Genotype Calculator: Genotype vs Phenotype, Ratios

A genotype is the pair of alleles an organism carries for a gene, written as AA, Aa or aa. Crossing Aa × Aa gives genotypes in a 1 AA : 2 Aa : 1 aa ratio and phenotypes in a 3 : 1 ratio.

The genotype is the genetic instruction. The phenotype is what that instruction produces in a real body and a real environment. Two genotypes, AA and Aa, can give the same phenotype when one allele is dominant, which is why you cannot always read a genotype from the outside.

Rule of thumb: a dihybrid cross AaBb × AaBb gives 9:3:3:1. If your counts give a chi-square p-value of 0.05 or more (below 7.815 on 3 df), they fit Mendel’s ratio.

Genotype definition in biology: a genotype is the set of alleles an organism carries at one or more genes. An allele is one version of a gene. Because most animals and plants are diploid, they carry two alleles for each gene, one from each parent. We write them as two letters: a capital letter for the dominant allele and a small letter for the recessive allele.

The word was coined by the Danish botanist Wilhelm Johannsen in 1909 and explained in English in 1911, together with its partner word, phenotype. Johannsen wanted to separate what an organism carries from what it shows. That split is still the heart of genetics.

What is the difference between genotype and phenotype? The genotype is the genetic code, such as Aa. The phenotype is the observable result, such as purple flowers, blood group A, or a disease. Phenotype depends on the genotype and on the environment. A plant with tall genotype grown in poor soil can still be short.

A worked one-liner: in peas, R (round seed) is dominant over r (wrinkled). Crossing two Rr plants gives 1/4 RR, 2/4 Rr and 1/4 rr. RR and Rr both look round, so 3/4 of seeds are round and 1/4 wrinkled. Mendel counted 5,474 round and 1,850 wrinkled seeds, a ratio of 2.96 to 1.

Types of genotype. For one gene with two alleles there are three genotypes. Homozygous dominant (AA) carries two dominant alleles. Heterozygous (Aa) carries one of each, so it is made of two different alleles. Homozygous recessive (aa) carries two recessive alleles, so it is made of the same allele twice. Only aa shows the recessive phenotype under complete dominance.

Who uses genotypes: plant and animal breeders predict which crosses give the trait they want. Genetic counsellors give couples the risk of a recessive disease. Clinical labs use SNP genotyping to choose drug doses. Ecologists and wildlife researchers genotype hair, scat or tissue samples to identify individuals and measure genetic diversity.

Genotype vs phenotype vs allele vs haplotype

4.1 What a genotype result actually tells you

Each genotype in the table is a probability for one offspring. If P(aa) = 25%, each child, seed or chick has a one in four chance of being aa. It does not mean that exactly one in every four offspring will be aa. In a litter of four puppies, all four can be aa, and that happens with probability (1/4)⁴ = 0.39%.

4.2 Genotypic ratio vs phenotypic ratio

The genotypic ratio counts distinct allele combinations. The phenotypic ratio counts distinct visible traits. Under complete dominance the phenotypic ratio is shorter, because AA and Aa collapse into one class: 1:2:1 becomes 3:1, and the dihybrid 1:2:1:2:4:2:1:2:1 becomes 9:3:3:1. Under incomplete dominance or codominance the heterozygote has its own phenotype, so the two ratios are the same.

4.3 Homozygous and heterozygous genotypes in the output

The zygosity column labels every gene. A homozygous genotype has two copies of the same allele (AA or aa) and always breeds true for that gene. A heterozygous genotype has two different alleles (Aa) and passes each one to half its gametes. Chart 4 shows the share of each zygosity class per gene. For any Aa × Aa gene it is 25% / 50% / 25%.

4.4 Reading the chi-square result

The chi-square test asks one question: are the gaps between observed and expected counts small enough to be chance? A p-value of 0.05 or more says yes, so the data fit the ratio. A p-value below 0.05 says the gaps are larger than chance usually gives. A good fit does not prove the model. It means your data give no reason to reject it.

4.5 Why p is not the probability that Mendel was right

p = 0.925 does not mean a 92.5% chance that the 9:3:3:1 model is true. It means that if the model is true, about 92.5% of experiments of this size would show deviations at least this large. Very high p-values across many experiments can even be suspicious, which is exactly the point Fisher raised in 1936 about Mendel’s own data.

4.6 What a failed fit usually means

In a two-gene cross the most common cause is linkage: the genes sit near each other on one chromosome, so parental combinations appear too often and recombinant ones too rarely. Other causes are a lethal genotype (for example a 2:1 ratio instead of 3:1), epistasis (9:3:4 or 9:7 instead of 9:3:3:1), mis-scored phenotypes, or a parent whose genotype was wrong.

4.7 Expected counts below 5

The chi-square p-value is an approximation that works when every expected count is 5 or more. Below that, it can be badly off. The calculator flags these classes. Count more offspring, merge rare classes, or use an exact binomial test, which for a two-class cross is easy to run in R or Python (Section 10.5 and 11.5).

4.8 Genotype vs phenotype in the real world

A Punnett square assumes the phenotype follows from the genotype alone. Real traits are messier. Penetrance below 100% means some carriers never show the trait. Variable expressivity means the trait shows at different strengths. Most human traits, such as height, are polygenic and depend on hundreds of genes plus the environment, so one Punnett square cannot predict them.

4.9 Allele frequencies from genotyped offspring

When you enter counts per genotype, the tool also gives the allele frequency p = (2 × AA + Aa) ÷ (2 × n). For an Aa × Aa cross the expected value is 0.5. A big gap can point to unequal gamete success, called segregation distortion, or to a genotyping error such as allele dropout that turns real heterozygotes into homozygotes.

4.10 Practical judgement

Start with the expected ratio. Then look at the observed vs expected chart before the p-value: it shows which class is off. If one class is short in every replicate, suspect lower survival of that genotype. If the two parental classes are both high, suspect linkage. Only then report the chi-square value, with n and df.

▶ Run the analysis above to auto-fill all five examples with your results.

This flowchart shows how the genotype and phenotype ratios of a cross are calculated, step by step, from two parent genotypes to a chi-square verdict. Every box uses the same worked example as the Excel, R and Python sections: Mendel’s dihybrid pea cross RrYy × RrYy, with 556 F2 seeds counted.

Check any box yourself: the Excel sheet in Section 9, the R script in Section 10 and the Python script in Section 11 all return χ² = 0.4700 and p = 0.9254 for these counts.

1. Enter Parent 1's genotype. Type one letter per gene, two letters per gene, capital for the dominant allele, for example RrYy. Up to 4 genes are supported.
2. Enter Parent 2's genotype. Use the same genes in the same order, for example RrYy or rryy for a test cross.
3. Set the dominance for each gene. Choose complete dominance, incomplete dominance or codominance. This decides whether the heterozygote has its own phenotype.
4. Name the traits. Edit the dominant, recessive and heterozygote trait names, such as Round and Wrinkled. They flow into the Punnett square, tables, charts and report.
5. Choose a sample dataset if you want to explore. Ten built-in crosses cover Mendel's peas, a test cross, incomplete dominance, codominance, a carrier couple, SNP genotyping and a failed fit caused by linked genes.
6. Add observed counts. Type offspring counts per phenotype or genotype class, comma separated, in the order shown under the box. Or set observed counts to None for expected ratios only.
7. Or upload genotype calls. On the Upload tab, choose a CSV or Excel file with one offspring genotype per row. The tool tallies each class and fills the counts for you.
8. Ask for one genotype's probability. Type a target genotype such as rryy to get its probability as a fraction and a percentage.
9. Press Calculate Genotypes. Nothing is computed until you press the button. Changing any input clears the old results so you never read stale numbers.
10. Read, copy and download the results. Check the Punnett square, the genotype and phenotype tables, the chi-square verdict and the four charts. Section 5 fills five report styles, and the buttons save a text report or a PDF.

Excel has no Punnett square function, but everything you need is basic arithmetic: expected counts, the chi-square sum and one p-value function. The easy part is the ratio. The parts that trip people up are absolute references when you fill a formula down, and Excel turning gene names into dates.

9.1 The functions you need

9.2 Step by step

9.3 The awkward cases

Dihybrid Punnett squares. For two genes, join the gene-by-gene results: =IF(EXACT(LEFT($A2),UPPER(LEFT($A2))),LEFT($A2)&LEFT(B$1),LEFT(B$1)&LEFT($A2))&IF(EXACT(RIGHT($A2),UPPER(RIGHT($A2))),RIGHT($A2)&RIGHT(B$1),RIGHT(B$1)&RIGHT($A2)). The classic mistake is forgetting that a gamete like Ry has one letter per gene, so LEFT and RIGHT pick the right allele.

Counting genotypes. COUNTIF ignores case, so it counts Rr, RR and rr as the same text. Use =SUMPRODUCT(--EXACT(B2:E5,"Rr")) instead. This is the most common silent error in student spreadsheets.

Percentages instead of counts. Chi-square must use raw counts. Feeding it 56.7%, 19.4% and so on gives a statistic that depends on how you scaled the numbers, not on your sample size.

9.4 Why your Excel answer might differ from R or Python

With the same counts, Excel, R and Python all give χ² = 0.4700 and p = 0.9254. Rounding shown on screen is not rounding stored: widen the column or use more decimals before you compare.

9.5 Charting observed vs expected in Excel

1. Select the class names in A1:A5, then hold Ctrl and select B1:B5 and D1:D5.
2. Choose Insert → Clustered Column.
3. Colour observed and expected differently and add data labels so readers see the gaps.
4. Put χ², df and p in the chart title, for example χ²(3) = 0.47, p = 0.93.
5. For a Punnett square figure, use Home → Conditional Formatting → Text that Contains to colour each phenotype.

9.6 Excel errors you will probably hit

In R the one-liner is chisq.test(obs, p = c(9, 3, 3, 1) / 16), but that skips the part people get wrong: building the Punnett square and the ratio from the parent genotypes. The script below does the whole job in base R. It builds the gametes and the Punnett square, counts genotypes and phenotypes, tests the counts, prints a plain-English verdict and saves one 300 dpi figure.

10.1 The complete script

R, base only, saves genotype_plot.png

# ============================================================
# Genotype calculator: Punnett square, genotype and phenotype
# ratios, and a chi-square goodness-of-fit test
# StatsUnlock - https://statsunlock.com/genotype-calculator/
# Base R only. No packages to install.
# ============================================================

# ---- 1. Your cross ------------------------------------------
# Parent genotypes: one letter per gene, capital = dominant allele
p1 <- "RrYy"
p2 <- "RrYy"

# Trait names for each gene: dominant first, then recessive
traits <- list(R = c("Round", "Wrinkled"),
               Y = c("Yellow", "Green"))

# Observed offspring per phenotype class, in the order printed below.
# Or read them from a file:
#   obs <- read.csv("counts.csv")$count
#   obs <- obs[!is.na(obs)]
obs <- c(315, 108, 101, 32)          # Mendel (1866), 556 F2 seeds

# ---- 2. Split each parent into genes ------------------------
split_genes <- function(g) {
  a <- strsplit(g, "")[[1]]
  sapply(seq(1, length(a), 2), function(i) paste(a[i], a[i + 1], sep = ""))
}
g1 <- split_genes(p1)
g2 <- split_genes(p2)
k  <- length(g1)

# Put the capital (dominant) allele first: "rR" becomes "Rr"
norm_pair <- function(x) {
  s <- strsplit(x, "")[[1]]
  paste(s[order(s != toupper(s))], collapse = "")
}

# ---- 3. Gametes: one allele from every gene -----------------
gametes <- function(genes) {
  opts <- lapply(genes, function(x) unique(strsplit(x, "")[[1]]))
  apply(expand.grid(rev(opts), stringsAsFactors = FALSE), 1,
        function(r) paste(rev(r), collapse = ""))
}
gam1 <- gametes(g1)
gam2 <- gametes(g2)

# ---- 4. Punnett square --------------------------------------
offspring <- function(a, b) {
  pa <- strsplit(a, "")[[1]]; pb <- strsplit(b, "")[[1]]
  paste(sapply(seq_len(k), function(i) norm_pair(paste0(pa[i], pb[i]))),
        collapse = "")
}
grid <- outer(gam1, gam2, Vectorize(offspring))
dimnames(grid) <- list(gam1, gam2)
cat("Punnett square:\n"); print(noquote(grid))

# ---- 5. Genotype and phenotype probabilities ----------------
geno_tab <- table(grid)
geno_p   <- geno_tab / sum(geno_tab)

phenotype <- function(gt) {
  s <- substring(gt, seq(1, 2 * k, 2), seq(2, 2 * k, 2))
  paste(sapply(seq_len(k), function(i) {
    dom <- any(strsplit(s[i], "")[[1]] == toupper(substr(s[i], 1, 1)))
    traits[[i]][ifelse(dom, 1, 2)]
  }), collapse = ", ")
}
phen <- table(sapply(grid, phenotype))
# order classes dominant-first, gene by gene
lvl <- apply(expand.grid(rev(traits))[, k:1, drop = FALSE], 1, paste, collapse = ", ")
phen <- phen[intersect(lvl, names(phen))]
phen_p <- phen / sum(phen)

ratio <- function(x) { g <- Reduce(function(a, b) if (b == 0) a else Recall(b, a %% b), x)
                       paste(x / g, collapse = ":") }
cat("\nGenotypic ratio :", ratio(as.vector(geno_tab)), "\n")
cat("Phenotypic ratio:", ratio(as.vector(phen)), "\n")
print(round(phen_p, 4))

# ---- 6. Chi-square goodness of fit --------------------------
expected <- sum(obs) * as.vector(phen_p)
test <- chisq.test(obs, p = as.vector(phen_p))
crit <- qchisq(0.95, df = length(obs) - 1)
cat(sprintf("\nn = %d, chi-square = %.4f, df = %d, critical = %.3f, p = %.4f\n",
            sum(obs), test$statistic, test$parameter, crit, test$p.value))
cat("Expected:", round(expected, 2), "\n")
cat(if (test$p.value >= 0.05) {
  "Verdict: the counts FIT the expected ratio (p >= 0.05).\n"
} else {
  "Verdict: the counts do NOT fit the expected ratio (p < 0.05).\n"
})

# ---- 7. One figure, 300 dpi ---------------------------------
set.seed(1)                          # nothing random here, kept for habit
cols <- c("#16a34a", "#2563eb", "#d97706", "#9333ea",
          "#0d9488", "#c2410c", "#0891b2", "#475569")[seq_along(obs)]
png("genotype_plot.png", width = 2700, height = 1800, res = 300)
par(mar = c(6.5, 5, 4, 1))
bp <- barplot(rbind(obs, expected), beside = TRUE,
              col = c("#2563eb", "#fbbf24"), border = NA,
              names.arg = names(phen), las = 2, cex.names = 0.8,
              ylab = "Offspring (count)", ylim = c(0, max(obs, expected) * 1.15))
legend("topright", c("Observed", "Expected"), fill = c("#2563eb", "#fbbf24"),
       border = NA, bty = "n")
title(main = sprintf("%s x %s: phenotypic ratio %s", p1, p2, ratio(as.vector(phen))),
      adj = 0, line = 2.2)
title(main = sprintf("chi-square = %.3f, df = %d, p = %.4f",
                     test$statistic, test$parameter, test$p.value),
      adj = 0, line = 0.8, font.main = 1, cex.main = 0.95)
invisible(dev.off())
cat("Saved genotype_plot.png\n")

Run it with Rscript genotype.R from a terminal, or paste it into the RStudio console. It prints the Punnett square, the 1:2:1:2:4:2:1:2:1 and 9:3:3:1 ratios, χ² = 0.4700, df = 3, p = 0.9254, and saves genotype_plot.png in your working directory (check it with getwd()).

10.2 What each part does

10.3 What the figure shows

Blue bars are the seeds Mendel counted and amber bars are what 9:3:3:1 predicts for 556 seeds. Each pair is almost the same height, and the subtitle gives χ² = 0.470, df = 3, p = 0.9254. That is the picture of a near-perfect fit.

10.4 Common R problems and fixes

10.5 Useful one-liners

# Monohybrid 3:1 test in one line
chisq.test(c(5474, 1850), p = c(3, 1) / 4)
# Exact binomial test when expected counts are small
binom.test(4, 7, p = 0.75)
# Probability of a genotype by the product rule
(1/2) * (1/4)            # P(Rr) x P(yy) = 0.125
# Recombination frequency from a test cross
(80 + 85) / 1000 * 100   # 16.5 percent, 16.5 cM
# Monte Carlo p-value for small samples
chisq.test(c(4, 3), p = c(3, 1) / 4, simulate.p.value = TRUE, B = 1e5)
# Allele frequency from genotype counts GG, Gg, gg
(2 * 28 + 49) / (2 * 100)

Python builds the Punnett square with itertools.product and tests it with scipy.stats.chisquare. The script below mirrors the R version line for line: same cross, same class order, same answer.

Install once with pip install numpy scipy matplotlib. Biopython is not needed, because no DNA sequence is read.

11.1 The complete script

Python 3, saves genotype_plot.png

# ============================================================
# Genotype calculator: Punnett square, genotype and phenotype
# ratios, and a chi-square goodness-of-fit test
# StatsUnlock - https://statsunlock.com/genotype-calculator/
# ============================================================
from collections import Counter
from functools import reduce
from itertools import product
from math import gcd

import matplotlib.pyplot as plt
import numpy as np
from scipy.stats import chi2, chisquare

# ---- 1. Your cross ------------------------------------------
p1, p2 = "RrYy", "RrYy"            # capital letter = dominant allele
traits = {"R": ("Round", "Wrinkled"),
          "Y": ("Yellow", "Green")}
obs = np.array([315, 108, 101, 32])  # Mendel (1866), 556 F2 seeds
# Or read the counts from a file:
#   import pandas as pd
#   obs = pd.read_csv("counts.csv")["count"].dropna().to_numpy()

np.random.seed(1)                    # nothing random here, kept for habit

# ---- 2. Split each parent into genes ------------------------
def genes(g):
    return [g[i:i + 2] for i in range(0, len(g), 2)]

def norm_pair(a, b):
    # capital (dominant) allele first: "rR" becomes "Rr"
    return "".join(sorted(a + b, key=lambda c: c.islower()))

g1, g2 = genes(p1), genes(p2)
k = len(g1)

# ---- 3. Gametes ---------------------------------------------
def gametes(gs):
    opts = [sorted(set(x), key=lambda c: c.islower()) for x in gs]
    return ["".join(c) for c in product(*opts)]

gam1, gam2 = gametes(g1), gametes(g2)

# ---- 4. Punnett square --------------------------------------
grid = [["".join(norm_pair(a[i], b[i]) for i in range(k)) for b in gam2] for a in gam1]
print("Punnett square:")
print("     " + "  ".join(f"{g:>6}" for g in gam2))
for a, row in zip(gam1, grid):
    print(f"{a:>4} " + "  ".join(f"{c:>6}" for c in row))

# ---- 5. Genotype and phenotype probabilities ----------------
cells = [c for row in grid for c in row]
total = len(cells)

def geno_key(g):
    return [-sum(ch.isupper() for ch in g[i:i + 2]) for i in range(0, len(g), 2)]

geno = Counter(cells)
geno_order = sorted(geno, key=geno_key)

def phenotype(gt):
    out = []
    for i, letter in enumerate(traits):
        pair = gt[2 * i:2 * i + 2]
        out.append(traits[letter][0] if any(c.isupper() for c in pair) else traits[letter][1])
    return ", ".join(out)

phen = Counter(phenotype(c) for c in cells)
order = [", ".join(t) for t in product(*traits.values())]
phen_names = [p for p in order if p in phen]
phen_counts = np.array([phen[p] for p in phen_names])

def ratio(x):
    g = reduce(gcd, x)
    return ":".join(str(v // g) for v in x)

print("\nGenotypic ratio :", ratio([geno[g] for g in geno_order]))
print("Phenotypic ratio:", ratio(list(phen_counts)))
for p, c in zip(phen_names, phen_counts):
    print(f"  {p:<20} {c}/{total} = {c / total:.4f}")

# ---- 6. Chi-square goodness of fit --------------------------
expected = obs.sum() * phen_counts / total
stat, pval = chisquare(obs, expected)
df = len(obs) - 1
crit = chi2.ppf(0.95, df)
print(f"\nn = {obs.sum()}, chi-square = {stat:.4f}, df = {df}, "
      f"critical = {crit:.3f}, p = {pval:.4f}")
print("Expected:", np.round(expected, 2))
print("Verdict: the counts", "FIT" if pval >= 0.05 else "do NOT fit",
      "the expected ratio.")

# ---- 7. One figure, 300 dpi ---------------------------------
x = np.arange(len(obs))
fig, ax = plt.subplots(figsize=(9, 6))
ax.bar(x - 0.2, obs, 0.4, label="Observed", color="#2563eb")
ax.bar(x + 0.2, expected, 0.4, label="Expected", color="#fbbf24")
ax.set_xticks(x, phen_names, rotation=30, ha="right")
ax.set_ylabel("Offspring (count)")
ax.set_title(f"{p1} x {p2}: phenotypic ratio {ratio(list(phen_counts))}\n"
             f"chi-square = {stat:.3f}, df = {df}, p = {pval:.4f}", loc="left")
ax.legend(frameon=False)
ax.spines[["top", "right"]].set_visible(False)
fig.tight_layout()
fig.savefig("genotype_plot.png", dpi=300, bbox_inches="tight")
print("Saved genotype_plot.png")

11.2 What each part does

11.3 What the figure shows

The same observed vs expected bars as the R figure, with the ratio and the test result in the title. Matching bar heights mean a good fit; one bar far from its partner tells you which class to investigate.

11.4 Common Python problems and fixes

11.5 Useful one-liners

# Monohybrid 3:1 test
from scipy.stats import chisquare, binomtest
chisquare([5474, 1850], [7324 * 3 / 4, 7324 / 4])
# Exact binomial test for small samples
binomtest(4, 7, 0.75).pvalue          # 0.3771
# Product rule
0.5 * 0.25                            # P(Rr) x P(yy) = 0.125
# Recombination frequency from a test cross
(80 + 85) / 1000 * 100                # 16.5 cM
# Allele frequency from genotype counts GG, Gg, gg
(2 * 28 + 49) / (2 * 100)             # 0.525

Run it with python genotype.py from a terminal. In Jupyter, drop the savefig line and add plt.show().

12.1 All six one-gene crosses

What to conclude: only Aa × Aa gives three genotypes, and only the test cross shows the genotype of the unknown parent directly in the offspring phenotypes. Source: Mendel’s laws as restated by Fisher (1936).

12.2 Classic Mendelian and modified ratios

What to conclude: if a dihybrid cross fails 9:3:3:1 but fits 9:3:4 or 9:7, the genes interact (epistasis) rather than being linked. Test each candidate ratio with the matching df.

12.3 How the numbers grow with each heterozygous gene

What to conclude: each extra gene multiplies the squares by 4, which is why the product rule (Formula 3) is faster than drawing the grid for three or more genes.

12.4 Chi-square critical values

What to conclude: if your χ² is below the 0.05 value, the data fit the ratio. Source: the chi-square distribution of Pearson (1900); values computed exactly.

12.5 Genotype terms and examples

What to conclude: only homozygous recessive genotypes can be read from the phenotype alone under complete dominance; every other genotype needs a test cross, a pedigree or a DNA test.

12.6 Mendel’s seven pea traits, F2 counts

What to conclude: every trait sits close to 3:1, and the larger the sample, the closer the ratio. Fisher (1936) argued these fits are, if anything, too good to be chance.

Eight genotype examples, from Mendel’s own peas to a genetic counselling case. Every number below was produced by this calculator, and you can load most of them from the sample dataset list.

Study design: a genotype ratio test needs parents of known genotype, offspring scored without bias, and every offspring counted, including the ones that look odd. The protocol below covers a classic phenotype-scored cross and a SNP genotyping run, because most labs now confirm genotypes with DNA.

Prepare

1. Confirm the parent genotypes first. Use true-breeding lines or a test cross to aa. For fruit flies, collect virgin females within 8 hours of eclosion at 25 °C, because older females may already have mated.
2. Choose and label the genotype letters before crossing. Write the dominance model for each gene (complete, incomplete, codominant) in the notebook. Changing it after scoring is how ratios get fitted to the data.
3. Plan the offspring number for a genotype ratio test. Aim for every expected class to be at least 5; for 9:3:3:1 that means at least 80 offspring (1/16 × 80 = 5), and 200 or more to detect linkage reliably.

Set up

4. Set up the cross. In peas, emasculate the female flower before anthesis and dust pollen from the male parent with a fine brush; bag the flower. In flies, put 3 to 5 virgin females with 3 to 5 males per vial on standard cornmeal food.
5. Collect DNA for genotyping. Take 20 to 30 mg of leaf or tissue, or a single fly, and extract with a silica column kit such as the Qiagen DNeasy Blood & Tissue Kit; elute in 100 µL and check A260/A280 near 1.8.
6. Set up the SNP genotyping reaction. For a TaqMan assay use 10 µL per well: 5 µL 2× genotyping master mix, 0.25 µL 40× assay, 10 to 20 ng genomic DNA and water to volume. Include two no-template controls and one known control of each genotype per plate.

Run

7. Run the genotyping PCR. 95 °C for 10 min, then 40 cycles of 95 °C for 15 s and 60 °C for 1 min, followed by an end-point read. The software clusters samples into AA, Aa and aa groups on an allelic discrimination plot.
8. Score phenotypes blind. The person scoring should not know which class is "expected" to be rare. Score flies 10 to 12 days after the cross is set, before the F1 generation can start breeding.
9. Re-test unclear calls. Samples between clusters, or with call rate below 95% across the plate, are repeated rather than guessed.

Record

10. Record the genotype of every offspring in one column, exactly as the layout below shows, ready to upload to the calculator.
11. Record dead, sterile and unscorable offspring separately. A missing class in the data is often a lethal genotype, which is a result, not noise.

Replicates: run at least 3 independent crosses (different parent pairs) and test each one before pooling. If they disagree with each other (a heterogeneity chi-square), do not pool them.

Lab notebook: parent IDs and genotypes, date set, date scored, temperature, scorer, kit and lot numbers, plate map, and the raw tally sheet.

Common mistakes that make genotype data unusable:

• Non-virgin females, which add offspring from an unknown father.
• Scoring only the "nice" offspring and leaving out damaged or late ones.
• Allele dropout in degraded DNA, which turns heterozygotes into false homozygotes.
• Typing gene names into Excel without formatting the column as Text.
• Pooling crosses that differ from each other.

Use it when:

• ✓ You know, or want to test, the genotypes of two parents.
• ✓ The trait is controlled by 1 to 4 genes with two alleles each.
• ✓ The genes assort independently (different chromosomes, or far apart).
• ✓ You want genotype and phenotype probabilities, ratios, or a chi-square test of real counts.
• ✓ You have SNP genotyping calls for offspring of a known cross.

Use something else when:

• ✗ The genes are linked: use a recombination frequency and map distance analysis.
• ✗ You have a random population sample, not a family: use a Hardy-Weinberg test.
• ✗ The gene has three or more alleles, such as ABO blood groups: list the alleles by hand or use a multi-allele Punnett square.
• ✗ The trait is sex-linked: draw the cross with X and Y chromosomes.
• ✗ The trait is polygenic, such as height or yield: use heritability and quantitative genetics.

Real-world examples:

1. Teaching genetics: checking a student’s dihybrid cross against 9:3:3:1.
2. Dog and cattle breeding: the chance that two carriers of a coat colour produce the recessive colour.
3. Genetic counselling: a 1 in 4 risk of a recessive disease for each pregnancy of a carrier couple.
4. Plant breeding: how many F2 plants to grow to find one rryyaa triple recessive (expected 1 in 64).

Experimental design: decide the expected ratio, the significance level and the sample size before you score anything. For a 3:1 test, about 100 offspring give a clear answer; for 9:3:3:1, aim for 200 or more.

Decision tree: Do you have a family cross? No → Hardy-Weinberg test. Yes → Are the genes linked? Yes → recombination frequency. No → Is it sex-linked? Yes → X-linked Punnett square. No → use this genotype calculator, then the chi-square goodness-of-fit test.

Assumptions

1. Independent assortment. Requires genes on different chromosomes or more than about 50 cM apart. Check with a dihybrid test cross. When it fails, parental classes are over-counted and recombinant classes under-counted, so the 9:3:3:1 test fails with a large χ².
2. Equal gamete success (no segregation distortion). Requires each allele to reach 50% of gametes. Check with a test cross or genotype counts. When it fails, one allele is over-represented, so its phenotype class is inflated.
3. Equal viability of every genotype. Requires no genotype to die before scoring. Check by counting dead embryos or unhatched eggs. When it fails, the lethal class is too small, for example 2:1 instead of 3:1.
4. Correct parent genotypes. Requires true-breeding or tested parents. Check by test crossing the parents. When it fails, every expected ratio is wrong, often in the direction of too few recessive offspring.
5. Full penetrance and no epistasis. Requires each genotype to always give the same phenotype. Check by genotyping a sample of phenotyped offspring. When it fails, recessive classes look too small or classes merge.
6. Independent, unbiased counts. Requires every offspring scored once by a blind scorer. When it fails, the bias usually favours the expected ratio, which makes p too high.

Limitations

1. Two alleles per gene only. Multiple alleles (ABO, MHC) need a multi-allele cross. Use a hand-drawn grid with every allele listed.
2. Autosomal genes only. X-linked traits differ between sons and daughters. Use an X-linked Punnett square.
3. Up to 4 genes. Beyond that, use the product rule (Formula 3) or a script.
4. No linkage model. For linked genes, use a recombination frequency calculator with Haldane or Kosambi mapping.
5. Large-sample chi-square. With expected counts below 5, use an exact binomial or multinomial test.
6. Not for populations. Genotype frequencies in a wild population follow Hardy-Weinberg, not a Punnett square.

A genotype is the allele pair behind every trait, and the phenotype is how that genotype shows up in a living organism. This calculator turns two parent genotypes into every possible offspring genotype, the Punnett square, the genotypic and phenotypic ratios, and a chi-square test of how well real offspring match.

What was found

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What it means at the bench

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Cautions

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Next steps

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Whether you are learning the difference between genotype and phenotype or checking a breeding experiment, the same rule holds: write the genotype first, then predict the phenotype, then let the counts test the genotype.

For a deeper test of observed against expected counts, see the chi-square goodness of fit test calculator.

The genotype concept, the genotype vs phenotype distinction and the genotyping methods on this page rest on the peer-reviewed literature below. Every DOI was checked before publication.

1. Alberch, P. (1991). From genes to phenotype: Dynamical systems and evolvability. Genetica, 84(1), 5–11. https://doi.org/10.1007/BF00123979
2. Browning, S. R., & Browning, B. L. (2011). Haplotype phasing: Existing methods and new developments. Nature Reviews Genetics, 12(10), 703–714. https://doi.org/10.1038/nrg3054
3. Churchill, F. B. (1974). William Johannsen and the genotype concept. Journal of the History of Biology, 7(1), 5–30. https://doi.org/10.1007/BF00179291
4. Fisher, R. A. (1918). The correlation between relatives on the supposition of Mendelian inheritance. Transactions of the Royal Society of Edinburgh, 52(2), 399–433. https://doi.org/10.1017/S0080456800012163
5. Fisher, R. A. (1936). Has Mendel’s work been rediscovered? Annals of Science, 1(2), 115–137. https://doi.org/10.1080/00033793600200111
6. Hardy, G. H. (1908). Mendelian proportions in a mixed population. Science, 28(706), 49–50. https://doi.org/10.1126/science.28.706.49
7. Houle, D., Govindaraju, D. R., & Omholt, S. (2010). Phenomics: The next challenge. Nature Reviews Genetics, 11(12), 855–866. https://doi.org/10.1038/nrg2897
8. Johannsen, W. (1911). The genotype conception of heredity. The American Naturalist, 45(531), 129–159. https://doi.org/10.1086/279202
9. Kacser, H., & Burns, J. A. (1981). The molecular basis of dominance. Genetics, 97(3–4), 639–666. https://doi.org/10.1093/genetics/97.3-4.639
10. Lander, E. S., & Schork, N. J. (1994). Genetic dissection of complex traits. Science, 265(5181), 2037–2048. https://doi.org/10.1126/science.8091226
11. Mackay, T. F. C., Stone, E. A., & Ayroles, J. F. (2009). The genetics of quantitative traits: Challenges and prospects. Nature Reviews Genetics, 10(8), 565–577. https://doi.org/10.1038/nrg2612
12. Nielsen, R., Paul, J. S., Albrechtsen, A., & Song, Y. S. (2011). Genotype and SNP calling from next-generation sequencing data. Nature Reviews Genetics, 12(6), 443–451. https://doi.org/10.1038/nrg2986
13. Orgogozo, V., Morizot, B., & Martin, A. (2015). The differential view of genotype–phenotype relationships. Frontiers in Genetics, 6, 179. https://doi.org/10.3389/fgene.2015.00179
14. Pearson, K. (1900). On the criterion that a given system of deviations from the probable in the case of a correlated system of variables is such that it can be reasonably supposed to have arisen from random sampling. Philosophical Magazine, 50(302), 157–175. https://doi.org/10.1080/14786440009463897
15. Ragoussis, J. (2009). Genotyping technologies for genetic research. Annual Review of Genomics and Human Genetics, 10, 117–133. https://doi.org/10.1146/annurev-genom-082908-150116
16. Syvänen, A.-C. (2001). Accessing genetic variation: Genotyping single nucleotide polymorphisms. Nature Reviews Genetics, 2(12), 930–942. https://doi.org/10.1038/35103535
17. Visscher, P. M., Hill, W. G., & Wray, N. R. (2008). Heritability in the genomics era: Concepts and misconceptions. Nature Reviews Genetics, 9(4), 255–266. https://doi.org/10.1038/nrg2322
18. Wigginton, J. E., Cutler, D. J., & Abecasis, G. R. (2005). A note on exact tests of Hardy-Weinberg equilibrium. American Journal of Human Genetics, 76(5), 887–893. https://doi.org/10.1086/429864
19. Wilkie, A. O. M. (1994). The molecular basis of genetic dominance. Journal of Medical Genetics, 31(2), 89–98. https://doi.org/10.1136/jmg.31.2.89
20. Zuk, O., Hechter, E., Sunyaev, S. R., & Lander, E. S. (2012). The mystery of missing heritability: Genetic interactions create phantom heritability. Proceedings of the National Academy of Sciences, 109(4), 1193–1198. https://doi.org/10.1073/pnas.1119675109