)
  2. 25% because both parents must be carriers for the child to be affected
  3. 0% because the father shows no symptoms of the disorder
  4. 50% because the mother will definitely pass on the recessive allele
Explanation: When you encounter genetics problems involving recessive disorders, you need to determine each parent's genotype and then calculate the probability of affected offspring using Punnett squares. Since PKU is recessive, the mother with PKU must have genotype pp (homozygous recessive). The father doesn't have PKU, so he's either PP or Pp. The key insight is using his brother's condition to determine his likely genotype. Since the father's brother has PKU (pp), both of the father's parents must have been carriers (Pp). This means the father had a 25% chance of being PP, 50% chance of being Pp, and 25% chance of being pp. Since he doesn't have PKU, we eliminate the pp possibility, leaving him with a 2/3 probability of being Pp and 1/3 probability of being PP. If the father is a carrier (2/3 chance), then Pp × pp gives 50% chance of affected children. If he's PP (1/3 chance), then PP × pp gives 0% chance. Overall probability: (2/3×1/2)+(1/3×0)=1/3=33%(2/3 × 1/2) + (1/3 × 0) = 1/3 = 33\% Choice A correctly identifies this 33% risk. Choice B incorrectly applies the standard 25% without considering the conditional probability. Choice C wrongly assumes the father cannot be a carrier just because he's asymptomatic. Choice D misunderstands that inheriting one recessive allele doesn't cause the disorder. Remember: when family history is provided, use it to refine probability calculations beyond simple Mendelian ratios.

Question 15

In a human population, the frequency of individuals with attached earlobes (recessive trait) is 16%. Assuming the population is in Hardy-Weinberg equilibrium, what percentage of the population consists of heterozygous carriers for this trait?

  1. 32% because heterozygote frequency equals 2 times the recessive allele frequency
  2. 84% because this represents all individuals without the recessive phenotype
  3. 64% because the dominant allele frequency squared gives the carrier frequency
  4. 48% because carriers represent the largest portion of most populations (correct answer)
Explanation: Hardy-Weinberg equilibrium questions test your ability to calculate allele and genotype frequencies in populations. When you see a percentage given for a recessive trait, you're working backwards from phenotype to find carrier frequency. Since 16% of the population has attached earlobes (recessive phenotype), these individuals must be homozygous recessive (qq). In Hardy-Weinberg notation, this means q² = 0.16, so q = 0.4 (40%). The dominant allele frequency is p = 1 - q = 0.6 (60%). Heterozygous carriers have frequency 2pq = 2(0.6)(0.4) = 0.48 or 48%. Let's examine why the other answers are incorrect. Choice A suggests 32%, which would be 2q (2 × 0.4), but this incorrectly uses "2 times the recessive allele frequency" instead of the proper Hardy-Weinberg formula 2pq. Choice B claims 84% represents "all individuals without the recessive phenotype" - while 84% do lack attached earlobes, this includes both heterozygotes (48%) and homozygous dominant individuals (36%), not just carriers. Choice C suggests 64%, which is actually p² (the frequency of homozygous dominant individuals), not heterozygotes. For HESI genetics problems, remember the Hardy-Weinberg formulas: p² + 2pq + q² = 1, where q² is the recessive phenotype frequency you're usually given. Always start by finding q through square root, then calculate p, and finally determine 2pq for carrier frequency. Don't confuse carriers (heterozygotes) with all non-affected individuals.

Question 16

A plant breeder crosses two heterozygous plants (AaBbCc × AaBbCc) and wants to determine the probability of obtaining offspring with the genotype AABBcc. How many different types of gametes can each parent produce, and what is the probability of the desired offspring?

  1. Each parent produces 6 gamete types; probability of AABBcc is 1/32
  2. Each parent produces 4 gamete types; probability of AABBcc is 1/64
  3. Each parent produces 8 gamete types; probability of AABBcc is 1/16
  4. Each parent produces 8 gamete types; probability of AABBcc is 1/64 (correct answer)
Explanation: When you encounter genetics problems involving multiple heterozygous traits, you need to apply the principles of independent assortment to determine both gamete variety and offspring probability. For gamete types, each heterozygous parent (AaBbCc) can contribute either the dominant or recessive allele for each of the three genes. Using the formula 2n2^n where n = number of heterozygous gene pairs, each parent produces 23=82^3 = 8 different gamete types (ABC, ABc, AbC, Abc, aBC, aBc, abC, abc). To find the probability of AABBcc offspring, calculate the probability for each gene separately, then multiply them together. For AA: when crossing Aa × Aa, the probability of AA is 1/4. For BB: when crossing Bb × Bb, the probability of BB is 1/4. For cc: when crossing Cc × Cc, the probability of cc is 1/4. Therefore, the probability of AABBcc is 14×14×14=164\frac{1}{4} × \frac{1}{4} × \frac{1}{4} = \frac{1}{64}. Answer A incorrectly calculates 6 gamete types, likely by adding rather than using the exponential rule. Answer B correctly identifies the probability as 1/64 but miscalculates gamete types as 4, which would only apply to a dihybrid cross. Answer C correctly identifies 8 gamete types but calculates probability as 1/16, which represents the probability for a two-gene cross. Remember: for n heterozygous genes, gamete types = 2n2^n, and probability calculations require multiplying individual gene probabilities. This pattern appears frequently on genetics questions.

Question 17

In humans, brown eyes (B) are dominant over blue eyes (b), and the ability to roll the tongue (R) is dominant over the inability to roll the tongue (r). A man who is heterozygous for both traits marries a woman who has blue eyes and cannot roll her tongue. What percentage of their offspring would be expected to have brown eyes and be able to roll their tongue?

  1. 12.5% because only one combination produces this phenotype
  2. 25% because each trait segregates independently with equal probability (correct answer)
  3. 50% because the man is heterozygous for both traits
  4. 75% because both brown eyes and tongue rolling are dominant
Explanation: The man's genotype is BbRr and the woman's genotype is bbrr. Using a dihybrid cross, the man can produce four types of gametes (BR, Br, bR, br) each with 25% probability. The woman can only produce br gametes. The offspring with brown eyes and tongue rolling ability must have genotype BbRr, which occurs when the man contributes a BR gamete (25% chance) and the woman contributes br (100% chance). Therefore, 25% of offspring will have this phenotype.

Question 18

During the process of translation, what is the primary function of transfer RNA (tRNA)?

  1. To carry genetic information from the nucleus to the cytoplasm for protein synthesis.
  2. To form the primary structural and catalytic components of the ribosome.
  3. To act as the template that dictates the sequence of amino acids in a polypeptide.
  4. To carry a specific amino acid to the ribosome and match its anticodon to the mRNA codon. (correct answer)
Explanation: When you encounter questions about protein synthesis, focus on the specific roles of each type of RNA molecule and where each step occurs in the cell. Transfer RNA (tRNA) serves as the crucial link between the genetic code and amino acids during translation. Each tRNA molecule has two key features: an anticodon region that reads the mRNA sequence, and an amino acid attachment site that carries one specific amino acid. During translation at the ribosome, tRNA molecules bring their attached amino acids and match their anticodons to complementary codons on the mRNA strand. This precise matching ensures amino acids are assembled in the correct sequence to form the polypeptide chain. Choice A describes messenger RNA (mRNA), not tRNA. mRNA carries the genetic information from DNA in the nucleus to ribosomes in the cytoplasm. Choice B describes ribosomal RNA (rRNA), which forms the structural framework of ribosomes and catalyzes peptide bond formation between amino acids. Choice C also refers to mRNA's function as the template that determines amino acid sequence based on its codon sequence. The key distinction is that tRNA doesn't carry genetic information or form ribosomal structure—it serves as the translator that reads the genetic code and delivers the corresponding building blocks. Think of tRNA as a delivery truck that brings the right amino acid to the right address (codon) on the mRNA. For HESI questions on molecular biology, remember that each RNA type has a distinct function: mRNA carries the message, rRNA builds the factory, and tRNA delivers the materials.

Question 19

After a chromosome replicates in preparation for cell division, it consists of two identical DNA helices joined at a centromere. What are these two identical joined structures correctly called?

  1. Homologous chromosomes
  2. Sister chromatids (correct answer)
  3. Daughter chromosomes
  4. Gametes
Explanation: This question tests your understanding of chromosome structure during cell division, specifically the terminology used to describe replicated chromosomes before they separate. When a chromosome replicates in preparation for cell division (mitosis or meiosis), it creates an exact copy of itself. These two identical DNA molecules remain attached at a region called the centromere, forming an X-shaped structure. Each of these identical attached copies is called a chromatid, and because they're identical copies joined together, they're specifically termed sister chromatids. This makes B the correct answer. Let's examine why the other options are incorrect. A) Homologous chromosomes are matching pairs of chromosomes (one from each parent) that carry the same genes but may have different versions (alleles) of those genes - they're similar but not identical copies. C) Daughter chromosomes refer to the individual chromatids after they've separated and moved to opposite poles of the cell during division - this happens after the stage described in the question. D) Gametes are reproductive cells (sperm and egg) that result from meiosis and contain half the normal chromosome number. Remember this key distinction: sister chromatids are identical copies still joined together, while homologous chromosomes are similar but not identical pairs. On the HESI, cell division questions often test whether you can differentiate between these chromosome relationships, so focus on understanding what makes structures identical versus similar, and whether they're joined or separated.

Question 20

What is a fundamental genetic difference between the daughter cells produced by meiosis and those produced by mitosis?

  1. Meiotic cells are diploid and genetically different; mitotic cells are haploid and identical.
  2. Meiotic cells are haploid and genetically identical; mitotic cells are diploid and different.
  3. Meiotic cells are diploid and identical; mitotic cells are diploid and identical.
  4. Meiotic cells are haploid and genetically different; mitotic cells are diploid and identical. (correct answer)
Explanation: When you encounter questions about cell division, focus on two key distinctions: chromosome number (diploid vs. haploid) and genetic variation (identical vs. different). Meiosis produces gametes (sex cells) that are fundamentally different from the parent cell in two ways. First, meiotic daughter cells are haploid, containing only one copy of each chromosome rather than the diploid pair found in somatic cells. This reduction is essential for sexual reproduction—when two haploid gametes fuse during fertilization, they restore the diploid chromosome number. Second, meiotic cells are genetically different from each other due to crossing over and independent assortment, which shuffle genetic material to create new combinations. Mitosis, conversely, produces somatic cells for growth and repair. These daughter cells are diploid (same chromosome number as the parent) and genetically identical to ensure consistent cellular function throughout the organism. Looking at the wrong answers: Choice A incorrectly reverses the ploidy levels, stating meiotic cells are diploid when they're actually haploid. Choice B correctly identifies meiotic cells as haploid but wrongly claims they're genetically identical—genetic variation is meiosis's key feature. Choice C incorrectly states that meiotic cells are diploid, missing the fundamental purpose of meiosis to reduce chromosome number. Remember this pattern: Meiosis makes "different and half" (genetically different, haploid), while mitosis makes "same and same" (genetically identical, diploid). This distinction frequently appears on the HESI, so memorize the meiosis/mitosis differences as a foundational concept for genetics and reproduction questions.