Mendelian Genetics and Inheritance Patterns

Mendel's laws, dominance, co-dominance, sex-linkage, epistasis, chi-squared analysis

# Mendelian Genetics and Inheritance Patterns

Gregor Mendel established the foundation of genetics through his work with pea plants. AP Biology requires understanding of Mendel's laws, non-Mendelian inheritance patterns, pedigree analysis, and statistical testing with chi-squared.


1. Mendel's Laws

Law of Segregation

  • Each organism has two alleles for each gene (one from each parent)
  • Alleles segregate during gamete formation — each gamete receives only ONE allele
  • Basis: separation of homologous chromosomes in meiosis I

Law of Independent Assortment

  • Genes on different chromosomes assort independently during meiosis
  • Basis: random orientation of bivalents in metaphase I
  • Exception: linked genes (on the same chromosome) tend to be inherited together

2. Patterns of Inheritance

Complete Dominance

  • Heterozygote shows dominant phenotype only
  • Monohybrid cross (Aa × Aa): 3:1 phenotypic ratio
  • Test cross (Aa × aa): 1:1 ratio confirms heterozygosity

Incomplete Dominance

  • Heterozygote shows intermediate phenotype
  • Example: Red × White snapdragons → Pink (CRCWC^RC^W)
  • F₂ ratio: 1:2:1 (red : pink : white)

Codominance

  • Both alleles fully expressed in heterozygote
  • Example: ABO blood groups — IAIBI^AI^B → Type AB (both A and B antigens present)
  • Sickle cell: HbAHbSHb^AHb^S → both normal and sickle haemoglobin produced

Multiple Alleles

  • More than two alleles exist in the population (individuals still have only 2)
  • Example: ABO system has three alleles: IAI^A, IBI^B, IOI^O

3. Sex-Linked Inheritance

  • Genes on the X chromosome show different patterns in males and females
  • Males (XYXY) express X-linked recessive alleles because they have only one X
  • Example: Colour blindness (XbYX^bY = affected male; XBXbX^BX^b = carrier female)
  • Reciprocal crosses give different results for sex-linked traits

4. Beyond Mendel

Epistasis

  • One gene masks or modifies the expression of another
  • Produces modified ratios (9:3:4, 12:3:1, 9:7, etc.)
  • Example: Labrador coat colour — E gene controls pigment deposition; B gene determines colour

Polygenic Inheritance

  • Multiple genes contribute to one phenotype
  • Produces continuous variation (bell curve)
  • Examples: height, skin colour, eye colour

Pleiotropy

  • One gene affects multiple phenotypes
  • Example: Sickle cell allele → sickle-shaped RBCs, anaemia, pain crises, malaria resistance

Environmental Effects

  • Environment can influence gene expression
  • Example: Hydrangea flower colour depends on soil pH; fur colour in Siamese cats depends on temperature

5. Pedigree Analysis

Autosomal Recessive

  • Affected individuals often have unaffected parents (carriers)
  • Appears in both sexes equally
  • Skip generations pattern

Autosomal Dominant

  • Affected individuals have at least one affected parent
  • Does not skip generations (usually)

X-linked Recessive

  • More males affected than females
  • Affected father cannot pass to sons (sons get Y)
  • Carrier mothers can have affected sons

6. Chi-Squared Test

χ2=(OE)2E\chi^2 = \sum \frac{(O - E)^2}{E}

Steps:

  1. State null hypothesis (H0H_0): no significant difference between observed and expected
  2. Calculate expected values from predicted ratio
  3. Calculate χ2\chi^2
  4. Determine degrees of freedom = categories − 1
  5. Compare to critical value (p = 0.05)
  6. If χ2\chi^2 < critical value → accept H0H_0 (results fit expected ratio)
  7. If χ2\chi^2 > critical value → reject H0H_0 (significant difference)

Example

Cross Aa × Aa, observe: 850 dominant, 150 recessive (total 1000) Expected (3:1): 750 dominant, 250 recessive χ2=(850750)2750+(150250)2250=10000750+10000250=13.33+40=53.33\chi^2 = \frac{(850-750)^2}{750} + \frac{(150-250)^2}{250} = \frac{10000}{750} + \frac{10000}{250} = 13.33 + 40 = 53.33 df = 1; critical value = 3.84. Since 53.33 > 3.84 → reject H0H_0. Results do NOT fit a 3:1 ratio.


Worked Example

Question: In fruit flies, eye colour is X-linked. Red (R) is dominant over white (r). Cross a carrier female with a red-eyed male. What phenotypic ratio is expected? (3 points)

Solution:

Parents: XRXrX^RX^r × XRYX^RY

XRX^R XrX^r
XRX^R XRXRX^RX^R (red ♀) XRXrX^RX^r (red ♀, carrier)
YY XRYX^RY (red ♂) XrYX^rY (white ♂)

Expected ratio: 3 red : 1 white. All females are red-eyed. 50% of males are red-eyed, 50% are white-eyed.


Practice Questions

    1. Distinguish between incomplete dominance and codominance. (2 points)
    1. Explain why X-linked recessive traits are more common in males. (2 points)
    1. A dihybrid cross (AaBb × AaBb) yields 590 A_B_, 200 A_bb, 195 aaB_, 15 aabb. Perform a chi-squared test against 9:3:3:1 (critical value at df=3 is 7.82). (4 points)
    1. Define epistasis and give an example. (2 points)

    Answers

    1. In incomplete dominance, the heterozygote shows a blended/intermediate phenotype (e.g., red × white → pink). In codominance, the heterozygote shows both parental phenotypes simultaneously (e.g., IAIBI^AI^B → type AB blood with both A and B antigens).

Want to check your answers and get step-by-step solutions?

Get it on Google PlayDownload on the App Store

Summary

  • Mendel's laws: segregation (alleles separate in meiosis I) and independent assortment (genes on different chromosomes).
  • Inheritance patterns: complete dominance (3:1), incomplete dominance (1:2:1), codominance, sex-linkage.
  • Non-Mendelian: epistasis, polygenic inheritance, pleiotropy, environmental effects.
  • Chi-squared test: χ2=(OE)2/E\chi^2 = \sum(O-E)^2/E; compare to critical value to evaluate ratios.
  • Pedigree analysis: identify autosomal dominant/recessive and X-linked patterns.

Ready to Ace Your AP biology?

Get instant step-by-step solutions to any problem. Snap a photo and learn with Tutor AI — your personal exam prep companion.

Get it on Google PlayDownload on the App Store