Monohybrid Crosses: Study Notes

Monohybrid Crosses: Definition and Scope

• A monohybrid cross involves only one set of contrasting characteristics. - Example: Height in pea plants (tall vs. short).

Mendel's Experiments: Father of Genetics

• Gregor Mendel (Austrian monk) conducted pea plant breeding experiments from 1857 to 1864.

• Studied seven contrasting characteristics, including:- Plant height: tall vs. short

  • Flower colour: red vs. white

  • Seed shape: round vs. wrinkled

• Contrasting characteristics: each trait can be expressed in opposite forms.

• Mendel laid the foundation for inheritance understanding, even though he did not know about genes or chromosomes.

Law of Dominance

• Experiment: Cross a homozygous tall plant (TT) with a homozygous short plant (tt).

• F1 generation: all offspring were tall (genotype: TT or Tt; phenotype: tall).

• Interbreeding F1: F2 generation shows approximately 75% tall and 25% short.

• Conclusions:- The tall trait is dominant; the short trait is recessive.

  • In a cross between pure contrasting traits, only one form appears in the progeny of the F1 generation.

• Law statement: $In ext{ a cross of parents that are pure for contrasting traits, only one form of the trait will appear in the progeny.}$

Law of Segregation

• Every characteristic is controlled by two factors (alleles) in an organism.

• During gamete formation, these two factors separate (segregate) from each other.

• Each gamete receives only one allele for a given trait.

• Example (height): some gametes carry the tall allele, others the short allele.

• Law statement: $For ext{ each characteristic a plant possesses two factors which separate or segregate so that each gamete contains only one of these factors.}$

Rules for Genetic Crosses

1. Representing Alleles

• Identify dominant and recessive traits.

• Dominant allele: represented by a capital letter (e.g., $T$).

• Recessive allele: represented by the same letter in lowercase (e.g., $t$).

• Tip: Use the first letter of the dominant trait (e.g., $T$ for Tall, $t$ for short).

• Each gene has two alleles (two letters).

2. Genetic Generations

• P1: Parent generation (the original cross).

• F1: First filial generation (offspring from the P1 cross).

• F2: Second filial generation (offspring from crossing F1 individuals).

3. Representing Gametes

• Gametes carry only one allele for each trait.

• When writing gametes in a cross, separate the alleles with a comma (e.g., $T, t$).

4. Punnett Square Rules

• The first block (top-left) of the Punnett Square is left empty.

• The arrangement of male vs. female gametes does not affect the final outcome.

• Within the Punnett Square, write the dominant allele first (e.g., $Tt$, not $tT$).

5. Presenting Results (Genotype and Phenotype)

• Report results as percentages or ratios (e.g., $100\%\ Tt$; $100\%\ Tall$).

6. Key Terms to Remember

• Pure bred = homozygous (two identical alleles, e.g., $TT$ or $tt$).

• Hybrid = heterozygous (two different alleles, e.g., $Tt$).

• Autosomal: traits controlled by genes on autosomes (non-sex chromosomes).

• Gonosomal: traits controlled by genes on sex chromosomes (X or Y).

Predicting Monohybrid Cross Outcomes

• Understanding parental genotypes allows prediction of offspring genotype and phenotype.

1: Homozygous Dominant (e.g., $TT$) \times Homozygous Recessive (e.g., $tt$)

• Genotype result: $100\%\ Tt$

• Phenotype result: $100\%\ Tall$

2: Homozygous \times Heterozygous

• Scenario A: $TT \times Tt$- Genotype: $50\%\ TT,\ 50\%\ Tt$ (ratio $1:1$ for genotypes)

  • Phenotype: $100\%\ Tall$ (dominant)

• Scenario B: $tt \times Tt$- Genotype: $50\%\ Tt,\ 50\%\ tt$ (ratio $1:1$)

  • Phenotype: $50\%\ Tall,\ 50\%\ short$ (ratio $1:1$)

3: Heterozygous \times Heterozygous

• Genotype: $25\%\ TT,\ 50\%\ Tt,\ 25\%\ tt$ (ratio $1:2:1$)

• Phenotype: $75\%\ Tall,\ 25\%\ short$ (ratio $3:1$)

Diagramming Monohybrid Crosses

• Cross diagrams are shown step-by-step using a standard diagrammatic format:- Example: P1 phenotype: Homozygous Tall \times Homozygous Short

  • P1 genotype: $TT \times tt$

  • Meiosis: gametes $T, T$ and $t, t$

  • Fertilization: F1 genotype: $100\%\ Tt$

  • F1 phenotype: $100\%\ Tall$

• Another example (P2): Heterozygous Tall \times Heterozygous Tall- P2 genotype: $Tt \times Tt$

  • Meiosis: gametes $T, t\; T, t$

  • Fertilization: F2 genotype: $TT,\ Tt,\ Tt,\ tt\; (25\%\ TT, 50\%\ Tt, 25\%\ tt)$

  • F2 phenotype: $75\%\ Tall,\ 25\%\ short$

Monohybrid Cross: Examples

Example 1: Dog Ear Length

• Problem: Pure-breeding long-eared dog \times pure-breeding short-eared dog. All F1 are long-eared.

• Deduction: Long ears are dominant. Let $L$ = long ears, $l$ = short ears. Pure breeding means homozygous.

• P1: $LL \times ll$

• Meiosis: $L, L$ and $l, l$

• F1 genotype: $100\%\ Ll$

• F1 phenotype: $100\%\ Long\ Ears$

Example 2: Plant Seed Colour

• Problem: Green-seeded plant \times yellow-seeded plant; F1 all green; F1 inbred to produce F2.

• Deduction: Green is dominant. Let $G$ = green, $g$ = yellow. P1 must be $GG \times gg$

• F1 genotype: $Gg$; F1 phenotype: $100\%\ Green\ Seeds$

• P2 cross (F1 \times F1): $Gg \times Gg$

• Meiosis: $G, g\; G, g$

• Fertilization: F2 genotype: $GG,\ Gg,\ Gg,\ gg\; (25\%\ GG, 50\%\ Gg, 25\%\ gg)$

• F2 phenotype: $75\%\ Green,\ 25\%\ Yellow$

Example 3: Fly Body Colour

• Problem: Grey-bodied flies \times black-bodied flies; F1 all grey. Determine F2 genotypes if F1 are inbred.

• Deduction: Grey (G) is dominant over black (g); F1 are heterozygous: $Gg$

• P2 cross: $Gg \times Gg$

• Meiosis: $G, g\; G, g$

• F2 genotype: $25\%\ GG, 50\%\ Gg, 25\%\ gg$

• F2 phenotype: $75\%\ Grey,\ 25\%\ Black$

Practice Questions (Multiple Choice) and Answer Key

• Q1: 4 different phenotypes are possible in the F1 generation if the parents’ blood types are…

  • A. B and B

  • B. A and B

  • C. O and AB

  • D. AB and AB

• Q2: In humans, brown eye color is dominant over blue. A blue-eyed mother has two children: one brown-eyed boy and one blue-eyed girl. The father’s eye color is…

  • A. Brown, because brown is sex-linked

  • B. Brown, because at least one parent must have brown

  • C. Blue, because family history may include blue eyes

  • D. Blue, because at least one parent must be heterozygous

• Q3: Inheritance of blood groups involves…

  • A. Multiple alleles

  • B. Co-dominance

  • C. Both A and B

  • D. Neither A nor B

• Q4: A characteristic that is only expressed in the homozygous state is…

  • A. Dominant

  • B. Recessive

  • C. Both A and B

  • D. Neither A nor B

• Q5: Blood group AB is a result of…

  • A. Complete dominance

  • B. Polygenic inheritance

  • C. Incomplete dominance

  • D. Co-dominance

• Q6: The probability that two heterozygous parents will have a homozygous dominant offspring is…

  • A. 75%

  • B. 50%

  • C. 25%

  • D. 100%

• Q7: The probability that two homozygous parents will have a heterozygous dominant offspring is…

  • A. 75%

  • B. 50%

  • C. 25%

  • D. 100%

• Q8: The probability that a homozygous parent and a heterozygous parent will have a heterozygous dominant offspring is…

  • A. 75%

  • B. 50%

  • C. 25%

  • D. 100%

• Q9: If a mother and child have blood type AB, the father cannot have blood type…

  • A. B

  • B. A

  • C. AB

  • D. O

• Q10: If both parents are blood type A, the child can be blood type…

  • A. AB

  • B. B

  • C. O

  • D. Both A and B

• Q11: A cross where red flowers (RR) \times white flowers (rr) produce pink offspring illustrates…

  • A. Complete dominance

  • B. Incomplete dominance

  • C. Co-dominance

  • D. None of the above

• Q12: Roan fur in a cross between white (CC) and red (RR) horses illustrates…

  • A. Complete dominance

  • B. Incomplete dominance

  • C. Co-dominance

  • D. None of the above

• Q13: A homozygous long whiskered cat crossed with a homozygous short whiskered cat yields F1 that are all long whiskered. The F1 genotype is…

  • A. LL

  • B. Ll

  • C. ll

  • D. None of the above

• Q14: In a family, two parents with blood types are given; if the father is A, what blood type must the mother have given the child types A, O, AB, B? (Refer to the table in the transcript)

  • A. A

  • B. B

  • C. O

  • D. AB

• Q15: The allele that does not express itself in a heterozygous condition is…

  • A. Dominant

  • B. Recessive

  • C. Both A and B

  • D. Neither A nor B

• Answer Key:

  • 1: B

  • 2: B

  • 3: C

  • 4: B

  • 5: D

  • 6: C

  • 7: D

  • 8: B

  • 9: D

  • 10: C

  • 11: B

  • 12: C

  • 13: B

  • 14: B

  • 15: B

Connections to Foundational Principles and Real-World Relevance

• Mendelian genetics underpins modern understanding of inheritance patterns in humans, plants, and animals.

• Distinguishing between dominant/recessive and homozygous/heterozygous genotypes helps predict trait distribution in offspring.

• Punnett squares and branching ratios provide a framework for genetic counseling, breeding programs, and understanding genetic diseases.

• The concepts of autosomal versus gonosomal (sexual) inheritance explain why some traits show sex differences or linkage.

• Real-world relevance includes predicting disease risk, agricultural breeding, and interpreting family genetic histories.

Notation and Formulas Summary (Quick Reference)

• Alleles: dominant capital, recessive lowercase. For a gene with alleles $A$ and $a$, genotypes are $AA, Aa, aa$

• Common genotype-to-phenotype mappings:- AA, Aa \
  \rightarrow A phenotype (dominant)

  • aa \
    \rightarrow a phenotype (recessive)

• Key ratios in monohybrid crosses:- Genotype ratio (heterozygous cross): $1:2:1$ for $AA: Aa: aa$

  • Phenotype ratio (dominant-recessive cross): $3:1$ for dominant:recessive trait

• Common cross results often presented as percentages, e.g., $25\%$, $50\%$, $75\%$.

Practical Takeaways for Study

• Always identify the parental genotypes before predicting offspring.