Biology — Semester B Practice

Question 1 of 10

TEKS 1B-8BEasy Image

For AA × aa, what is the genotype of all offspring?

A1 AA : 2 Aa : 1 aa

BAll AA

CAll Aa

DAll aa

Explanation

Each parent contributes a different homozygous allele → 100% Aa heterozygous.

Question 2 of 10

TEKS 13A-13DEasy Word Image

Which process in the water cycle is shown by arrow X?

ARunoff

BPrecipitation

CEvaporation

DCondensation

Explanation

📌 Arrow X shows water rising from the surface to the clouds → Evaporation
Evaporation = liquid water → water vapor (gas) due to heat from the sun.

Complete cycle: Evaporation → Condensation (clouds form) → Precipitation (rain) → Runoff → Collection

Question 3 of 10

TEKS 1B-8BMedium Word Image

Based on the Punnett square below, what is the probability of the offspring being heterozygous?

A25%

B100%

C75%

D50%

Explanation

📌 From Bb × Bb cross:
BB = 1/4 (homozygous dominant)
Bb = 2/4 = 1/2 (heterozygous) ← THIS
bb = 1/4 (homozygous recessive)
Probability of heterozygous = 50%

Question 4 of 10

TEKS 12A-12BMedium

What is the role of guard cells?

AThey contain the pigments that capture light for photosynthesis

BThey change shape to open and close the stomata

CThey transport sugars from the leaves down to the roots

DThey absorb water and minerals from the surrounding soil

Explanation

A pair of guard cells flanks each stoma; when they take in water they bow apart and the pore opens, and when they lose water they straighten and it closes. That single mechanism governs the trade-off between taking in carbon dioxide and losing water.

Question 5 of 10

TEKS 9A-10DMedium Diagram

The graph shows population over time of a beetle species. Dark beetles increased while light beetles decreased on a tree with dark bark. What is this an example of?

AGenetic drift (the dark color spread by pure random chance)

BMutation (a new mutation created the dark color this generation)

CNatural selection (camouflage gives dark beetles a survival advantage)

DGene flow (dark beetles migrated in from a nearby population)

Explanation

On a dark tree, dark beetles are camouflaged from predators → survive and reproduce more. Light beetles are eaten more often → decrease. This is natural selection by predation. The shift is directional and tied to the dark bark, not the random chance of drift, not migration bringing beetles in, and not a single new mutation.

Question 6 of 10

TEKS 13A-13DMedium Image

The food web shown contains many species. Which type of organism is missing for this ecosystem to be sustainable?

AAnother top predator

BNothing — the web is complete

CDecomposers

DAnother producer

Explanation

A complete ecosystem needs decomposers to break down dead organisms and waste, returning nutrients to the soil for producers to reuse.

Question 7 of 10

TEKS 9A-10DMedium Diagram

In the phylogenetic tree shown, which two species are most closely related?

ASpecies A and Species B

BAll three are equally related

CSpecies B and Species C

DSpecies A and Species C

Explanation

In a phylogenetic tree, organisms sharing the most recent common ancestor (closest branch point on the right) are most closely related. A and B branch off from each other after C diverged.

Question 8 of 10

TEKS 1B-8BHard

A gene for coat color in a certain animal exhibits epistasis: one gene (B/b) determines whether pigment is black (B) or brown (b), while a separate gene (E/e) determines whether ANY pigment is deposited in the fur at all — ee individuals are albino (white) regardless of their genotype at the B gene. In a cross between two BbEe individuals, what phenotypic ratio is expected among the offspring?

A9 black : 3 brown : 4 white, since the standard 9:3:3:1 dihybrid ratio is modified because the bbee and Bbee genotypes (3+1=4 of 16) both appear white due to the epistatic ee genotype masking the B gene's effect

B9 black : 3 brown : 3 white : 1 unknown, following the completely unmodified standard dihybrid ratio with no epistatic interaction — a claim that treats a minor or secondary factor as if it were the primary explanation here

C1 black : 2 brown : 1 white, following an incomplete dominance pattern instead of an epistatic one — a claim that sounds reasonable but confuses this process with a different, unrelated mechanism

D12 black : 4 white, since epistasis always eliminates the brown phenotype category entirely from the cross — a plausible guess that does not match the specific mechanism at work in this exact case

Explanation

A standard BbEe x BbEe dihybrid cross gives the underlying 9 B_E_ : 3 bbE_ : 3 B_ee : 1 bbee genotypic ratio. Epistasis means ee (regardless of B genotype) always produces white/albino, so the 3 B_ee and 1 bbee categories (4 of 16 total) combine into one white phenotype group, while B_E_ remains black (9) and bbE_ remains brown (3) — yielding the modified 9:3:4 ratio.

Question 9 of 10

TEKS 12A-12BHard Word

The kidney filters glucose freely into the nephron and then reabsorbs it using carrier proteins that saturate. In one patient the glomerular filtration rate is a steady 125 mL of plasma per minute, and the reabsorption carriers can move at most 375 mg of glucose per minute back into the blood no matter how much arrives. Untreated diabetes pushes this patient's plasma glucose to 500 mg per deciliter. At that plasma concentration, at what rate does glucose appear in the urine?

A250 mg per minute

B625 mg per minute

C375 mg per minute

DAbout 156 mg per minute

Explanation

The filtered load is filtration rate times plasma concentration, but the units must agree first: 125 mL per minute is 1.25 dL per minute, so the load is 1.25 x 500 = 625 mg per minute. The carriers are saturated well below this, so they reabsorb their transport maximum of 375 mg per minute and nothing more. Whatever is not reabsorbed is excreted: 625 - 375 = 250 mg per minute. An equivalent route is to find the plasma threshold at which the carriers first saturate, 375 / 1.25 = 300 mg per deciliter, and then excrete the excess, (500 - 300) x 1.25 = 250 mg per minute. Reporting 625 stops at the filtered load and forgets that most of the glucose is still recaptured. Reporting 375 confuses the amount reabsorbed with the amount lost, which is the whole point of a saturable transporter. The value near 156 comes from subtracting before converting, taking 500 - 375 = 125 mg per deciliter and multiplying by 1.25; that subtracts a rate in milligrams per minute from a concentration in milligrams per deciliter, which are not the same kind of quantity.

Question 10 of 10

TEKS 9A-10DHard

Sickle cell allele frequency remains high in regions where malaria is common. This persistence is best explained by —

Agene flow bringing the allele in continuously from neighbouring populations

Bthe fact that natural selection acts too slowly to remove a harmful allele

Ca mutation rate that is high enough to replace every affected allele that is lost

Dheterozygote advantage, since one copy gives resistance without the disease

Explanation

Heterozygotes resist malaria while homozygotes suffer sickle cell disease, so selection maintains both alleles rather than eliminating either.

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