In pea plants, flower color is controlled by one gene with two alleles. The allele for purple flowers (P) is dominant over the allele for white flowers (p). A plant shows white flowers only if it has two copies of the recessive allele.
A gardener crossed a purple-flowered plant that has the genotype Pp with a white-flowered plant. The Punnett square below shows the possible combinations of alleles in the offspring. Each of the four boxes is equally likely.
Based on the Punnett square, what percent of the offspring are expected to have white flowers?
Hint
Which genotype gives white flowers? Count how many of the 4 boxes have that genotype.
Step-by-step solution
Where the answer comes from: the passage says "a plant shows white flowers only if it has two copies of the recessive allele," that is, pp. The Punnett square has four equally likely boxes: Pp, pp, Pp, pp.
Step by step:
Identify the white genotype: pp.
Count pp boxes: 2 out of 4.
Convert: \dfrac{2}{4} = \dfrac{1}{2} = 50\%.
The other 2 boxes (Pp) have one dominant P allele, so those plants are purple.
Answer: 50%.
Why 25% is tempting: many students memorize "1 in 4 recessive," but that is the result of crossing two heterozygous parents (Pp × Pp). Here one parent is pp, so half the offspring get two recessive alleles.
Tip: never apply a memorized ratio without checking the parents' genotypes. Count the boxes in the square that is actually given.
Question 2 of 36
Drop-downFlower Color in Pea Plants
In pea plants, flower color is controlled by one gene with two alleles. The allele for purple flowers (P) is dominant over the allele for white flowers (p). A plant shows white flowers only if it has two copies of the recessive allele.
A gardener crossed a purple-flowered plant that has the genotype Pp with a white-flowered plant. The Punnett square below shows the possible combinations of alleles in the offspring. Each of the four boxes is equally likely.
Select the words that correctly complete each sentence.
The purple-flowered parent is for the flower-color gene. The white-flowered offspring are . The expected ratio of purple-flowered to white-flowered offspring is .
Choose an answer from each menu.
Hint
Homo- means "same" and hetero- means "different." Then count purple boxes and white boxes in the square.
Step-by-step solution
Where the answer comes from: the passage gives the purple parent's genotype as Pp, and the Punnett square shows two Pp boxes and two pp boxes.
Step by step:
Pp has two different alleles, so the purple parent is heterozygous.
White offspring are pp: two copies of the same recessive allele, so they are homozygous recessive.
Purple boxes (Pp) = 2, white boxes (pp) = 2. Ratio purple : white = 2 : 2 = 1 : 1.
Genotype is the allele pair (Pp). Phenotype is the visible trait (purple).
Tip: a 3 : 1 ratio comes from Aa × Aa; a 1 : 1 ratio comes from Aa × aa. Recognizing these two patterns lets you work backward from offspring counts to parent genotypes.
Question 3 of 36
Drag and drop
Drag the terms into the boxes so that each item contains the item before it. Two terms are not used.
Smallest
Largest
Nucleotide (one DNA base unit)
Tap a tile, then tap a box to place it.
Hint
A gene is a section of DNA made of many nucleotides. Where are genes found, and where is that structure kept in a eukaryotic cell?
Step-by-step solution
How genetic material is packaged:
A nucleotide is one building block of DNA. It contains one base: A, T, C or G.
A gene is a section of DNA, often thousands of nucleotides long, that carries the instructions for one protein or trait.
A chromosome is one long, tightly coiled DNA molecule carrying hundreds to thousands of genes. Humans have 46 chromosomes in each body cell.
The nucleus is the organelle that holds all the chromosomes in a eukaryotic cell.
Step by step:
Nucleotides are strung together to make a gene, so gene comes next.
Many genes lie along one chromosome.
All the chromosomes are kept in the nucleus.
Unused terms: a ribosome is an organelle in the cytoplasm that builds proteins; it does not contain chromosomes. A protein is the product made from a gene's instructions, not a container of DNA.
Answer: Gene → Chromosome → Nucleus.
Tip: remember the chain "DNA bases spell genes; genes sit on chromosomes; chromosomes live in the nucleus."
Question 4 of 36
ABO Blood Types
Human ABO blood type is controlled by one gene with three alleles: I^A, I^B and i. Every person has two of these alleles, one from each parent.
Alleles I^A and I^B are codominant: when a person has both, both are fully expressed, and the person has type AB blood. Both I^A and I^B are dominant over i.
Blood type
Possible genotypes
A
I^A I^A or I^A i
B
I^B I^B or I^B i
AB
I^A I^B
O
i\,i
A person with the genotype I^A I^B has type AB blood. Which term describes the relationship between the alleles I^A and I^B?
Hint
In type AB blood, is the result a blend of A and B, or are A and B both fully present?
Step-by-step solution
Where the answer comes from: the passage says "Alleles I^A and I^B are codominant: when a person has both, both are fully expressed, and the person has type AB blood."
Step by step:
In complete dominance, one allele hides the other (like I^A hiding i in an I^A i person, who has type A).
In incomplete dominance, the result is a blend in between the two (red + white snapdragon alleles → pink).
In codominance, both alleles show fully and at the same time. A type AB person's red blood cells carry both the A marker and the B marker.
Answer: codominance.
Check: type AB is not a "halfway" type between A and B; it has all of both markers, so it is not incomplete dominance.
Tip: "co-" means "together": both alleles show together. Incomplete dominance = a mix; codominance = both at full strength.
Question 5 of 36
Guinea Pig Coat Color
In guinea pigs, one gene controls coat color. The allele for a black coat (B) is dominant over the allele for a white coat (b).
A breeder made three crosses. For each pair of parents, she recorded the coat colors of all the offspring they produced over several litters.
Cross
Parent 1
Parent 2
Black offspring
White offspring
1
Black
White
12
0
2
Black
White
7
6
3
Black
Black
15
5
Which two guinea pigs would have the same phenotype but different genotypes?
Hint
Phenotype = what the coat looks like. Genotype = the two letters. Which pair looks alike but has different letters?
Step-by-step solution
Where the answer comes from: the passage says "The allele for a black coat (B) is dominant over the allele for a white coat (b)."
Step by step:
Translate each genotype into a phenotype:
BB → black (two dominant alleles)
Bb → black (the dominant B hides b)
bb → white (no dominant allele)
Find the pair with the same coat color but different letters: BB and Bb are both black.
Check the others: Bb and bb look different (black vs. white); BB and bb look different; bb and bb have the same genotype.
Answer: a BB guinea pig and a Bb guinea pig.
Check: this is why the black parents in the table cannot be identified just by looking at them; both BB and Bb guinea pigs are black.
Tip: a dominant phenotype always has two possible genotypes (homozygous or heterozygous). A recessive phenotype has only one.
Question 6 of 36
The DNA Ladder
DNA is shaped like a twisted ladder, called a double helix. The sides of the ladder are made of sugar and phosphate. Each rung is a pair of bases. There are four bases: adenine (A), thymine (T), cytosine (C) and guanine (G).
The bases always pair the same way: A pairs with T, and C pairs with G. Because of this rule, the two strands are complementary: the order of bases on one strand sets the order on the other. When a cell copies its DNA, the two strands separate, and each strand serves as a pattern for building a new partner strand.
Because every A is paired with a T and every C is paired with a G, a double-stranded DNA molecule always contains equal amounts of A and T, and equal amounts of C and G.
One strand of a DNA molecule has the base sequence A T G C C A. Which sequence is on the complementary strand?
Hint
Use the pairing rule from the passage for each base, one at a time.
Step-by-step solution
Where the answer comes from: the passage states the pairing rule: "A pairs with T, and C pairs with G."
Step by step: pair each base in order.
Given strand
A
T
G
C
C
A
Partner
T
A
C
G
G
T
Answer: T A C G G T.
Check: pair the answer back with the original: A–T, T–A, G–C, C–G, C–G, A–T. Every rung is an A-T or C-G pair ✓
Why the traps fail: U (uracil) is found in RNA, not DNA; copying the same sequence ignores the pairing rule; pairing A with G and T with C breaks the rule.
Tip: remember the pairs by shape of the letters: the straight-line letters A and T go together, and the curvy letters C and G go together.
Question 7 of 36
Drop-downGuinea Pig Coat Color
In guinea pigs, one gene controls coat color. The allele for a black coat (B) is dominant over the allele for a white coat (b).
A breeder made three crosses. For each pair of parents, she recorded the coat colors of all the offspring they produced over several litters.
Cross
Parent 1
Parent 2
Black offspring
White offspring
1
Black
White
12
0
2
Black
White
7
6
3
Black
Black
15
5
Select the words that correctly complete each sentence.
A white guinea pig must have the genotype . This genotype is called . The white color of its coat is its .
Choose an answer from each menu.
Hint
White is the recessive trait. How many copies of a recessive allele must an animal have for the trait to show?
Step-by-step solution
Where the answer comes from: the passage says the black allele (B) "is dominant over the allele for a white coat (b)." A recessive trait shows only when there is no dominant allele to hide it.
Step by step:
A white guinea pig cannot carry any B, or it would be black. So its genotype is bb.
Two identical alleles = homozygous; both are the recessive allele, so bb is homozygous recessive.
The visible trait (white coat) is the phenotype; the letters (bb) are the genotype.
Answer: bb; homozygous recessive; phenotype.
Check: "homo-" = same, "hetero-" = different. Bb would be heterozygous and black, BB homozygous dominant and black ✓
Tip: genotype = gene letters; phenotype = what you can see (or measure).
Question 8 of 36
A Family Pedigree
A genetic counselor drew the pedigree below for a family in which some members have a rare inherited trait. The trait is controlled by a single gene with two alleles. The counselor writes the dominant allele as A and the recessive allele as a. The gene is not on a sex chromosome, so males and females inherit it in the same way.
In a pedigree, squares are males and circles are females. A shaded symbol means the person has the trait. A horizontal line joining a male and a female shows a couple, and the vertical line below it leads to their children.
Which statement about the trait is best supported by the pedigree?
Hint
Look at II-1. Do her parents have the trait? What does that tell you about whether the trait's allele can be hidden?
Step-by-step solution
Where the answer comes from: in the pedigree, II-1 is shaded (has the trait) but her parents, I-1 and I-2, are both unshaded. The same pattern repeats in the next generation: III-2 has the trait, but his parents II-3 and II-4 do not.
Step by step:
If the trait were dominant, anyone with the trait would have at least one allele A, and that allele must come from a parent, who would then also show the trait. Neither of II-1's parents shows it, so the trait cannot be dominant.
If the trait is recessive, both parents can be carriers (Aa) who do not show it, and each can pass a hidden a to a child, who is then aa.
So the trait is recessive, and the unaffected parents are carriers.
Eliminate: the trait skips generation I, so it does not appear in every generation; it appears in a male (III-2), so it is not female-only.
Answer: The trait is recessive, because II-1 has the trait even though neither of her parents does.
Tip: two unaffected parents with an affected child means recessive. Two affected parents with an unaffected child means dominant. Find one of these "key families" first.
Question 9 of 36
A Family Pedigree
A genetic counselor drew the pedigree below for a family in which some members have a rare inherited trait. The trait is controlled by a single gene with two alleles. The counselor writes the dominant allele as A and the recessive allele as a. The gene is not on a sex chromosome, so males and females inherit it in the same way.
In a pedigree, squares are males and circles are females. A shaded symbol means the person has the trait. A horizontal line joining a male and a female shows a couple, and the vertical line below it leads to their children.
What is the genotype of individual I-1?
Hint
I-1 does not have the trait, but one of his children does. What allele must he have passed to that child?
Step-by-step solution
Where the answer comes from: in the pedigree, I-1 (male, generation I) is unshaded, so he does not have the trait. His daughter II-1 is shaded, so she does.
Step by step:
The trait is recessive (an affected child has two unaffected parents), so II-1's genotype is aa.
A child gets one allele from each parent, so II-1 received an a from I-1.
I-1 does not show the trait, so he must also have a dominant A.
Therefore I-1 is Aa, a carrier. The same reasoning shows I-2 is also Aa.
Answer: Aa.
Why the others fail: AA is impossible, because then he could not pass an a to II-1. aa would mean he shows the trait, but he is unshaded. "Cannot be determined" is tempting, but II-1's genotype gives enough information.
Tip: in pedigree problems, work from the people whose genotypes you know for sure (affected people with a recessive trait are always aa), then reason to their parents and children.
Question 10 of 36
One Letter, One Disease
Hemoglobin is the protein in red blood cells that carries oxygen. The instructions for building one part of hemoglobin are stored in a gene, a sequence of DNA bases (A, T, C and G). A cell reads the bases in groups of three, called codons. Each codon stands for one amino acid (or a "stop" signal), and the chain of amino acids folds into the finished protein.
The table shows a short section of the normal gene and the same section in a person with sickle cell disease.
Codon number
1
2
3
4
5
6
Normal gene
CTG
ACT
CCT
GAG
GAG
AAG
Sickle cell gene
CTG
ACT
CCT
GTG
GAG
AAG
Codon
CTG
ACT
CCT
GAG
GAA
GTG
AAG
TAG or TGA
Amino acid
Leucine
Threonine
Proline
Glutamic acid
Glutamic acid
Valine
Lysine
Stop (end of protein)
In sickle cell disease, the changed hemoglobin makes red blood cells bend into a curved, sickle shape when oxygen levels are low. The stiff, sickled cells can block small blood vessels and cause pain.
Based on the passage, which type of mutation causes the change shown in the sickle cell gene?
Hint
Compare the two rows codon by codon. Did the number of bases change, or did one base change into another?
Step-by-step solution
Where the answer comes from: in the first table, every codon is the same except codon 4, which is GAG in the normal gene and GTG in the sickle cell gene. The second letter changed from A to T. The total number of bases (18) is unchanged.
Step by step:
Compare codon by codon: codons 1, 2, 3, 5 and 6 match.
Codon 4: G-A-G becomes G-T-G. One base was swapped for another.
A swap of one base for another is a substitution (point mutation).
Effect: in the codon table, GAG is glutamic acid and GTG is valine, so one amino acid in hemoglobin changes. That single change is enough to cause sickle cell disease.
Answer: a substitution, in which one base is replaced by another.
Why the others fail: an insertion would add a base (19 bases), and a deletion would remove one (17 bases); the count did not change. A change in chromosome number affects whole chromosomes, not one letter in a gene.
Tip: to classify a mutation, first count the bases. Same count with one letter changed is a substitution. A different count is an insertion or deletion.
Question 11 of 36
One Letter, One Disease
Hemoglobin is the protein in red blood cells that carries oxygen. The instructions for building one part of hemoglobin are stored in a gene, a sequence of DNA bases (A, T, C and G). A cell reads the bases in groups of three, called codons. Each codon stands for one amino acid (or a "stop" signal), and the chain of amino acids folds into the finished protein.
The table shows a short section of the normal gene and the same section in a person with sickle cell disease.
Codon number
1
2
3
4
5
6
Normal gene
CTG
ACT
CCT
GAG
GAG
AAG
Sickle cell gene
CTG
ACT
CCT
GTG
GAG
AAG
Codon
CTG
ACT
CCT
GAG
GAA
GTG
AAG
TAG or TGA
Amino acid
Leucine
Threonine
Proline
Glutamic acid
Glutamic acid
Valine
Lysine
Stop (end of protein)
In sickle cell disease, the changed hemoglobin makes red blood cells bend into a curved, sickle shape when oxygen levels are low. The stiff, sickled cells can block small blood vessels and cause pain.
Which change to codon 4 of the normal gene would not change the hemoglobin protein?
Hint
Use the codon table. Find a new codon that codes for the same amino acid as GAG.
Step-by-step solution
Where the answer comes from: the codon table shows that GAG codes for glutamic acid, and so does GAA. GTG codes for valine, AAG codes for lysine, and TAG is a stop signal.
Step by step:
Normal codon 4 is GAG, which is glutamic acid.
Check each choice in the table:
GAA is glutamic acid, the same amino acid, so there is no change in the protein.
GTG is valine, a different amino acid. This is the sickle cell change.
AAG is lysine, a different amino acid.
TAG is stop, so the protein would end early, after only three amino acids.
Only GAA leaves the protein unchanged.
Answer: GAG changed to GAA.
This is called a silent mutation. The DNA changes, but because more than one codon can code for the same amino acid, the protein stays the same.
Tip: not every mutation is harmful. Mutations can be harmful, neutral (silent) or occasionally helpful. The effect depends on what happens to the protein.
Question 12 of 36
Pink Snapdragons
Snapdragon flower color shows incomplete dominance: neither allele fully hides the other. A plant with two red alleles (RR) has red flowers. A plant with two white alleles (WW) has white flowers. A plant with one of each (RW) has pink flowers, a color in between.
A student crossed two pink-flowered snapdragons (RW × RW). She planted 400 of the seeds they produced and recorded the flower color of each new plant.
Flower color
Number of plants
Red
98
Pink
205
White
97
Total
400
Which statement best explains the student's results?
Hint
Divide each count by about 100. What ratio do you get, and which Punnett square produces it?
Step-by-step solution
Where the answer comes from: the table shows 98 red, 205 pink, 97 white out of 400. The passage says RR = red, RW = pink and WW = white.
Step by step:
Simplify: 98 : 205 : 97 is very close to 100 : 200 : 100, which is 1 : 2 : 1.
Punnett square for RW × RW: each parent passes R or W with equal chance. The boxes are RR, RW, RW, WW, giving 1 red : 2 pink : 1 white.
The observed counts match the predicted ratio closely. Small differences (205 instead of 200) are normal random variation.
Answer: Each pink parent passed on either R or W with equal chance, producing about 1 red : 2 pink : 1 white.
Why the 3 : 1 trap fails: if you lump red and pink together (98 + 205 = 303) against white (97), you get about 3 : 1. But red and pink are different phenotypes, so lumping them hides the real pattern. Red is not dominant, because RW plants are pink, not red.
Tip: with incomplete dominance, the heterozygote has its own in-between phenotype, so a cross of two heterozygotes shows three phenotypes in a 1 : 2 : 1 ratio, not two phenotypes in 3 : 1.
Question 13 of 36
Fill inPink Snapdragons
Snapdragon flower color shows incomplete dominance: neither allele fully hides the other. A plant with two red alleles (RR) has red flowers. A plant with two white alleles (WW) has white flowers. A plant with one of each (RW) has pink flowers, a color in between.
A student crossed two pink-flowered snapdragons (RW × RW). She planted 400 of the seeds they produced and recorded the flower color of each new plant.
Flower color
Number of plants
Red
98
Pink
205
White
97
Total
400
According to a Punnett square for the RW × RW cross, how many of the 400 offspring would be expected to have pink flowers?
plants
Type your answer in the box. Fractions like 3/4 are fine.
Hint
Draw the 2 × 2 Punnett square. What fraction of the boxes are RW? Multiply that fraction by 400.
Step-by-step solution
Where the answer comes from: the passage says pink plants are RW and the parents are RW × RW, with 400 offspring.
Step by step:
Punnett square:
R
W
R
RR
RW
W
RW
WW
RW boxes: 2 out of 4, so P(\text{pink}) = \dfrac{2}{4} = \dfrac{1}{2}.
Check: expected red = \tfrac{1}{4} \times 400 = 100 and expected white = 100, so 100 + 200 + 100 = 400 ✓
Answer: 200 plants. The student observed 205, close to the prediction. Real results vary a little by chance, and larger samples usually come closer to the expected ratio.
Tip: expected number = probability × total. Use the Punnett square for the probability, not the observed data.
Question 14 of 36
Fill inGuinea Pig Coat Color
In guinea pigs, one gene controls coat color. The allele for a black coat (B) is dominant over the allele for a white coat (b).
A breeder made three crosses. For each pair of parents, she recorded the coat colors of all the offspring they produced over several litters.
Cross
Parent 1
Parent 2
Black offspring
White offspring
1
Black
White
12
0
2
Black
White
7
6
3
Black
Black
15
5
The black parent in Cross 2 is heterozygous (Bb). If it is mated with a different white guinea pig, what is the probability that any one offspring will be white?
Enter your answer as a percent. %
Type your answer in the box. Fractions like 3/4 are fine.
Hint
What is the genotype of any white guinea pig? Set up a Punnett square for Bb × that genotype.
Step-by-step solution
Where the answer comes from: the passage says black (B) is dominant over white (b), so every white guinea pig is bb. The black parent is given as Bb.
Step by step:
Cross: Bb × bb.
Punnett square:
B
b
b
Bb
bb
b
Bb
bb
Two of the four boxes are bb (white): \dfrac{2}{4} = \dfrac{1}{2} = 50\%.
Answer: 50%.
Check: Cross 2 in the table (also Bb × white) gave 7 black and 6 white offspring, close to the 1 : 1 ratio that 50% predicts ✓
Tip: a cross between a heterozygote and a homozygous recessive always gives a 1 : 1 ratio, 50% of each phenotype.
Question 15 of 36
Cystic fibrosis is a disorder caused by a recessive allele (f). People with the genotype Ff do not have the disorder but are carriers. If two carriers (Ff × Ff) have a child, what is the probability that the child will also be a carrier?
Hint
Draw the Punnett square for Ff × Ff. Count only the boxes with exactly one F and one f.
Step-by-step solution
Where the answer comes from: the stem defines a carrier as a person with the genotype Ff, and both parents are Ff.
Step by step:
Punnett square for Ff × Ff:
F
f
F
FF
Ff
f
Ff
ff
Count the carrier boxes (Ff): 2 of 4.
Probability: \dfrac{2}{4} = 50\%.
Answer: 50%.
Check: all four boxes add to 100%: FF 25% (not a carrier), Ff 50% (carrier), ff 25% (has the disorder) ✓
Why the traps fail: 25% is the chance of having the disorder (ff); 75% is the chance of not having the disorder (FF + Ff); 100% assumes children copy their parents.
Tip: read the question for the exact genotype asked about. "Has the disorder," "is a carrier" and "does not have the disorder" are three different answers from the same square.
Question 16 of 36
Fill inCounting Corn Kernels
In corn, the allele for purple kernels (P) is dominant over the allele for yellow kernels (p). Each kernel on an ear of corn is a separate offspring, so one ear can show hundreds of results of the same cross.
A farmer crossed two corn plants that were both heterozygous (Pp × Pp). A student counted the kernels on one ear produced by this cross.
Kernel color
Number of kernels
Purple
381
Yellow
119
Total
500
The student wants to compare the ear with the ratio predicted for a Pp × Pp cross. Out of the 500 kernels on the ear, how many yellow kernels does the Punnett square predict?
kernels
Type your answer in the box. Fractions like 3/4 are fine.
Hint
First find the fraction of offspring expected to be pp. Then take that fraction of 500.
Step-by-step solution
Where the answer comes from: the passage says purple (P) is dominant over yellow (p), both parents are Pp, and the ear has 500 kernels in total.
Step by step:
Punnett square for Pp × Pp:
P
p
P
PP
Pp
p
Pp
pp
Only pp is yellow: 1 box of 4, so \dfrac{1}{4} of the kernels.
Expected number: \dfrac{1}{4} \times 500 = 125.
Answer: 125 kernels.
Check: expected purple = \dfrac{3}{4} \times 500 = 375, and 375 + 125 = 500 ✓ The student actually counted 119 yellow, close to 125.
Tip: expected number = probability × total. Use the total from the passage, not the number of boxes in the Punnett square.
Question 17 of 36
Color Blindness in a Family
Red-green color blindness is a recessive trait. The gene for it is on the X chromosome, so the trait is called X-linked (sex-linked). Females have two X chromosomes (XX). Males have one X and one Y chromosome (XY), and the Y chromosome does not carry this gene. A son always receives his X chromosome from his mother and his Y chromosome from his father.
The normal allele is written X^B and the color-blind allele X^b. A female with one of each (X^B X^b) has normal color vision but is a carrier: she can pass the color-blind allele to her children.
Based on the passage, which statement best explains why red-green color blindness is much more common in males than in females?
Hint
Count the X chromosomes in a male and in a female. How many copies of the recessive allele does each need for the trait to show?
Step-by-step solution
Where the answer comes from: the passage says the gene is "on the X chromosome," that males have "one X and one Y chromosome (XY), and the Y chromosome does not carry this gene," and that the trait is recessive.
Step by step:
A male (XY) has only one copy of the gene. If that one X carries X^b, nothing can hide it, so he is color-blind.
A female (XX) has two copies. She is color-blind only if both X chromosomes carry X^b. With one X^b she is only a carrier.
Getting one rare allele is far more likely than getting two, so many more males than females have the trait.
Answer: A male has only one X chromosome, so a single color-blind allele causes the trait, while a female needs two.
Check: in the pedigree, both color-blind people (II-1 and III-1) are males, and their mothers have normal vision, which fits.
Tip: for X-linked recessive traits, think "males: one strike and you're out; females: need two strikes."
Question 18 of 36
ABO Blood Types
Human ABO blood type is controlled by one gene with three alleles: I^A, I^B and i. Every person has two of these alleles, one from each parent.
Alleles I^A and I^B are codominant: when a person has both, both are fully expressed, and the person has type AB blood. Both I^A and I^B are dominant over i.
Blood type
Possible genotypes
A
I^A I^A or I^A i
B
I^B I^B or I^B i
AB
I^A I^B
O
i\,i
A woman with the genotype I^A i and a man with the genotype I^B i have a child. What is the probability that the child will have type O blood?
Hint
Which genotype gives type O? Set up a Punnett square with the woman's two alleles on one side and the man's on the other.
Step-by-step solution
Where the answer comes from: the table shows that type O has only one genotype, i\,i. The woman is I^A i and the man is I^B i.
Step by step:
Punnett square:
I^B
i
I^A
I^A I^B (AB)
I^A i (A)
i
I^B i (B)
i\,i (O)
Only one box of four is i\,i: \dfrac{1}{4} = 25\%.
Answer: 25%.
Check: the four boxes give four different blood types (AB, A, B, O), each 25%, and 4 \times 25\% = 100\% ✓
Why the traps fail: 0% forgets that both parents carry a hidden i; 50% would need two of the four boxes to be i\,i, but only one is; 75% is the chance of not having type O.
Tip: two parents with a dominant phenotype can still have a child with the recessive phenotype if both carry the recessive allele.
Question 19 of 36
Select allABO Blood Types
Human ABO blood type is controlled by one gene with three alleles: I^A, I^B and i. Every person has two of these alleles, one from each parent.
Alleles I^A and I^B are codominant: when a person has both, both are fully expressed, and the person has type AB blood. Both I^A and I^B are dominant over i.
Blood type
Possible genotypes
A
I^A I^A or I^A i
B
I^B I^B or I^B i
AB
I^A I^B
O
i\,i
A mother has type AB blood and a father has type O blood. Select all the blood types their children could have.
Select all that apply.
Hint
Write each parent's genotype from the table. Each child gets one allele from the mother and one from the father.
Step-by-step solution
Where the answer comes from: the table shows type AB has only one genotype, I^A I^B, and type O has only one genotype, i\,i.
Step by step:
The mother can pass on I^A or I^B. The father can pass on only i.
Possible children: I^A i → type A, and I^B i → type B.
Type AB would need both I^A and I^B, but the father has neither. Type O would need i from both parents, but the mother has no i.
Answer: type A and type B.
Check: none of the children would have the same blood type as either parent, a surprising but correct result that blood-type tests confirm.
Tip: list every allele each parent can pass on, then combine them one from each side. Do not assume children can match a parent.
Question 20 of 36
Fill inThe DNA Ladder
DNA is shaped like a twisted ladder, called a double helix. The sides of the ladder are made of sugar and phosphate. Each rung is a pair of bases. There are four bases: adenine (A), thymine (T), cytosine (C) and guanine (G).
The bases always pair the same way: A pairs with T, and C pairs with G. Because of this rule, the two strands are complementary: the order of bases on one strand sets the order on the other. When a cell copies its DNA, the two strands separate, and each strand serves as a pattern for building a new partner strand.
Because every A is paired with a T and every C is paired with a G, a double-stranded DNA molecule always contains equal amounts of A and T, and equal amounts of C and G.
A sample of double-stranded DNA is analyzed, and 30% of its bases are adenine (A). What percent of its bases are guanine (G)?
%
Type your answer in the box. Fractions like 3/4 are fine.
Hint
If A is 30%, what must T be? The rest of the bases are split equally between C and G.
Step-by-step solution
Where the answer comes from: the passage says a double-stranded DNA molecule "always contains equal amounts of A and T, and equal amounts of C and G."
Step by step:
A = 30%, so T = 30% too.
A + T = 30\% + 30\% = 60\%.
The remaining bases are C and G: 100\% - 60\% = 40\%.
C and G are equal, so each is half of 40%: 40\% \div 2 = 20\%.
Answer: 20%.
Check:A + T + C + G = 30 + 30 + 20 + 20 = 100\% ✓
Tip: a common error is to set G equal to A (30%) or to give 70% (forgetting T). Pair each base with its partner first, then split what is left.
Question 21 of 36
Guinea Pig Coat Color
In guinea pigs, one gene controls coat color. The allele for a black coat (B) is dominant over the allele for a white coat (b).
A breeder made three crosses. For each pair of parents, she recorded the coat colors of all the offspring they produced over several litters.
Cross
Parent 1
Parent 2
Black offspring
White offspring
1
Black
White
12
0
2
Black
White
7
6
3
Black
Black
15
5
What are the most likely genotypes of the two black parents in Cross 3?
Hint
Two black parents had some white offspring. Where did each white offspring's two b alleles come from?
Step-by-step solution
Where the answer comes from: the table shows that Cross 3 (Black × Black) produced 15 black and 5 white offspring.
Step by step:
A white offspring is bb, so it received one b from each parent.
Both parents are black, so each also has a B. Therefore both parents are Bb.
Test it: Bb × Bb gives \dfrac{3}{4} black and \dfrac{1}{4} white, a 3 : 1 ratio.
The data: 15 : 5 = 3 : 1 ✓
Answer: Bb × Bb.
Check: if either parent were BB, every offspring would receive at least one B and be black, so no white offspring could appear. The 5 white offspring rule that out.
Tip: when two parents that show the dominant trait have an offspring with the recessive trait, both parents must be heterozygous carriers.
Question 22 of 36
One Plant, Three Gardens
Researchers studied a wild flowering plant that grows from sea level high into the mountains. They took cuttings (pieces of stem that grow into new plants) from a single plant, so all the cuttings were genetically identical. They planted the cuttings in three gardens at different elevations. After two growing seasons, they measured the average height of the plants in each garden.
Garden elevation (meters above sea level)
Average plant height (cm)
30
62
1,400
38
3,050
15
The higher gardens had colder temperatures, stronger winds and a shorter growing season.
What do the results from the three gardens best show about plant height?
Hint
The cuttings all had the same genes. So what must have caused the differences in height?
Step-by-step solution
Where the answer comes from: the passage says the cuttings came "from a single plant, so all the cuttings were genetically identical." The table shows the average height dropping from 62 cm at 30 m to 38 cm at 1,400 m to 15 cm at 3,050 m.
Step by step:
Genotype was the same in all three gardens.
Phenotype (height) was very different.
The only thing that differed was the environment (temperature, wind, length of growing season). So the environment changed the phenotype.
The data cannot show that genes do not matter at all; they show only that the same genes can produce different heights in different conditions.
Answer: The environment can change a plant's phenotype even when its genotype stays the same.
Check: the conclusion uses both pieces of the study: same genotype (the design) and different heights (the data).
Tip: phenotype = genotype + environment. When genotype is held constant, any difference in phenotype points to the environment.
Question 23 of 36
Fill inA Family Pedigree
A genetic counselor drew the pedigree below for a family in which some members have a rare inherited trait. The trait is controlled by a single gene with two alleles. The counselor writes the dominant allele as A and the recessive allele as a. The gene is not on a sex chromosome, so males and females inherit it in the same way.
In a pedigree, squares are males and circles are females. A shaded symbol means the person has the trait. A horizontal line joining a male and a female shows a couple, and the vertical line below it leads to their children.
If II-3 and II-4 have another child, what is the probability that this child will have the trait?
Enter your answer as a percent. %
Type your answer in the box. Fractions like 3/4 are fine.
Hint
First use their child III-2 to work out the genotypes of II-3 and II-4. Then set up a Punnett square.
Step-by-step solution
Where the answer comes from: in the pedigree, II-3 and II-4 are both unshaded, but their son III-2 is shaded.
Step by step:
The trait is recessive (unaffected parents have affected children), so III-2 is aa.
III-2 got one a from each parent. Since II-3 and II-4 are unaffected, each must also have an A. Both are Aa.
Punnett square, Aa × Aa:
A
a
A
AA
Aa
a
Aa
aa
Only aa has the trait: 1 box of 4, so P = \dfrac{1}{4} = 25\%.
Each child is a separate event, so having one affected child already does not change the odds for the next child.
Answer: 25%.
Tip: probability "has no memory." A couple of carriers has a 25% chance of an affected child with every pregnancy, no matter what happened before.
Question 24 of 36
A Family Pedigree
A genetic counselor drew the pedigree below for a family in which some members have a rare inherited trait. The trait is controlled by a single gene with two alleles. The counselor writes the dominant allele as A and the recessive allele as a. The gene is not on a sex chromosome, so males and females inherit it in the same way.
In a pedigree, squares are males and circles are females. A shaded symbol means the person has the trait. A horizontal line joining a male and a female shows a couple, and the vertical line below it leads to their children.
Which individual's genotype can be determined with certainty to be heterozygous (Aa)?
Hint
For each person, ask: could they be AA, or is there evidence (an affected child) that forces them to carry a?
Step-by-step solution
Where the answer comes from: the pedigree shows that II-4 is unshaded and that his son III-2 is shaded.
Step by step:
II-4: III-2 is aa, so he received an a from his father, II-4. II-4 is unaffected, so he also has an A. II-4 is certainly Aa, even though he married into the family.
II-2: his parents are both Aa. As an unaffected child, he could be AA or Aa. We cannot be certain.
III-1: her parents are Aa × Aa and she is unaffected, so she could be AA or Aa. Not certain.
II-1: she has the trait, so she is certainly aa. Her genotype is certain, but it is homozygous, not heterozygous.
Answer: II-4.
Why this is subtle: it is easy to assume a person from outside the family is AA. But a child's genotype is evidence about both parents.
Tip: certainty in a pedigree comes from evidence: an affected child proves that both unaffected parents are carriers. Without such evidence, an unaffected person with carrier parents could be AA or Aa.
Question 25 of 36
Hot spotColor Blindness in a Family
Red-green color blindness is a recessive trait. The gene for it is on the X chromosome, so the trait is called X-linked (sex-linked). Females have two X chromosomes (XX). Males have one X and one Y chromosome (XY), and the Y chromosome does not carry this gene. A son always receives his X chromosome from his mother and his Y chromosome from his father.
The normal allele is written X^B and the color-blind allele X^b. A female with one of each (X^B X^b) has normal color vision but is a carrier: she can pass the color-blind allele to her children.
Click on a female in the pedigree who must be a carrier of the color-blind allele (X^B X^b).
Click on the figure to mark your answer (or focus it and use the arrow keys).
Hint
Find each color-blind male. His X chromosome came from his mother, and she has normal vision.
Step-by-step solution
Where the answer comes from: the passage says "A son always receives his X chromosome from his mother." In the pedigree, the color-blind sons are II-1 (mother I-2) and III-1 (mother II-4). Both mothers are unshaded (normal vision).
Step by step:
II-1 is color-blind, so his only X is X^b. It came from his mother, I-2.
I-2 has normal vision, so she must also have an X^B. Her genotype must be X^B X^b: a carrier.
The same reasoning applies to III-1 and his mother II-4: II-4 must be X^B X^b.
II-2 and III-2 might be carriers (each got an X from a carrier mother, with a 50% chance of X^b), but nothing proves it.
Answer: click I-2 or II-4.
Check: neither father's X can explain the sons' color blindness, because a father gives his son a Y, not an X.
Tip: in X-linked recessive pedigrees, the mother of every affected son with normal vision is a certain carrier.
Question 26 of 36
Drag and dropGuinea Pig Coat Color
In guinea pigs, one gene controls coat color. The allele for a black coat (B) is dominant over the allele for a white coat (b).
A breeder made three crosses. For each pair of parents, she recorded the coat colors of all the offspring they produced over several litters.
Cross
Parent 1
Parent 2
Black offspring
White offspring
1
Black
White
12
0
2
Black
White
7
6
3
Black
Black
15
5
Drag the most likely genotype of the black parent into each box. One genotype is not used.
Black parent in Cross 1
Black parent in Cross 2
Tap a tile, then tap a box to place it.
Hint
Each black parent was mated with a white (bb) guinea pig. What offspring would a BB parent produce? What would a Bb parent produce?
Step-by-step solution
Where the answer comes from: the table shows Cross 1 (Black × White) gave 12 black, 0 white, and Cross 2 (Black × White) gave 7 black, 6 white.
Step by step:
The white parent is always bb, so it gives every offspring a b.
If the black parent is BB: every offspring is Bb → all black. That matches Cross 1 (12 black, 0 white).
If the black parent is Bb: half the offspring get b from it → 1 black : 1 white. That matches Cross 2 (7 : 6).
bb is not used: a bb guinea pig would be white, not black.
Answer: Cross 1 = BB; Cross 2 = Bb.
Check: for Cross 1, if the parent were Bb, the chance of 12 black offspring in a row would be (\tfrac{1}{2})^{12}, about 1 in 4,000, so BB is far more likely. That is why the question says "most likely."
Tip: crossing an unknown dominant-looking individual with a homozygous recessive is called a test cross. Any recessive offspring reveal a hidden recessive allele.
Question 27 of 36
Color Blindness in a Family
Red-green color blindness is a recessive trait. The gene for it is on the X chromosome, so the trait is called X-linked (sex-linked). Females have two X chromosomes (XX). Males have one X and one Y chromosome (XY), and the Y chromosome does not carry this gene. A son always receives his X chromosome from his mother and his Y chromosome from his father.
The normal allele is written X^B and the color-blind allele X^b. A female with one of each (X^B X^b) has normal color vision but is a carrier: she can pass the color-blind allele to her children.
Which evidence from the pedigree best shows that the color-blind allele is recessive?
Hint
A recessive trait can skip a generation: it appears in a child even though the parents do not show it. Look for that.
Step-by-step solution
Where the answer comes from: in the pedigree, II-1 is shaded (color-blind), but his parents I-1 and I-2 are both unshaded (normal vision).
Step by step:
If the color-blind allele were dominant, anyone carrying it would be color-blind. II-1 got his X^b from his mother, so she would have to be color-blind too.
But I-2 has normal vision. So she carries the allele without showing it, which means it is recessive (hidden by X^B).
The same thing happens again with III-1 and his unaffected mother, II-4.
The other choices are true facts, but they do not distinguish dominant from recessive.
Answer: II-1 is color-blind even though neither of his parents, I-1 or I-2, is color-blind.
Check: if the trait were dominant, every color-blind child would have at least one color-blind parent. II-1 has none, so dominant is ruled out.
Tip: "unaffected parents → affected child" is the classic proof that a trait is recessive.
Question 28 of 36
Fill inA Trait in Three Generations
A genetic counselor is studying a rare trait in one family. The trait is controlled by a single gene with two alleles, T and t. The gene is not on a sex chromosome, so males and females inherit it in the same way. Two people who married into the family, II-4 and II-5, are shown at the ends of generation II.
In the pedigree, squares are males, circles are females, and shaded symbols show people who have the trait.
Assume the trait is caused by a dominant allele (T). II-3 and II-4 are planning one more child. What is the chance, as a percent, that this child will be born with the trait?
%
Type your answer in the box. Fractions like 3/4 are fine.
Hint
First work out II-3's genotype. He has the trait, but look at his mother, I-2. Then set up a Punnett square with II-4.
Step-by-step solution
Where the answer comes from: in the pedigree, II-3 is shaded (has the trait), his mother I-2 is unshaded, and his wife II-4 is unshaded.
Step by step:
The trait is dominant, so anyone without it is tt. I-2 and II-4 are both tt.
II-3 has the trait, so he has at least one T. He got one allele from his mother I-2, who could only give t. So II-3 is Tt.
Punnett square, Tt × tt:
T
t
t
Tt
tt
t
Tt
tt
Two boxes of four (Tt) have the trait: \dfrac{2}{4} = 50\%.
Answer: 50%.
Check: their two children so far, III-3 (trait) and III-4 (no trait), fit a 1 : 1 pattern, though each child is an independent 50% chance.
Tip: for a dominant trait, an unaffected person is always homozygous recessive, and an affected person with one unaffected parent must be heterozygous.
Question 29 of 36
Counting Corn Kernels
In corn, the allele for purple kernels (P) is dominant over the allele for yellow kernels (p). Each kernel on an ear of corn is a separate offspring, so one ear can show hundreds of results of the same cross.
A farmer crossed two corn plants that were both heterozygous (Pp × Pp). A student counted the kernels on one ear produced by this cross.
Kernel color
Number of kernels
Purple
381
Yellow
119
Total
500
A Punnett square predicts 375 purple and 125 yellow kernels, but the student counted 381 purple and 119 yellow. Which statement best explains the difference?
Hint
Is a Punnett square a guarantee, or a statement about probability? Think about flipping a coin 500 times.
Step-by-step solution
Where the answer comes from: the passage says "Each kernel on an ear of corn is a separate offspring." The table shows 381 purple and 119 yellow out of 500, while a 3 : 1 Punnett ratio predicts 375 and 125.
Step by step:
A Punnett square gives probabilities for each offspring: each kernel has a \tfrac{3}{4} chance of purple and a \tfrac{1}{4} chance of yellow.
Which pollen grain joins which egg is random, like a coin flip, so real counts wander a little around the expected numbers.
The difference here is small: 6 kernels out of 500 (about 1%). The observed ratio, 381 \div 119 \approx 3.2 : 1, is close to 3 : 1.
Larger samples (several ears) would usually come even closer to 3 : 1.
Answer: Which alleles each kernel receives depends on chance, so actual counts usually differ slightly from the expected numbers.
Check: if the parents were really PP × pp, all kernels would be purple; any yellow kernels at all rule that out.
Tip: small differences between expected and observed results are normal. A large, consistent difference would suggest the hypothesis is wrong.
Question 30 of 36
Fill inColor Blindness in a Family
Red-green color blindness is a recessive trait. The gene for it is on the X chromosome, so the trait is called X-linked (sex-linked). Females have two X chromosomes (XX). Males have one X and one Y chromosome (XY), and the Y chromosome does not carry this gene. A son always receives his X chromosome from his mother and his Y chromosome from his father.
The normal allele is written X^B and the color-blind allele X^b. A female with one of each (X^B X^b) has normal color vision but is a carrier: she can pass the color-blind allele to her children.
II-4 is a carrier (X^B X^b), and II-5 has normal vision (X^B Y). If they have another son, what is the probability that he will be color-blind?
Enter your answer as a percent. %
Type your answer in the box. Fractions like 3/4 are fine.
Hint
A son gets his Y from his father. The only question is which of his mother's two X chromosomes he receives.
Step-by-step solution
Where the answer comes from: the stem gives the genotypes X^B X^b (mother) and X^B Y (father), and the passage says "A son always receives his X chromosome from his mother and his Y chromosome from his father."
Step by step:
Punnett square:
X^B
Y
X^B
X^B X^B (girl)
X^B Y (boy, normal)
X^b
X^B X^b (girl, carrier)
X^b Y (boy, color-blind)
The question is about a son, so look only at the two boy boxes: X^B Y and X^b Y.
One of the two sons is color-blind: \dfrac{1}{2} = 50\%.
Answer: 50%.
Check: if the question asked about any child (son or daughter), the answer would be 1 of 4 boxes, or 25%. Because we already know the child is a son, only half the square applies.
Tip: when a question names the child's sex, shrink the Punnett square to only the boxes for that sex before counting.
Question 31 of 36
The table shows the first five codons of a gene and part of a codon chart.
Codon number
1
2
3
4
5
Normal gene
ATG
GGC
TGG
CAG
TTC
Codon
ATG
GGC
TGG
CAG
TTC
TGA
Meaning
Methionine (start)
Glycine
Tryptophan
Glutamine
Phenylalanine
Stop (end of protein)
The normal protein made from this gene is 300 amino acids long. A mutation changes codon 3 from TGG to TGA. Which statement best describes the most likely effect?
Hint
Look up the new codon, TGA, in the chart. What does a cell do when it reaches that codon?
Step-by-step solution
Where the answer comes from: the chart shows that TGA means "Stop (end of protein)." The mutation changes codon 3 (TGG, tryptophan) into TGA.
Step by step:
The cell reads codon 1 (methionine) and codon 2 (glycine) normally.
At codon 3 it now finds a stop signal, so it ends the protein there.
The protein is only 2 amino acids long instead of 300. A protein that short cannot fold into its working shape, so it almost certainly cannot do its job.
This is still a substitution (one base, G → A, swapped), but because it creates a stop codon it has a huge effect.
Answer: The protein would end after only 2 amino acids, so it would be far too short to work.
Check: the number of bases did not change, so it is not a frameshift; the codons after codon 3 are simply never read.
Tip: always look up the new codon. A one-letter change can do nothing (same amino acid), swap one amino acid, or, as here, stop the protein early.
Question 32 of 36
One Plant, Three Gardens
Researchers studied a wild flowering plant that grows from sea level high into the mountains. They took cuttings (pieces of stem that grow into new plants) from a single plant, so all the cuttings were genetically identical. They planted the cuttings in three gardens at different elevations. After two growing seasons, they measured the average height of the plants in each garden.
Garden elevation (meters above sea level)
Average plant height (cm)
30
62
1,400
38
3,050
15
The higher gardens had colder temperatures, stronger winds and a shorter growing season.
Why was it important that the researchers used cuttings from a single plant instead of seeds collected from many different plants?
Hint
What would be different among seeds from many plants that is the same in cuttings from one plant?
Step-by-step solution
Where the answer comes from: the passage says the cuttings came "from a single plant, so all the cuttings were genetically identical."
Step by step:
The researchers wanted to know whether the environment (elevation) affects height.
Seeds from many plants would each have a different mix of alleles. Then a short plant might be short because of its genes or because of its garden, and there would be no way to tell.
Cuttings from one plant all have the same genotype, so genes are a controlled variable. Any difference in height must come from the environment.
Answer: Identical genotypes meant that differences in height could be explained by the environment, not by genes.
Check: the independent variable is elevation, the dependent variable is height, and genotype is held constant, which is exactly what a fair test needs.
Tip: in a good experiment, only the variable being tested should differ between groups. Using clones is a way to control genetic differences.
Question 33 of 36
One Letter, One Disease
Hemoglobin is the protein in red blood cells that carries oxygen. The instructions for building one part of hemoglobin are stored in a gene, a sequence of DNA bases (A, T, C and G). A cell reads the bases in groups of three, called codons. Each codon stands for one amino acid (or a "stop" signal), and the chain of amino acids folds into the finished protein.
The table shows a short section of the normal gene and the same section in a person with sickle cell disease.
Codon number
1
2
3
4
5
6
Normal gene
CTG
ACT
CCT
GAG
GAG
AAG
Sickle cell gene
CTG
ACT
CCT
GTG
GAG
AAG
Codon
CTG
ACT
CCT
GAG
GAA
GTG
AAG
TAG or TGA
Amino acid
Leucine
Threonine
Proline
Glutamic acid
Glutamic acid
Valine
Lysine
Stop (end of protein)
In sickle cell disease, the changed hemoglobin makes red blood cells bend into a curved, sickle shape when oxygen levels are low. The stiff, sickled cells can block small blood vessels and cause pain.
Suppose a different mutation inserts one extra C between codon 1 and codon 2 of the normal gene. Based on the passage, which outcome is most likely?
Hint
The cell reads bases in groups of three, starting from the beginning. Rewrite the sequence with the extra C and regroup it into threes.
Step-by-step solution
Where the answer comes from: the passage says "a cell reads the bases in groups of three, called codons." The normal sequence is CTG ACT CCT GAG GAG AAG.
Step by step:
Insert a C after codon 1: CTG CACT CCT GAG GAG AAG.
Regroup into threes from the start: CTG | CAC | TCC | TGA | GGA | GAA | G.
Codon 1 (CTG, leucine) is unchanged. Every codon after the insertion is different, because the reading frame has shifted by one base.
The fourth new codon is TGA, which the codon table lists as stop. The protein would end after only three amino acids, two of them wrong.
Answer: The reading frame shifts, so the codons after the insertion are read differently and the protein is changed from that point on.
This is a frameshift mutation. Insertions and deletions of one or two bases are usually far more damaging than a single substitution, because they change every codon downstream instead of just one.
Why the distractors fail: A treats the insertion like a substitution. C ignores the shift (C is a normal base, but its position matters). D would be true only if three bases were inserted, adding a whole codon without shifting the frame.
Tip: for any insertion or deletion, rewrite the sequence and regroup by threes. Do not assume that only one codon changes.
Question 34 of 36
A Trait in Three Generations
A genetic counselor is studying a rare trait in one family. The trait is controlled by a single gene with two alleles, T and t. The gene is not on a sex chromosome, so males and females inherit it in the same way. Two people who married into the family, II-4 and II-5, are shown at the ends of generation II.
In the pedigree, squares are males, circles are females, and shaded symbols show people who have the trait.
Which observation from the pedigree proves that the trait is caused by a dominant allele rather than a recessive allele?
Hint
Suppose the trait were recessive. What genotype would every person with the trait have, and what children could two such people have?
Step-by-step solution
Where the answer comes from: in the pedigree, II-1 and II-5 are both shaded (have the trait), but their daughter III-1 is unshaded.
Step by step:
Test the recessive idea: if the trait were recessive, everyone with it would be tt.
Two tt parents can pass on only t, so every child would be tt and have the trait.
But II-1 × II-5 had an unaffected daughter, III-1. That is impossible if the trait is recessive.
With a dominant allele it works: II-1 and II-5 can both be Tt, and each can pass on t, giving a tt (unaffected) child.
The other choices describe patterns that could also happen with a recessive trait (for example, if I-2 were a carrier), so they do not prove anything.
Answer: II-1 and II-5 both have the trait, but their daughter III-1 does not.
Check: this is the mirror image of the recessive rule: affected × affected → unaffected child proves dominant, just as unaffected × unaffected → affected child proves recessive.
Tip: "proves" means the observation must be impossible under the other explanation. Test each choice against the alternative.
Question 35 of 36
ABO Blood Types
Human ABO blood type is controlled by one gene with three alleles: I^A, I^B and i. Every person has two of these alleles, one from each parent.
Alleles I^A and I^B are codominant: when a person has both, both are fully expressed, and the person has type AB blood. Both I^A and I^B are dominant over i.
Blood type
Possible genotypes
A
I^A I^A or I^A i
B
I^B I^B or I^B i
AB
I^A I^B
O
i\,i
A child has type O blood. Her mother has type A blood. Which blood type could the child's biological father not have?
Hint
What genotype must the child have? Then ask: which blood type has no way to pass on that allele?
Step-by-step solution
Where the answer comes from: the table shows type O has only the genotype i\,i. Type AB has only the genotype I^A I^B.
Step by step:
The child is i\,i, so she received one i from her mother and one i from her father.
The mother (type A) must be I^A i, which is possible.
The father must carry at least one i. Check each type:
Type A: I^A i is possible ✓
Type B: I^B i is possible ✓
Type O: i\,i ✓
Type AB: I^A I^B has noi ✗
Answer: type AB.
Check: a type AB father would give every child either I^A or I^B, so none of his children could have type O.
Tip: in work-backwards problems, start with the offspring whose genotype is certain (a recessive phenotype like type O), then ask what each parent must carry.
Question 36 of 36
Himalayan rabbits have white fur over most of the body but black fur on the ears, nose, feet and tail, the coolest parts of the body. The black color depends on an enzyme, made from instructions in a gene, that works only at cooler temperatures.
A researcher shaves a patch of white fur from a rabbit's back and keeps an ice pack on the patch while the fur grows back. The new fur on the patch is black.
Which conclusion is best supported?
Hint
Did the ice pack change the rabbit's DNA, or did it change the conditions in which the enzyme works? Only DNA in eggs and sperm is passed to offspring.
Step-by-step solution
Where the answer comes from: the stem says the black color comes from "an enzyme, made from instructions in a gene, that works only at cooler temperatures," and that the fur on the cooled patch grew back black.
Step by step:
The rabbit's genes did not change; the same gene was present in the patch before and after.
Cooling the skin let the temperature-sensitive enzyme work there, so the fur grew in black. The environment changed the phenotype in that patch.
Offspring inherit genes from the parents' eggs and sperm. Since no gene changed, the rabbit's offspring would inherit the normal Himalayan pattern, not a black back.
The choices about mutations or a new genotype go against the evidence; the claim that genes play no role ignores that the enzyme comes from a gene.
Answer: The cold changed how the fur-color gene was expressed in the patch, not the gene itself, so the rabbit's offspring would not inherit a black patch.
Check: the pattern of the rabbit's normal fur (black only on cool body parts) fits the same idea without any experiment.
Tip: traits gained from the environment during an organism's life are not inherited, because they do not change the DNA in eggs or sperm.
