Mastering genetics requires more than just memorizing definitions; it demands a solid grasp of probability, logical deduction, and visualizing genetic crosses. This guide provides comprehensive, step-by-step solutions to advanced problems, common pitfalls, and visual aids to help you excel.
A botanist finds a pea plant with axial flowers (a dominant trait, A) but does not know if it is homozygous (AA) or heterozygous (Aa). To find out, she crosses it with a plant that has terminal flowers (aa). The cross yields 120 offspring, out of which 58 have axial flowers and 62 have terminal flowers.
Determine the genotype of the botanist's plant and deduce the phenotypic ratio.
-
Analyze the given data:
- Dominant trait: Axial flowers (A)
- Recessive trait: Terminal flowers (a)
- Test plant genotype: Unknown (A_)
- Crossed with: Terminal flowers (aa)
- Offspring: ~50% Axial (58), ~50% Terminal (62).
-
Formulate Hypotheses:
- Hypothesis 1: If the unknown plant is AA (homozygous dominant), crossing with aa will yield 100% Aa (Axial) offspring. This does not match our data.
- Hypothesis 2: If the unknown plant is Aa (heterozygous), the cross is Aa × aa.
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Perform the Cross (Aa × aa):
- Gametes from Aa: A and a
- Gametes from aa: a and a
| Cross: Aa (Axial) × aa (Terminal) | Gamete a (from Terminal parent) |
|---|
Gamete A (from Axial parent) | Aa (Axial) |
Gamete a (from Axial parent) | aa (Terminal) |
- Calculate Ratios:
- Genotypic Ratio: 1 Aa : 1 aa
- Phenotypic Ratio: 1 Axial : 1 Terminal (or 50% : 50%)
Conclusion: The nearly 1:1 ratio in the offspring (58:62) confirms that the unknown plant is heterozygous (Aa). This specific type of cross, used to determine an unknown genotype, is called a Test Cross.
In pea plants, tall stems (T) are dominant to dwarf stems (t), and purple flowers (P) are dominant to white flowers (p). A plant heterozygous for both traits is crossed with a dwarf, white-flowered plant.
Calculate the probability of obtaining offspring that are tall with white flowers.
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Identify Parent Genotypes:
- Heterozygous for both (Dihybrid): TtPp
- Dwarf, white-flowered (Double recessive): ttpp
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Determine Gametes:
- From TtPp: TP, Tp, tP, tp
- From ttpp: tp (only one type of gamete produced)
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Construct the Punnett Square Diagram:
| Gametes from TtPp (Side) \ Gamete from ttpp (Top) | tp |
|---|
| TP | TtPp (Tall, Purple) |
| Tp | Ttpp (Tall, White) |
| tP | ttPp (Dwarf, Purple) |
| tp | ttpp (Dwarf, White) |
(Note: In a test cross of a dihybrid, the Punnett square can be visualized as a flowchart since one parent only produces one gamete type.)
- Calculate the Probability:
- The possible genotypes are: 1 TtPp : 1 Ttpp : 1 ttPp : 1 ttpp
- The phenotypes are:
- Tall, Purple (TtPp) = 1/4 (25%)
- Tall, White (Ttpp) = 1/4 (25%)
- Dwarf, Purple (ttPp) = 1/4 (25%)
- Dwarf, White (ttpp) = 1/4 (25%)
Conclusion: The probability of obtaining offspring that are tall with white flowers is 25% or 1/4.
A man with Blood Group A and a woman with Blood Group B have a child with Blood Group O.
- What are the exact genotypes of the parents?
- What is the probability that their next child will have Blood Group AB?
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Understand Blood Group Genetics:
- Blood types are determined by three alleles: IA, IB, and i.
- IA and IB are co-dominant, while i is recessive to both.
- Blood Group O requires a homozygous recessive genotype: ii.
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Deduce Parent Genotypes:
- Since the child is Blood Group O (ii), the child must have inherited one i allele from each parent.
- The father has Blood Group A, so his genotype must be IAi (heterozygous), not IAIA.
- The mother has Blood Group B, so her genotype must be IBi (heterozygous), not IBIB.
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Perform the Cross (IAi × IBi):
- Gametes from Father: IA, i
- Gametes from Mother: IB, i
| IB (Mother) | i (Mother) |
|---|
| IA (Father) | IAIB (Group AB) | IAi (Group A) |
| i (Father) | IBi (Group B) | ii (Group O) |
- Analyze the Probabilities:
- Group AB (IAIB): 1/4 (25%)
- Group A (IAi): 1/4 (25%)
- Group B (IBi): 1/4 (25%)
- Group O (ii): 1/4 (25%)
Conclusion:
- The genotypes of the parents are IAi and IBi.
- The probability of their next child having Blood Group AB is 25%.
Color blindness is a recessive sex-linked trait located on the X chromosome (Xc). A man with normal vision marries a woman who is a carrier for color blindness.
What is the probability of them having a color-blind daughter? Explain why.
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Define the Genotypes:
- Normal vision man: XCY (He only has one X chromosome, so it must carry the normal dominant allele XC).
- Carrier woman: XCXc (She has normal vision but carries the recessive allele Xc).
-
Perform the Cross (XCY × XCXc):
- Gametes from Father: XC, Y
- Gametes from Mother: XC, Xc
Probability of Offspring Phenotypes:
- Normal Female (XCXC): 25%
- Carrier Female (XCXc): 25%
- Normal Male (XCY): 25%
- Color-blind Male (XcY): 25%
| XC (Mother) | Xc (Mother) |
|---|
| XC (Father) | XCXC (Normal Daughter) | XCXc (Carrier Daughter) |
| Y (Father) | XCY (Normal Son) | XcY (Color-blind Son) |
- Analyze the Female Offspring:
- To have a color-blind daughter, the daughter must inherit two recessive alleles (XcXc).
- She must get one Xc from her mother (which is possible).
- She must get another Xc from her father. However, the father is XCY; he only has a dominant XC to give to his daughters.
Conclusion: The probability of them having a color-blind daughter is 0%. Any daughter will inherit the dominant XC allele from the father, guaranteeing normal vision.
These questions test if you are actually reading the problem carefully or just relying on memorized ratios.
Question: In a standard dihybrid cross of two heterozygous pea plants for seed shape and color (RrYy × RrYy), what fraction of the Round Yellow offspring are completely homozygous for both traits (RRYY)?
- Common Mistake: Students instantly recall the dihybrid Punnett square and answer 1/16.
- The Trap: The question asks for the fraction out of the Round Yellow offspring, NOT out of the total offspring!
- Correct Solution:
- Total Round Yellow offspring in the F2 generation = 9 (out of 16).
- Number of completely homozygous Round Yellow (RRYY) = 1.
- Therefore, the correct fraction is 1/9.
Question: A heterozygous dominant brown-eyed man (Bb) and a blue-eyed woman (bb) have three children, all of whom have brown eyes. What is the probability that their fourth child will have blue eyes?
- Common Mistake: Thinking that because they already have 3 brown-eyed children, the "ratio" must balance out, so the next one must be blue-eyed. Alternatively, thinking the probability is 0% because all previous kids were brown.
- The Trap: Genetic events (like coin flips) are independent. Past pregnancies do not affect future ones.
- Correct Solution: The cross is Bb × bb. The probability of a blue-eyed child (bb) is always 50% for every single pregnancy, regardless of the phenotypes of previous children.
BioNotes