Mendelian Genetics: Dihybrid Crosses, Pedigrees, and Probability Rules

Attendance and course overview

• Roll call occurred: Abigail, Jessica, Ryan, Miana, Mercedes, Olivia, Mikaela (all present).
• This week’s plan: continue Mendelian genetics with dihybrid crosses; next class will cover pedigree analyses; then an extension of Mendelian genetics with more complex scenarios.
• Homework: due at the beginning of next class; print and show work; electronic copy available via Canvas or email if needed.

Mendelian genetics: quick recap and key ideas

• Mendel’s fourth postulate (independent assortment): During gamete formation, segregating pairs of unit factors (alleles) assort independently of one another.
• Dihybrid crosses examine two traits simultaneously (e.g., color and texture in pea plants).
• Classic result: When crossing true-breeding yellow-round peas (dominant yellow Y and dominant round R) with true-breeding green-wrinkled peas (recessive y and r), the F2 generation shows a phenotypic ratio of $9:3:3:1$ for yellow-round : yellow-wrinkled : green-round : green-wrinkled.
• Key takeaway: Each trait behaves independently in the dihybrid cross; the overall ratio is the product of the individual single-trait ratios when genes assort independently.

Dihybrid cross: the nine-to-three-to-three-to-one pattern

• For a dihybrid cross (YyRr x YyRr), the F2 phenotypic ratio is:
  • Yellow & Round: $9$ parts
  • Yellow & Wrinkled: $3$ parts
  • Green & Round: $3$ parts
  • Green & Wrinkled: $1$ part
• Example interpretation: If you look at a single trait, say yellow, the phenotype distribution is 9+3 = 12 yellow out of 16 total, i.e., $12/16 = 3/4$, which matches the single-trait ratio of yellow vs green.
• The same logic applies to the other trait (round vs wrinkled).

Worked example: cats with curled ears and fur color (two genes, independent)

• Genes and alleles:
  • Ears: curled ears (dominant C) vs normal ears (recessive c)
  • Fur color: black (dominant G) vs gray (recessive g)
• Given cross setup:
  • Gray cat with curled ears (gg and CC or Cc; the problem uses a heterozygous F1 for both traits): genotype CcGg
  • Black cat with normal ears: genotype ccGG
  • F1 generation: all black with curled ears (CcGg)
• Part a: If two F1 cats mate, F2 phenotypes and proportions
  • Classic dihybrid cross (CcGg x CcGg) yields the $9:3:3:1$ phenotypic ratio: Black curled : Black normal : Gray curled : Gray normal
• Part b: F1 cat mates with a stray gray fur, normal ears cat (ggcc)
  • F1 genotype: CcGg; stray cat genotype: ggcc
  • Possible gametes from F1: {Cg, Cc, cG, cg}; from stray: {cg} only
  • Offspring phenotypes: any with at least one C (curled) and at least one G (black) will be black; but since the stray contributes only g and c alleles, the resulting phenotypic distribution across the two traits becomes a 1:1:1:1 ratio for the four phenotypes: Black curled, Black normal, Gray curled, Gray normal
  • This illustrates how cross context (one parent homozygous recessive for both traits) can collapse the dihybrid proportions to equal-probability phenotypes in the F2 generation
• Practical note: When one parent is homozygous recessive for both genes, only the dominant phenotypes from the other parent can appear, but the combination across two loci yields four equally likely phenotype categories in this particular cross

Test crosses: determining unknown genotypes

• Purpose: to determine whether an unknown phenotype is homozygous dominant or heterozygous for a gene, or to infer dihybrid genotypes
• Classic approach: cross the unknown with a homozygous recessive individual for the same two genes (a test cross)
• Example discussion: If a dihybrid cross yields the expected 9:3:3:1 phenotypic ratio, you can use test crosses to deduce which parents were homozygous vs heterozygous for each gene
• Takeaway: Punnett squares visualize outcomes; test crosses help infer genotype information that phenotype alone cannot reveal

Probability rules in genetics: product rule and sum rule

• Product rule: The probability of two independent events A and B both occurring is $P(A ext{ and } B) = P(A) imes P(B)$
  • Example: In a monohybrid cross $Yy imes Yy$, probability of YY offspring is $P(Y) imes P(Y) = frac{1}{2} imes frac{1}{2} = frac{1}{4}$, which matches the Punnett square
• Sum rule: The probability of either of two mutually exclusive events A or B occurring is $P(A ext{ or } B) = P(A) + P(B)$
  • Example: Probability of obtaining a heterozygous offspring (Yy) from $Yy imes Yy$ is $P(Yy) = P( ext{Y from sperm and y from ovule}) + P( ext{y from sperm and Y from ovule}) = frac{1}{4} + frac{1}{4} = frac{1}{2}$
• Product and sum rules align with Punnett-square results and underpin multi-gene cross calculations

Applying probability rules to dihybrid and multi-gene crosses

• Example: Dihybrid cross with pea color (yellow vs green) and texture (round vs wrinkled)
  • Probability yellow: $P( ext{yellow}) = 3/4$; probability green: $1/4$
  • Probability round: $P( ext{round}) = 3/4$; probability wrinkled: $1/4$
  • Probability of yellow and round together (independent traits): $P( ext{yellow and round}) = (3/4) imes (3/4) = 9/16$
  • Probability yellow and wrinkled: $(3/4) imes (1/4) = 3/16$
  • Probability green and round: $(1/4) imes (3/4) = 3/16$
  • Probability green and wrinkled: $(1/4) imes (1/4) = 1/16$
• This mirrors the Punnett-square-derived $9:3:3:1$ ratio for phenotypes

Counting gametes and multigene crosses

• Number of gametes produced when counting heterozygous genes: $2^n$, where $n$ is the number of heterozygous genes
  • Example: 4 heterozygous genes → $2^4 = 16$ possible gametes from one parent
  • If both parents are heterozygous for the same four genes, there are $16 imes 16 = 256$ possible genotypes in the offspring (this is cumbersome to do with a full Punnett square)
• Practical approach: break into independently assorting monohybrid crosses and multiply results; or use branched line diagrams for quicker visualization

Branched line diagrams: a quick-probability alternative

• Concept: Break problem into independent gene events and multiply probabilities along branches, then combine outcomes
• Example with two genes (A/a and B/b, both heterozygous in the cross):
  • Gene A: dominant phenotype occurs with probability $3/4$, recessive with $1/4$
  • Gene B: dominant phenotype occurs with probability $3/4$, recessive with $1/4$
  • Probability of offspring showing dominant phenotype for A and B: $(3/4) imes (3/4) = 9/16$
  • Other phenotype combinations follow similarly: $(3/4) imes (1/4) = 3/16$, $(1/4) imes (3/4) = 3/16$, $(1/4) imes (1/4) = 1/16$
• Branched line diagrams scale to more genes by adding more branches; they’re generally faster than expanding large Punnett squares
• You can also use branched lines to answer questions about combining multiple genotype or phenotype conditions (e.g., dominant across several genes, recessive in one gene, etc.)

Worked example: probability for multiple traits with branched diagrams

• Example: A cross with four genes, all heterozygous in the parents. We want the fraction of progeny with:
  • Dominant traits for A, C, D, and a recessive trait for B
• Probabilities for each gene (assuming standard Aa x Aa):
  • A: dominant phenotype probability = $3/4$
  • B: recessive phenotype probability = $1/4$
  • C: dominant phenotype probability = $3/4$
  • D: dominant phenotype probability = $3/4$
• Combined probability (product rule, assuming independence): $P = (3/4) imes (1/4) imes (3/4) imes (3/4) = 27/256$
• For genotypes with the same four genes, but focusing on the exact genotype (e.g., AA bb Cc Dd with specific zygosity), probabilities adjust accordingly (example below)
• Genotype example (AA bb Cc Dd):
  • P(AA) = 1/4, P(bb) = 1/4, P(Cc) = 1/2, P(Dd) = 1/2
  • Combined genotype probability: $(1/4) imes (1/4) imes (1/2) imes (1/2) = 1/64$
• Key point: always check whether the question asks about phenotypes or genotypes, and apply the appropriate probabilities accordingly

Practice problems and worked solutions (selected from the session)

• PKU inheritance (human) practice problem:
  • PKU is recessive; normal allele is dominant. Let k be PKU allele, K be normal.
  • Two parents both heterozygous (Kk) have a child with PKU (kk). From this:
  • (a) Probability that the father’s sperm carries the PKU allele (k): 50% (half of his gametes carry k)
  • (b) Probability that the mother’s egg carries the PKU allele: 50%
  • (c) Probability that their next child will have PKU: 25%
  • (d) Probability that their next child will be heterozygous (Kk): 50%
• Another problem: “an individual with genotype shown here” – probability of producing a gamete containing dominant alleles for all five genes
  • Approach: treat each gene separately; if a gene is heterozygous, chance of dominant allele in a gamete is 1/2; if homozygous dominant, chance is 1 (1/1)
  • Example calculation yields $1/16$ for five genes with four heterozygous and one homozygous dominant condition as described in the slide set
• Quick check: the probability that a gamete contains dominant alleles for all five genes in a given heterozygous context can be 1/16 depending on the exact zygosity pattern used in the example

Real-world and exam-oriented takeaways

• Punnett squares are powerful visualization tools for one- or two-gene crosses but become unwieldy for many genes; probability laws and branched diagrams provide efficient alternatives
• The product rule and sum rule are foundational for solving dihybrid and multi-gene problems without enumerating every single Punnett square
• Remember the distinction between analyzing phenotypes vs. genotypes; probabilities differ depending on what the question asks you to count
• The number of possible gametes from a genotype is $2^n$, where $n$ is the number of heterozygous loci; when crossing two such individuals, the potential genotype space can be enormous, which motivates modular solution strategies

Connections, implications, and practical notes

• Pedigree analyses (next class) expand these concepts to family trees and inheritance patterns across generations
• Extensions of Mendelian genetics may include incomplete dominance, codominance, polygenic traits, gene linkage, epistasis, and environmental interactions; these add complexity beyond the dihybrid ideals but build on the same probabilistic framework
• Ethical and practical implications in human genetics include genetic counseling, interpretation of probabilistic risk, and the limitations of predicting phenotypes from genotypes in real populations

Homework reminder and next class focus

• Bring completed homework to next class for submission; if unavailable, contact the instructor for an electronic copy
• Next class topics: pedigree analyses, then extension/expansion of Mendelian genetics with more complex scenarios
• Be prepared to apply Punnett squares, probability rules, and branched-line diagrams to new problems