Genetics, Heredity and Evolution Lesson Test

Answer: B

Correct answer: DNA. Genes are stretches of DNA, and the order of its four bases (A, T, C and G) works like a code that cells read to build proteins. Almost all human DNA sits in the nucleus as 46 chromosomes, with a small amount in the mitochondria. ATP is tempting because it also contains adenine, a sugar and phosphate, but it is a short-lived energy carrier, not a long-term information store. A useful hook: DNA is the instruction manual, and ATP is the battery.

• A: Starch is an energy-storage carbohydrate in plants, not the hereditary molecule.
• B (correct): DNA stores hereditary information in its base sequence.
• C: ATP mainly transfers usable energy; it is not the main hereditary archive.
• D: Cholesterol is a lipid with structural and other roles, not the hereditary sequence.

Answer: D

Correct answer: A version of a gene or DNA sequence at a locus. A locus is a specific address on a chromosome, and the gene at that address can come in different versions, such as the A, B and O alleles of the ABO blood-group gene. Because chromosomes come in pairs, a person carries two alleles for most genes, one from each parent. Alleles are often confused with chromosomes; one chromosome holds many genes, while an allele is just one variant of a single gene.

• A: A ribosome is a structure that performs protein synthesis, not a version of a gene.
• B: A nucleus contains genetic material but is not a version of a gene.
• C: Homologous chromosomes are whole chromosomes carrying many loci; an allele is one variant at a single locus.
• D (correct): An allele is a variant at a genetic location.

Answer: D

Correct answer: Thymine. The two DNA strands are held together by hydrogen bonds between bases, and the bases fit only in specific pairs: adenine with thymine (two hydrogen bonds) and guanine with cytosine (three). A larger base (A or G) always pairs with a smaller one (T or C), keeping the helix an even width. Uracil is the common trap: it pairs with adenine in RNA, but DNA uses thymine instead. A memory hook is that the straight-lined letters A and T go together.

• A: Cytosine pairs with guanine.
• B: Uracil replaces thymine in RNA, not normal DNA pairing.
• C: Guanine pairs with cytosine in DNA.
• D (correct): DNA follows A–T complementary pairing.

Answer: A

Correct answer: Bb. Hetero- means different and homo- means same, so a heterozygous genotype has two different alleles at one locus, usually written as one capital and one lowercase letter. Each parent contributed one of the two. AA, BB and bb are all homozygous because both letters match, whether both alleles are dominant or both are recessive. A common slip is thinking the particular letter matters; any matched pair is homozygous, and any mixed pair is heterozygous.

• A (correct): Two different alleles at the locus make the genotype heterozygous.
• B: Two A alleles are identical: homozygous.
• C: Two b alleles are identical: homozygous.
• D: Two B alleles are identical: homozygous.

Answer: C

Correct answer: bb. Under complete dominance, a single dominant allele is enough to produce the dominant phenotype because it masks the recessive one. So BB and Bb look the same, and the recessive trait appears only when no dominant allele is present, which requires two copies of b. The tempting mistake is choosing Bb because it contains the recessive allele; carrying an allele is not the same as expressing it, and Bb individuals are called carriers.

• A: Both alleles are dominant, producing the dominant phenotype.
• B: A dominant B allele masks the recessive phenotype in this model.
• C (correct): The recessive phenotype requires two recessive alleles under complete dominance.
• D: This is not a specified diploid genotype at the modeled autosomal locus.

Answer: D

Correct answer: 23. Human body cells are diploid, with 46 chromosomes arranged in 23 pairs, one set from each parent. Meiosis halves this number, so each sperm or egg is haploid with 23 single chromosomes, one from each pair. At fertilization, 23 + 23 restores the 46 a new person needs. Choosing 46 is the usual error: if gametes carried 46, the chromosome count would double every generation.

• A: 46 is the ordinary diploid human somatic count.
• B: A gamete is haploid and does not carry 46 pairs.
• C: 92 is not the normal count in a human sperm cell.
• D (correct): Human gametes normally contain 23 chromosomes.

Answer: B

Correct answer: Meiosis. Meiosis has two rounds of division after a single round of DNA copying, so one diploid cell yields four haploid cells, each with one set of chromosomes. It also shuffles chromosomes and allows crossing over, which makes each gamete genetically unique. Mitosis is the tempting distractor because it is also cell division, but it makes two cells identical to the parent cell, as needed for growth and tissue repair.

• A: Mitosis usually preserves chromosome set rather than creating haploid gametes.
• B (correct): Meiosis reduces chromosome sets during formation of reproductive cells.
• C: Photosynthesis captures light energy, not chromosome reduction.
• D: Osmosis is water movement across a semipermeable membrane.

Answer: D

Correct answer: Phenotype. Genotype is the set of alleles an organism carries, while phenotype is what can actually be observed or measured, such as eye color, height or blood type. Phenotype comes from genotype interacting with the environment, so two people with the same genes can differ in weight because of diet. The common mix-up is between these two terms; a memory hook is that phenotype is the physical result, while genotype refers to the genes.

• A: A nucleotide is a molecular building block of nucleic acids.
• B: A chromosome packages genetic material.
• C: Genotype identifies the genetic variants present.
• D (correct): Phenotype is an observable trait and may depend on genes and environment.

Answer: D

Correct answer: Homologous structures. Features with the same underlying bone plan that now serve different purposes are homologous. They point to a common ancestor, which is why they count as evidence for evolution.

• A: Analogous structures do similar jobs but evolved separately, such as bird and butterfly wings; these limbs do different jobs and share ancestry.
• B: Acquired traits develop during one organism's life and are not inherited, unlike these inherited bone patterns.
• C: Codominance describes two alleles both showing in a heterozygote, not shared body structures among species.
• D (correct): These share an underlying design inherited from a common ancestor, even though their jobs now differ.

Answer: B

Correct answer: Chromosome. Each chromosome is one long DNA molecule wound around proteins called histones, which lets roughly two meters of DNA fit inside a tiny nucleus. Genes are segments along that molecule, and a single human chromosome can carry hundreds to thousands of them. Ribosome is tempting because it works with genetic information, but it only reads mRNA copies to build proteins and holds no genes. Think of a chromosome as a book and its genes as chapters.

• A: A cell wall is a supportive boundary, not a gene carrier.
• B (correct): Chromosomes package long DNA molecules with many genes.
• C: Vacuoles are cellular compartments, not packages of many genes.
• D: Ribosomes translate RNA into protein.

Answer: A

Correct answer: PP × pp. Each parent passes one allele to each offspring. A homozygous PP parent can give only P, and a homozygous pp parent can give only p, so every offspring must be Pp and purple. Any cross with a heterozygous Pp parent leaves room for chance, because that parent gives P half the time and p half the time. The trap is assuming any cross containing both letters works; only two opposite homozygous parents guarantee all heterozygous offspring.

• A (correct): One parent always supplies P and the other always supplies p.
• B: This yields PP and Pp, not 100% Pp.
• C: This produces PP, Pp and pp in a 1:2:1 genotype ratio.
• D: This yields only pp offspring.

Answer: B

Correct answer: 25%. Each Aa parent passes A or a with equal chance, so the chance that both pass a is

A Punnett square shows the same thing: of four equally likely boxes, one is aa. The common mistake is mixing up genotype and phenotype probabilities: 75% is the chance of showing the dominant trait, and 50% is the chance of being an Aa carrier. Each child is an independent event, so the 25% applies to every birth.

• A: 50% is the Aa genotype probability.
• B (correct): The four equiprobable combinations are AA, Aa, Aa and aa.
• C: Each parent can pass a, so aa is possible.
• D: 75% is the dominant phenotype probability in this model.

Answer: D

Correct answer: 50%. This is a test cross: the aa parent can pass only a, so each offspring's phenotype depends entirely on which allele the Aa parent gives. That parent gives A or a with equal chance, so half the offspring are Aa and show the dominant trait, and half are aa and show the recessive trait, a 1:1 ratio. The tempting answer of 25% comes from the Aa × Aa cross, where both parents must give a.

• A: 25% is the aa chance in Aa × Aa, a different cross.
• B: The heterozygous parent can transmit a.
• C: Half can be Aa and express the dominant phenotype.
• D (correct): Aa supplies A or a while aa always supplies a; half are aa.

Answer: B

Correct answer: A–C–G–T. Base pairing is fixed by shape and hydrogen bonding: A always pairs with T, and G always pairs with C. To find the opposite strand, replace each base with its partner one position at a time, so T becomes A, G becomes C, C becomes G and A becomes T. Frequent mistakes are copying the original letters or slipping in uracil, which appears only in RNA. In a real double helix, the two strands also run in opposite directions.

• A: Those letters do not complement the stated positions.
• B (correct): T pairs with A, G with C, C with G and A with T.
• C: Uracil belongs to RNA, and several pairs are wrong.
• D: Copying the same bases does not apply complementarity.

Answer: C

Correct answer: RNA uses uracil instead of thymine. DNA and RNA share adenine, guanine and cytosine, but RNA uses uracil where DNA uses thymine, and its sugar is ribose rather than deoxyribose. During transcription, each adenine in the DNA template is matched with uracil in the new RNA. The claim that every gene codes for a protein is tempting but false: some genes produce RNAs, such as transfer RNA and ribosomal RNA, that do their jobs without being translated.

• A: Ribosomes conduct translation, not the cell wall.
• B: DNA replication is performed by cellular enzymes, not the Golgi apparatus.
• C (correct): Standard RNA uses A, U, C and G.
• D: Some genes generate functional noncoding RNAs.

Answer: D

Correct answer: One-half. A Punnett square for Dd × Dd gives four equally likely outcomes: DD, Dd, Dd and dd. Two of the four are heterozygous, so 2/4, or one-half, are carriers who have one d allele but do not show the condition. One-fourth are DD with no d allele, and one-fourth are dd and affected. The most tempting wrong answer is three-fourths, which counts all unaffected children, including DD children who are not carriers.

• A: Both DD and dd outcomes remain possible.
• B: One-fourth is the probability of each homozygous genotype.
• C: Three-fourths refers to the unaffected phenotype under full dominance, including DD.
• D (correct): Dd appears in two of four Punnett combinations and is an unaffected carrier in the stated model.

Answer: D

Correct answer: His daughters, not his sons. A father has one X and one Y chromosome, and each sperm carries one or the other. A daughter must receive his X to be XX, while a son must receive his Y to be XY. So every daughter gets his X-linked variant, and no son gets it from him. The common error is assuming all children have equal odds, as with autosomal genes; an X-linked variant never passes directly from father to son.

• A: A son ordinarily receives the father’s Y rather than X.
• B: Daughters normally receive a paternal X.
• C: Paternal X inheritance differs from autosomal patterns.
• D (correct): Fathers generally transmit X to daughters and Y to sons.

Answer: C

Correct answer: Crossing over. Early in meiosis I, homologous chromosomes pair up and trade matching segments between their chromatids. A chromosome that came from the mother can then carry some alleles from the father's copy, creating combinations that neither original chromosome had. Independent assortment, the random lining up of homologous pairs, also mixes alleles, but none of the other choices involve chromosomes at all: osmosis and digestion are unrelated processes, and learned behaviors do not change DNA.

• A: Osmosis transports water and does not reshuffle homologous DNA.
• B: Learning does not directly rearrange allele combinations in gametes.
• C (correct): Exchange between homologous chromosome chromatids creates new combinations.
• D: Digestion breaks down nutrients and is unrelated to meiotic recombination.

Answer: A

Correct answer: 50%. In incomplete dominance, neither allele fully masks the other, so the heterozygote has an in-between phenotype. The Rr × Rr cross gives a 1 RR : 2 Rr : 1 rr genotype ratio, and because each genotype has its own color, the phenotype ratio is 1 red : 2 pink : 1 white. A tempting error is applying the 3:1 ratio from complete dominance and answering 75%; that ratio holds only when the heterozygote looks like one parent.

• A (correct): Half the offspring are Rr and show the intermediate pink phenotype.
• B: Both homozygous outcomes also occur.
• C: One-fourth RR are red and one-fourth rr white.
• D: Three-fourths is not the pink proportion under incomplete dominance.

Answer: C

Correct answer: A and B blood-group alleles both contribute to type AB. In codominance, both alleles in a heterozygote are fully expressed side by side rather than blended. A person with type AB blood makes both the A and the B markers on red blood cells. This is easy to confuse with incomplete dominance, where the heterozygote is intermediate, like pink flowers from red and white parents. A memory hook: co- means together, so both traits show at once.

• A: That describes complete dominance.
• B: This is not an inherited two-allele interaction.
• C (correct): AB blood group illustrates coexpression of both alleles.
• D: That is a simplified incomplete-dominance pattern.

Answer: B

Correct answer: Many genes and environmental factors contribute to variation. Height is polygenic: a large number of gene variants each add a small effect, which is why adult heights form a smooth, bell-shaped range rather than a few distinct categories. Environment matters too, because childhood nutrition and health influence growth; average heights in many countries rose during the 1900s as diets improved. The idea that height follows one simple recessive gene fails because single-gene traits produce clear-cut categories, not continuous variation.

• A: Inheritance and environment vary among siblings.
• B (correct): Height is polygenic and influenced by environment.
• C: Nutrition and health can affect growth.
• D: Real height does not follow one simple recessive locus.

Answer: A

Correct answer: It is usually a somatic variant and not directly inherited. Only changes in germ-line cells, the cells that give rise to eggs or sperm, can be passed to children. Ultraviolet light damages DNA in skin cells, and a resulting mutation is copied only into the skin cells that descend from the damaged one. Such somatic changes still matter for the person, for example by contributing to skin cancer. The common misconception is that any DNA change in a parent will appear in offspring.

• A (correct): A change limited to skin tissue normally does not enter offspring’s inherited DNA.
• B: Somatic acquired DNA variation alone does not prove heredity.
• C: Skin-cell changes are somatic and not routinely transmitted by gametes.
• D: An individual somatic variant does not alter an entire population.

Answer: B

Correct answer: A-A-C-U-G. In transcription each template base pairs with its partner, except that RNA uses uracil in place of thymine: T gives A, T gives A, G gives C, A gives U and C gives G. The messenger RNA is A-A-C-U-G.

• A: This pairs the bases correctly but uses thymine, which RNA does not contain.
• B (correct): Pairing each template base gives A, A, C, U, G, with uracil placed opposite adenine.
• C: This simply repeats the template strand rather than building its complementary RNA.
• D: This copies the template and only swaps T for U, instead of pairing each base with its partner.

Answer: A

Correct answer: Genetic drift through a bottleneck. Ask whether the trait affected survival. Because the fire killed beetles at random, the rise of the green allele is a chance effect of a sudden population crash, which is the bottleneck form of genetic drift.

• A (correct): A disaster cut the population sharply and killed beetles at random, so the survivors' allele mix changed by chance.
• B: Mutations do not appear because organisms need them, and the green allele already existed before the fire.
• C: No beetles moved in from elsewhere; the population is isolated and the change came from who survived.
• D: Selection would require green shells to improve survival, but the fire killed beetles regardless of color.

Answer: C

Correct answer: A few bacteria already carried resistance, survived the drug and multiplied. This is natural selection. Resistance mutations were already present in a few cells by chance; the antibiotic killed the susceptible bacteria, and the resistant survivors reproduced until they made up most of the population.

• A: This reverses the process; the drug kills susceptible bacteria and leaves resistant ones.
• B: Evolution is a change in a population over generations, not a change within one individual.
• C (correct): The drug removed susceptible cells, so resistant cells that existed by chance left most of the offspring.
• D: Mutations arise randomly, not on demand; the antibiotic sorted existing variants rather than creating them.

Answer: A

Correct answer: Over 30 generations, dark-winged moths rise from 10 to 60 percent of a population. Evolution is a shift in a population's inherited traits from one generation to the next. Only the moth example tracks a heritable trait becoming more common in a population over many generations; the others are changes within a single individual's life.

• A (correct): This is a change in the inherited makeup of a population across generations, which is what evolution means.
• B: Muscles built by training are acquired during one life and are not inherited.
• C: This is a short-term change in one individual, and individuals do not evolve.
• D: Tanning is a short-term adjustment in one person, not an inherited change in a population.

Answer: C

Correct answer: A mutation can be neutral, harmful or advantageous depending on context. A mutation is any change in a DNA sequence. Many occur in regions that do not affect proteins or swap a codon for one that codes for the same amino acid, so they are neutral. Others disrupt a needed protein and cause harm, while a few help, such as the sickle-cell allele, which gives one-copy carriers resistance to malaria. The misconception is that mutations are always bad or appear because an organism needs them.

• A: Mutations are not necessarily beneficial.
• B: Some variants are harmless, neutral or context-dependent.
• C (correct): The effect of a DNA change varies by location and environment.
• D: Variants are not created by intention.

Answer: B

Correct answer: 5. Diploid cells carry two sets of chromosomes, one from each parent, while haploid cells carry one set. Meiosis separates the members of each homologous pair, so the gamete count is always half the diploid number:

Fertilization then joins two gametes of 5 to restore 10. A common slip is keeping the count at 10, as mitosis would, or doubling it; the human version of the same rule is 46 halved to 23.

• A: This doubles rather than halves the diploid count.
• B (correct): The haploid set is half the diploid count of 10.
• C: This preserves the diploid count and ignores reduction.
• D: Haploid does not mean two chromosomes.

Answer: D

Correct answer: Meiosis I. In meiosis I, homologous chromosomes pair up and are pulled to opposite poles, so each daughter cell gets one member of each pair and becomes haploid; this is why meiosis I is called the reduction division. Meiosis II then splits the sister chromatids of each chromosome, much like mitosis. The tempting error is mixing up the two stages; remember that homologs separate first and sister chromatids second, matching the order of the numerals.

• A: Replication copies DNA before division, not separates chromosomes.
• B: Sister chromatids normally separate in meiosis II.
• C: Translation reads mRNA to build proteins.
• D (correct): Homologous chromosomes separate during meiosis I.

Answer: B

Correct answer: Its fitness effect may differ by environment. Fitness is not a fixed property of an allele; it depends on how the resulting trait performs under local conditions such as temperature, food or predators. A protein that works better in the cold may be neutral or even costly in warm habitats. The tempting error is to treat a benefit seen in one setting as universal. Sickle-cell trait is a classic example: it helps against malaria where the disease is common but offers no advantage elsewhere.

• A: Germline variants can be transmitted to offspring.
• B (correct): A variant helpful in one context can be neutral or harmful elsewhere.
• C: Dominance describes allelic expression, not environmental fitness.
• D: Selective advantage can depend on the environment.

Answer: A

Correct answer: 2/3. This is a conditional probability: once you know the offspring shows the dominant phenotype, the aa outcome is ruled out. The Punnett square gives AA, Aa, aA and aa, each equally likely; removing aa leaves three equal outcomes, two of which are heterozygous. The common mistake is answering 1/2, the chance of Aa among all offspring, which ignores the information that the phenotype is already known. Always shrink the sample space to what is observed.

• A (correct): Among AA, Aa and Aa dominant outcomes, two of three are heterozygous.
• B: Half is the unconditional Aa probability before conditioning on phenotype.
• C: Three-fourths is the unconditional dominant-phenotype chance.
• D: One-fourth is the unconditional AA or aa chance, not the conditional Aa probability.

Answer: C

Correct answer: 1/16. When genes assort independently, you can solve each gene as its own monohybrid cross and then multiply. Aa × Aa gives aa one time in four, and Bb × Bb gives bb one time in four, so both happen together with probability 1/4 × 1/4. This matches the single double-recessive box in the 16-box dihybrid Punnett square. A frequent slip is stopping at 1/4 for one gene or confusing a genotype probability with the 9:3:3:1 phenotype classes.

• A: This corresponds to one single-recessive-class phenotype in a basic dihybrid cross.
• B: One-fourth is the aa chance alone or bb chance alone.
• C (correct): Under independence, P(aa)×P(bb) = $\frac{1}{4}$×$\frac{1}{4}$ = $\frac{1}{16}$.
• D: This is not the product of the two separate $\frac{1}{4}$ probabilities.

Answer: C

Correct answer: 55%. Allele frequency counts gene copies, not individuals. Each diploid organism carries two copies, so 20 organisms hold 40 alleles. AA individuals contribute two A copies each (16), Aa individuals contribute one each (6), and aa contribute none, giving 22 of 40. The tempting shortcut is counting the share of individuals that carry A, which treats heterozygotes as if they had two A copies. Always multiply by two for homozygotes and by one for heterozygotes.

• A: AA+Aa fraction is not the same as allele-A frequency.
• B: 40% is the fraction of individuals with AA genotype, not allele-A frequency.
• C (correct): There are 40 allele copies and 22 A copies: (2×8+6)/40 = 55%.
• D: 30% counts Aa individuals, not A alleles.

Answer: A

Correct answer: 1/16. Each pregnancy is a separate, independent event: the parents' gametes do not remember earlier outcomes. For two Aa parents, each child has a 1/4 chance of being aa, and the probability that two independent events both occur is the product of their probabilities. The common mistake is adding probabilities, which applies to either-or questions, not both-and questions. A helpful rule: 'and' usually means multiply, while 'or' for mutually exclusive outcomes means add.

• A (correct): Multiply $\frac{1}{4}$ times $\frac{1}{4}$ for two independent aa outcomes.
• B: This is not the probability of two independent quarter-probability events.
• C: This is the chance of one aa outcome, not both.
• D: Adding the two separate $\frac{1}{4}$ probabilities is not correct for both events.

Answer: B

Correct answer: Pp. This is the logic of a testcross: crossing an unknown dominant-phenotype individual with a homozygous recessive reveals its hidden allele. Every white offspring is pp, and since the white parent can only give p, the other p must have come from the purple parent. A PP parent could give only P, so all offspring would be purple. The tempting idea that the genotype stays unknown fails because a single recessive offspring is decisive evidence of heterozygosity.

• A: pp itself expresses white, not purple, under the model.
• B (correct): A white pp child requires one p allele from each parent, so the purple parent is Pp.
• C: PP cannot pass a p allele, so cannot produce pp with a pp parent.
• D: The observed pp child resolves the purple parent’s second allele in the simplified model.

Answer: A

Correct answer: 75 percent. Count offspring for each group, then find the green share of the total.

Green rises from 60 to 75 percent in one generation through natural selection.

• A (correct): Green offspring are 60 times 3, or 180, out of 180 plus 60, or 240 in all, and 180 divided by 240 is 75 percent.
• B: This is the starting share of green beetles, which ignores the difference in offspring numbers.
• C: This compares only the per-beetle rates, 3 out of 4.5, and ignores how many beetles of each color there were.
• D: This is the brown share of the next generation, 60 out of 240, not the green share.

Answer: C

Correct answer: Incorrect: dominance describes heterozygote phenotype, not fitness or frequency. Dominance only tells you which allele's trait shows when an individual carries one copy of each. Whether an allele becomes common depends on selection, mutation, migration and chance events such as genetic drift. Huntington disease is caused by a dominant allele yet is rare and harmful, while many recessive alleles are common. The tempting belief that dominant means stronger or more widespread confuses how a trait is expressed with how well it helps survival and reproduction.

• A: Dominance does not establish harm either; many dominant alleles are neutral or beneficial.
• B: Autosomal or sex-chromosome location does not turn dominance into evidence of fitness.
• C (correct): Dominance tells which allele's trait shows in a heterozygote; it says nothing about survival advantage or how common the allele is.
• D: Dominance does not guarantee higher reproductive success, so a dominant allele need not spread.

Answer: D

Correct answer: Frogs and humans share a more recent common ancestor than either shares with sharks. Find where branches join. Frogs and humans meet at the most recent split, sharks join them one split earlier, and lampreys join at the oldest split. Species alive today are cousins on different branches, so none is the ancestor of another.

• A: Living species are relatives on separate branches, not ancestors of one another.
• B: Modern lampreys sit on their own branch; the tree shows a shared ancestor, not lampreys as ancestors.
• C: Frogs and humans join the shark line at the same split, so sharks are equally related to both; habitat does not determine relatedness on a tree.
• D (correct): Frogs and humans join at the last split, so their shared ancestor is more recent than the one they share with sharks.

Answer: A

Correct answer: 8 chromosomes; separation of homologous chromosomes in meiosis I. A diploid cell has two sets of chromosomes, and a gamete should carry one set, so 2n = 16 gives n = 8. The halving happens in meiosis I, when each homologous pair splits and one member goes to each daughter cell. Meiosis II separates sister chromatids, which resembles mitosis and does not change the set number. A common trap is thinking DNA replication or meiosis II causes the reduction.

• A (correct): Half of 16 is 8; reduction occurs as homologous chromosome pairs segregate.
• B: Fertilization restores a diploid count and does not halve 16 to 4.
• C: Replication doubles DNA amount but does not reduce chromosome sets.
• D: A gamete should be haploid, and homologous segregation reduces sets.

Answer: C

Correct answer: Linked loci may not assort independently, so the classic ratio may not apply. The 9:3:3:1 ratio depends on Mendel's law of independent assortment, which holds for genes on different chromosomes or far apart on one chromosome. Genes very close together tend to be inherited as a unit because crossing over between them is rare, so parental combinations appear more often than expected. The tempting fix of switching to another fixed ratio fails; the actual ratio depends on recombination frequency.

• A: Linkage alone does not guarantee this alternative ratio.
• B: Dominance alone does not establish independent assortment.
• C (correct): The standard ratio assumes independence and complete dominance for both traits.
• D: Many genes occur on the same chromosome.

Answer: B

Correct answer: By how many surviving offspring it leaves in its environment. Fitness means reproductive success in a particular environment, counted by surviving offspring. A small, plain animal that raises many young has higher fitness than a large, strong one that raises none.

• A: Fitness in evolution is not physical strength or speed; a strong, fast animal that leaves no offspring has low fitness.
• B (correct): Evolutionary fitness is reproductive success: the number of surviving offspring compared with others in the same setting.
• C: Size can help or hurt depending on the environment, so it is not the measure of fitness itself.
• D: Living long without reproducing adds nothing to the next generation, so lifespan alone does not measure fitness.

Answer: A

Correct answer: G-C-T-A-A-C. In DNA, A pairs with T and G pairs with C. Working base by base: C gives G, G gives C, A gives T, T gives A, T gives A, G gives C, so the partner strand is G-C-T-A-A-C.

• A (correct): Each base is replaced by its partner: C with G, G with C, A with T and T with A.
• B: This uses uracil, which appears only in RNA; a DNA strand pairs A with T.
• C: This simply repeats the original strand instead of pairing each base with its partner.
• D: This pairs C with A and G with T, mixing up the two base-pairing rules.

Answer: C

Correct answer: 1/2, because he gets one of his mother's two X chromosomes. A son receives his Y from his father and one X from his mother. The carrier mother has one normal X and one X with the allele, so each son has a 1/2 chance of getting the allele, and with only one X he will be color-blind.

• A: A son gets his X from his mother, not his father, so the father's normal vision does not protect him.
• B: One-fourth is the chance that a child of either sex is an affected son, not the chance among sons only.
• C (correct): Each son receives one of the mother's two X chromosomes, and one of the two carries the allele; with only one X, he shows it.
• D: The mother passes her normal X just as often as the X with the allele, so not every son is affected.

Answer: C

Correct answer: 1/8. Work each gene separately, then multiply. Yy × yy gives 1/2 Yy (yellow), and Rr × Rr gives 1/4 rr (wrinkled).

• A: This is the yellow, wrinkled share from YyRr crossed with YyRr; the second parent here is yy, not Yy.
• B: This is the fraction that are yellow and round: 1/2 times 3/4.
• C (correct): Yy times yy gives 1/2 yellow, and Rr times Rr gives 1/4 wrinkled; 1/2 times 1/4 equals 1/8.
• D: This treats yellow as a 1/4 outcome as well, but Yy crossed with yy gives yellow half the time.

Answer: A

Correct answer: Two identical cells, each with 12 chromosomes. Mitosis is used for growth and repair. The DNA is copied first and then split evenly, so one cell becomes two identical cells that keep the full diploid number of 12.

• A (correct): Mitosis keeps the chromosome number the same and produces two genetically identical daughter cells.
• B: This describes meiosis, which halves the number and makes gametes, not repair cells.
• C: Mitosis does not halve the chromosome number; halving happens only in meiosis I.
• D: Mitosis is a single division producing two cells, not four.

Answer: D

Correct answer: Gene flow. When individuals migrate and breed in another population, they carry their alleles with them. This movement of alleles between populations is gene flow, and it tends to make neighboring populations more alike.

• A: Drift is a random change in allele frequencies within a population, not alleles arriving from another population.
• B: Mutation creates brand-new alleles by changing DNA; here existing alleles are simply moving in.
• C: Selection depends on a trait affecting survival or reproduction, which the stem does not describe.
• D (correct): Alleles moving from one population into another through migrating, breeding individuals is gene flow.

Answer: B

Correct answer: Founder effect. When a few individuals start a new, isolated population, their alleles, including rare ones, can be overrepresented by chance. This is the founder effect, a form of genetic drift.

• A: A bottleneck needs a disaster that sharply cuts an existing population, and the stem rules that out.
• B (correct): A small group starting a new, isolated population can carry rare alleles at a high frequency purely by chance.
• C: The allele gives no survival advantage, so selection is not what raised its frequency.
• D: Gene flow needs alleles moving between populations, but this community stayed isolated.

Answer: D

Correct answer: Analogous structures, showing similar solutions evolved independently. Ask whether the similarity comes from shared ancestry or from a shared job. Bat and fly wings share only their function, not their construction, so they are analogous structures.

• A: Homologous structures share an inherited underlying design; a bat wing and a fly wing do not.
• B: Similar function without shared design does not point to recent common ancestry.
• C: Traits gained through use during life are not inherited, and this does not describe how wings arose.
• D (correct): Features that do a similar job but evolved separately, without a shared design, are analogous.

Answer: D

Correct answer: 45 percent. There are 100 plants, so 200 gene copies. Each rr plant has two r alleles and each Rr plant has one.

• A: This is the share of individuals that are rr, a genotype frequency rather than an allele frequency.
• B: This is the frequency of the R allele, not the r allele.
• C: This counts every plant carrying at least one r, instead of counting r alleles out of all gene copies.
• D (correct): There are 90 r alleles out of 200 gene copies, which is 0.45.

Answer: A

Correct answer: 1/2. The cross is Hh × hh, so half the boxes are Hh and half are hh. Every fertilization is independent, so earlier children do not change the odds: each new child has a 1/2 chance of being Hh and affected.

• A (correct): Each child independently receives either H or h from the father, so the chance stays 1/2 no matter how earlier children turned out.
• B: This multiplies 1/2 for three births in a row, but the question asks only about the next child.
• C: One-fourth is the chance of an affected child when two carriers of a recessive allele have children, not for this dominant cross.
• D: Three-fourths is the dominant-phenotype share from Hh crossed with Hh; here the mother is hh.

Answer: C

Correct answer: Transcription. Gene expression runs DNA to RNA to protein. The first step, making messenger RNA from DNA, is transcription; the second, building a protein from the RNA, is translation.

• A: Translation is the next step, when a ribosome reads the RNA codons and builds a chain of amino acids.
• B: Replication copies DNA into new DNA before a cell divides, not into RNA.
• C (correct): Transcription copies a gene's DNA sequence into messenger RNA, in the nucleus of eukaryotic cells.
• D: Crossing over is an exchange of segments between homologous chromosomes during meiosis.