Interleaved — Phase 3

Biology interleaved practice

printable — key stays hidden on paper

Instructions: Solve each problem showing your reasoning. Watch carefully — consecutive problems draw from different topics, so decide which genetic principle or method applies before you start. Show Punnett squares, probability calculations, and pedigree logic where relevant. Total: 50 marks.

1. In a cross between two heterozygous tall pea plants ( $Tt \times Tt$ ), where $T$ (tall) is dominant, work out the expected phenotypic ratio of the offspring. Then state the probability that a randomly chosen offspring is homozygous recessive. (5 marks)

2. A couple, both with blood type AB, have children. Using the codominance model for the ABO system, determine all possible blood types of their children and the ratio in which they occur. (5 marks)

3. Define each of the following precisely and give one example of how they differ: gene, allele, genotype, phenotype. (4 marks)

4. A red-flowered snapdragon ( $C^R C^R$ ) is crossed with a white one ( $C^W C^W$ ), and all F1 are pink. Two pink F1 plants are then crossed. State which inheritance pattern this is, and give the F2 genotypic and phenotypic ratios. (5 marks)

5. Hemophilia is an X-linked recessive disorder. A carrier woman marries a man who does NOT have hemophilia. Calculate the probability that (a) a son has hemophilia, and (b) a daughter is a carrier. (5 marks)

6. In a dihybrid cross $RrYy \times RrYy$ (round/wrinkled, yellow/green peas; $R$ and $Y$ dominant), use the product rule to find the probability of an offspring being round AND green. (5 marks)

7. Briefly summarize how the Hershey–Chase experiment provided evidence that DNA, not protein, is the genetic material. (4 marks)

8. You have a fruit fly showing the dominant phenotype for body color but do not know its genotype. Describe the test cross you would perform and explain how the results tell you whether it is homozygous or heterozygous. (5 marks)

9. In the pedigree below, filled symbols = affected individuals. Determine whether the trait is most likely autosomal recessive or X-linked recessive, and justify with two pieces of evidence.

   I     ●───────□
         │       │
   II  ┌─┴─┬─────┴─┐
       □   ■       ○

(Circle = female, square = male, filled = affected.) (6 marks)

10. Two genes are linked on the same chromosome. In a test cross of 1000 offspring, 460 + 440 are parental types and 55 + 45 are recombinants. Calculate the recombination frequency and state the approximate map distance in map units. (6 marks)

Answer keyMark scheme & solutions

1. (Tests 3.1.6 Monohybrid cross + 3.1.9 probability) $Tt \times Tt$ → Punnett square gives $1\,TT : 2\,Tt : 1\,tt$ . Phenotypic ratio = 3 tall : 1 short. P(homozygous recessive $tt$ ) = $\tfrac{1}{4}$ . Why: single-trait cross with both parents heterozygous → standard monohybrid, recognized by one gene with dominant/recessive alleles.

2. (Tests 3.2.2 Codominance / ABO) $I^A I^B \times I^A I^B$ → offspring: $\tfrac14 I^A I^A$ , $\tfrac12 I^A I^B$ , $\tfrac14 I^B I^B$ . Phenotypes: 1 type A : 2 type AB : 1 type B. Why: A and B alleles are codominant (both expressed in AB), so heterozygotes show a distinct third phenotype — not a 3:1 dominance ratio.

3. (Tests 3.1.1 Key terms)

Gene: a segment of DNA coding for a functional product/trait (e.g., flower-color gene).
Allele: an alternative version of a gene (e.g., red vs. white).
Genotype: the genetic makeup / allele combination (e.g., $Rr$ ).
Phenotype: the observable trait resulting from genotype + environment (e.g., red flowers). Why: pure definitional recall; distinguishing genotype (alleles) from phenotype (appearance) is the key contrast.

4. (Tests 3.2.1 Incomplete dominance) Pattern = incomplete dominance (heterozygote intermediate). $C^R C^W \times C^R C^W$ → $1\,C^R C^R : 2\,C^R C^W : 1\,C^W C^W$ . Genotypic AND phenotypic ratio = 1 red : 2 pink : 1 white. Why: F1 blend (pink) signals incomplete dominance, so genotype ratio = phenotype ratio (each genotype has a unique appearance).

5. (Tests 3.2.8 X-linked recessive + 3.1.9 probability) Carrier mother $X^H X^h$ × normal father $X^H Y$ . Sons: $X^H Y$ (normal) or $X^h Y$ (hemophiliac) → P(son affected) = ½. Daughters: $X^H X^H$ (normal) or $X^H X^h$ (carrier) → P(daughter carrier) = ½. Why: X-linked recessive — sons inherit X only from mother, so affected sons appear even with a normal father; daughters need two copies so are unaffected but can be carriers.

6. (Tests 3.1.7 Dihybrid + 3.1.10 product rule) P(round) = $\tfrac34$ (from $Rr\times Rr$ ); P(green) = $\tfrac14$ (from $Yy\times Yy$ ). Independent genes → multiply: $\tfrac34 \times \tfrac14 = \tfrac{3}{16}$ . P(round and green) = 3/16. Why: use the product rule on two independently assorting genes rather than building a full 16-box square.

7. (Tests 3.3.1 DNA is genetic material) Hershey–Chase labeled phage DNA with $^{32}$ P and protein coats with $^{35}$ S. After infecting bacteria and blending/centrifuging, the $^{32}$ P (DNA) entered the bacterial cells and directed production of new phages, while $^{35}$ S (protein) stayed outside. This showed DNA, not protein, is the hereditary material injected into the host. Why: the differential radioactive labeling isolates which molecule carries genetic information.

8. (Tests 3.1.8 Test cross) Cross the unknown dominant individual with a homozygous recessive (e.g., $?B \times bb$ ).

If all offspring show dominant phenotype → unknown is homozygous ( $BB$ ).
If ~½ offspring show recessive phenotype → unknown is heterozygous ( $Bb$ ). Why: the recessive tester reveals hidden alleles because recessive offspring only appear if the unknown carried a recessive allele.

9. (Tests 3.1.11 Pedigree interpretation) Most likely autosomal recessive. Evidence: (1) An affected child (II) is born to unaffected parents → the trait skips a generation, indicating recessive inheritance. (2) Affected individuals include a male whose mother is unaffected; if X-linked recessive from an unaffected non-carrier appearing parent it's less consistent — and affected offspring can appear from two carrier parents regardless of sex, pointing to autosomal. Why: "skips generations + appears from unaffected parents" = recessive; equal likelihood across sexes argues autosomal over X-linked.

10. (Tests 3.2.10 Recombination frequency + 3.2.11 map distance) Recombinants = $55 + 45 = 100$ ; total = 1000. $RF = \dfrac{100}{1000} = 0.10 = 10\%$ . Map distance ≈ 10 map units (centimorgans). Why: RF = recombinants/total; 1% RF ≈ 1 map unit, so distance = 10 cM.

[
  {"claim":"P(tt) in Tt x Tt = 1/4","code":"from sympy import Rational\np=Rational(1,4)\nresult = (p==Rational(1,4))"},
  {"claim":"round and green = 3/16","code":"from sympy import Rational\nres=Rational(3,4)*Rational(1,4)\nresult = (res==Rational(3,16))"},
  {"claim":"recombination frequency = 10%","code":"rf=(55+45)/1000\nresult = (abs(rf-0.10)<1e-9)"}
]