D3.2 Inheritance

Inheritance explains how alleles, chromosomes, meiosis, pedigrees, linkage, variation and statistical tests predict genetic outcomes across generations in inheritance problems.

Learning objectives

D3.2.1Haploid gametes + fusion = diploid zygote• Haploid gametes fuse during fertilization to form a diploid zygote• Diploid organisms usually carry two alleles for each autosomal gene D3.2.2Genetic crosses in flowering plants• Genetic crosses track parental, F1, and F2 generations using Punnett grids• Flowering plant crosses control pollen transfer to study inheritance ratios D3.2.3Genotype• Genotype is the allele combination inherited for a gene or genes• Homozygous genotypes have matching alleles; heterozygous genotypes have different alleles D3.2.4Phenotype• Phenotype is the observable characteristic or trait• Phenotype can be determined by genotype, environment, and their interaction D3.2.5Dominant and recessive alleles• Dominant alleles are expressed in heterozygotes• Recessive alleles are expressed when no dominant allele masks them D3.2.6Phenotypic plasticity• Phenotypic plasticity is environment-driven phenotype change without genotype change• It depends on altered gene expression and can be adaptive D3.2.7Phenylketonuria (PKU)• PKU is an autosomal recessive disorder affecting phenylalanine metabolism• Low-phenylalanine diet and newborn screening reduce harmful effects D3.2.8SNPs and multiple alleles• SNPs are single-base differences that can create different alleles• Populations can have multiple alleles, but diploid individuals carry at most two D3.2.9ABO blood groups• ABO blood group is controlled by IA, IB, and i alleles• IA and IB are codominant; i is recessive, producing four blood phenotypes D3.2.10Incomplete dominance and codominance• Codominance expresses both heterozygous alleles, as in AB blood type• Incomplete dominance gives an intermediate heterozygote phenotype D3.2.11Sex determination• Human chromosomal sex is usually determined by XX or XY chromosome combination• The SRY/TDF region on the Y chromosome directs testis development D3.2.12Haemophilia• Haemophilia is an X-linked recessive blood-clotting disorder• Carrier females and affected males are represented with X-linked allele notation D3.2.13Pedigree charts• Pedigree charts show family inheritance across generations• Patterns help infer autosomal dominant, autosomal recessive, or sex-linked inheritance D3.2.14Continuous variation• Continuous variation often results from polygenic inheritance plus environment• Human skin colour, height, and body mass show many intermediate phenotypes D3.2.15Box-and-whisker plots• Box-and-whisker plots summarize non-normal continuous data• They show median, quartiles, interquartile range, maximum/minimum, and outliers D3.2.16(HL)—Segregation and independent assortment• Alleles segregate as homologous chromosomes separate during meiosis• Unlinked genes assort independently through random bivalent orientation D3.2.17(HL)—Dihybrid crosses• Dihybrid crosses track inheritance of two genes simultaneously• Unlinked autosomal genes can produce 9:3:3:1 F2 or 1:1:1:1 test-cross ratios D3.2.18(HL)—Human gene loci• Genome databases identify human gene loci and polypeptide products• Genes on different chromosomes are unlinked; close loci on one chromosome can be linked D3.2.19(HL)—Autosomal gene linkage• Linked autosomal genes are on the same non-sex chromosome• Linked genes tend to be inherited together unless crossing over occurs between loci D3.2.20(HL)—Recombinants• Recombinants have allele combinations different from parental chromosomes• Crossing over between linked genes produces fewer recombinants than parental types D3.2.21(HL)—Chi-squared test• Chi-squared tests whether observed genetic data fit expected ratios• Use observed/expected values, degrees of freedom, and p = 0.05 significance

diploid parent (two homologous copies of each autosome) → meiosis → haploid gametes (one copy) → fertilization → diploid zygote (two copies again)

Cell	Chromosome sets	Alleles possible at one autosomal locus
body cell in a diploid parent	2n	two alleles, one on each homolog
gamete	n	one allele
zygote	2n	one allele from each parent

Two alleles does not mean two different alleles. A diploid genotype may be homozygous, with matching alleles, or heterozygous, with different alleles.

Term	Meaning	Example
gene	DNA sequence with a functional product or role	a gene affecting pea height
locus	the gene's position on a chromosome	the height-gene position
allele	one sequence version at that locus	T or t
genotype	allele combination carried	TT, Tt or tt

homozygous: two matching alleles, such as TT or tt
heterozygous: two different alleles, such as Tt

Use genotype for the allele combination, not for the visible trait. A genotype can be written for one locus, several loci or the whole genome, so state the scope when it matters.

A phenotype is an observable or measurable characteristic produced by the genotype, the environment, or an interaction between them.

Main cause	Example	Why
genotype	ABO blood group	inherited alleles determine A and B antigens
environment	an acquired scar	injury changes tissue without changing inherited alleles
genotype × environment	adult height	growth potential interacts with nutrition and health

The same phenotype can come from different genotypes, and one genotype can produce different phenotypes in different environments. Phenotype alone therefore does not always identify genotype.

Under complete dominance, a dominant allele determines the heterozygous phenotype. A recessive phenotype appears only when no dominant allele is present.

Genotype	Allele state	Phenotype when T is dominant
TT	homozygous dominant	tall
Tt	heterozygous	tall
tt	homozygous recessive	dwarf

Dominant does not mean better, stronger or more common. It only describes expression in a heterozygote; a harmful or rare allele can be dominant.

choose parents with known, contrasting phenotypes and preferably true-breeding genotypes
remove immature anthers from the recipient flower before they release pollen
transfer pollen from the chosen donor to the receptive stigma
prevent stray pollen from reaching the flower, then label the cross
collect seeds and score enough offspring to compare with a predicted ratio

Emasculation and isolation prevent self-pollination, so the experimenter knows the parental sources of both gametes. Reciprocal crosses can also test whether switching the pollen and ovule parents changes the result.

A controlled flowering-plant cross removes self-pollinating anthers, transfers pollen from a chosen parent, grows F1 offspring and crosses the F1 to obtain an F2 generation.

P generation: true-breeding tall TT × true-breeding dwarf tt → every gamete carries T or t, respectively.

F1 generation: every offspring is Tt and tall. The t allele has not blended away; its phenotypic effect is masked in the heterozygote.

F1 self-cross: Tt × Tt → F2 genotypes 1 TT : 2 Tt : 1 tt → phenotypes 3 tall : 1 dwarf in a large sample.

Reappearance of dwarf offspring shows that hereditary factors remain discrete across generations. The observed ratio approaches the probability prediction only when many offspring are counted.

define allele symbols and the dominance relationship
write both parental genotypes
derive the allele carried by each possible gamete
place one parent's gametes across the grid and the other's down the side
combine alleles in each cell, then convert genotypes to phenotypes

Tt × Tt	T gamete	t gamete
T gamete	TT, tall	Tt, tall
t gamete	Tt, tall	tt, dwarf

Each cell has probability 1/4 because each parent produces T and t gametes with probability 1/2. The grid predicts probabilities for each fertilization, not a guarantee that every four offspring will contain one of each cell.

A tall plant could be TT or Tt because both contain the dominant T allele. Cross the unknown plant with a homozygous recessive tester, tt, whose gametes reveal what the unknown parent contributes.

Unknown parent	Cross with tt	Expected offspring evidence
TT	TT × tt	all Tt, all tall
Tt	Tt × tt	about 1 Tt tall : 1 tt dwarf

A small all-tall sample does not prove the parent is TT because a heterozygote can produce tall offspring repeatedly by chance. Confidence increases as more offspring are scored.

Phenotypic plasticity is the capacity of one genotype to develop different phenotypes in different environments by changing patterns of gene expression.

environmental cue → signalling pathway → transcription or translation changes → different proteins, cell activity or growth → phenotype better matched to the experienced environment

Plastic response	Genetic evolution
genotype is unchanged within the individual	allele frequencies change across generations
can be reversible during a lifetime	inherited population change is not reversed by one individual's environment
response range itself can be inherited	selection acts on heritable differences

A suntan or training response is acquired plasticity; it does not mean the altered phenotype is encoded as a new allele and passed directly to offspring.

two loss-of-function PAH alleles → little or no phenylalanine hydroxylase → phenylalanine is not converted efficiently to tyrosine → phenylalanine and harmful derivatives accumulate → untreated neural development is damaged

Intervention	Why it helps
newborn screening	detects the metabolic problem before severe symptoms appear
early low-phenylalanine diet	limits substrate accumulation while still supplying controlled essential phenylalanine
continued monitoring	keeps blood concentrations within a safer range as diet and growth change

The genotype does not change with treatment, but the environment does. Dietary control can greatly change the phenotype, making PKU a clear genotype–environment interaction.

Genotype	PKU status	Reproductive meaning
PP	unaffected, not a carrier	passes P
Pp	unaffected carrier	can pass P or p
pp	affected	passes p

Pp × Pp	P gamete	p gamete
P gamete	PP unaffected	Pp carrier
p gamete	Pp carrier	pp affected

For each pregnancy: 1/4 affected, 1/2 unaffected carrier, 1/4 unaffected non-carrier. The probability resets for every fertilization; previous children do not change the next gamete combination.

Carrier and affected are not synonyms. A heterozygous carrier has a working dominant allele and usually does not show PKU, but can transmit the recessive allele.

A single-nucleotide polymorphism (SNP) is a common difference at one nucleotide position among genomes. A SNP within or near a gene can create or mark a different allele.

Scale	What may exist at one locus
one haploid gamete	one allele
one diploid individual	at most two alleles
a population gene pool	two, three or many alternative alleles

Multiple alleles means several versions of one gene in the population. It does not mean that one ordinary diploid person carries every version.

Phenotype	Possible genotype(s)	Red-cell antigen(s)
A	IᴬIᴬ or Iᴬi	A
B	IᴮIᴮ or Iᴮi	B
AB	IᴬIᴮ	A and B
O	ii	neither A nor B

Iᴬ and Iᴮ are codominant because both antigen products appear in IᴬIᴮ. Each is dominant to i, which produces neither A nor B antigen.

Blood group is a phenotype; Iᴬ, Iᴮ and i are alleles. A person with group A or B may hide an i allele, so phenotype does not always identify genotype.

Red blood cells show A antigen for blood group A, B antigen for group B, both antigens for AB and neither antigen for O, alongside the corresponding ABO genotypes.

A group-A parent could be IᴬIᴬ or Iᴬi, and a group-B parent could be IᴮIᴮ or Iᴮi. State the genotype assumption before predicting children.

Iᴬi × Iᴮi	Iᴮ gamete	i gamete
Iᴬ gamete	IᴬIᴮ, group AB	Iᴬi, group A
i gamete	Iᴮi, group B	ii, group O

With these heterozygous parents, A, B, AB and O phenotypes each have probability 1/4. If either parent is homozygous, the outcome changes.

ABO data can sometimes exclude a proposed parent–child relationship, but common blood groups cannot uniquely prove parentage because many people share the same phenotype and genotype.

Allele relationship	Heterozygote phenotype	Example
complete dominance	matches the dominant homozygote	Tt pea is tall
codominance	both allele products are detectably expressed	IᴬIᴮ has A and B antigens
incomplete dominance	intermediate between the two homozygotes	red × white Mirabilis gives pink F1

Heterozygote × heterozygote	Genotype ratio	Phenotype ratio
complete dominance	1:2:1	3:1
codominance or incomplete dominance	1:2:1	1:2:1 when all three genotypes are distinguishable

Codominance is not blending: both products remain identifiable. In incomplete dominance, the heterozygote is intermediate rather than showing two separate products side by side.

P generation: FᴿFᴿ red × FᵂFᵂ white → all FᴿFᵂ pink F1 offspring.

FᴿFᵂ × FᴿFᵂ	Fᴿ gamete	Fᵂ gamete
Fᴿ gamete	FᴿFᴿ red	FᴿFᵂ pink
Fᵂ gamete	FᴿFᵂ pink	FᵂFᵂ white

F2 genotype ratio = phenotype ratio = 1 red : 2 pink : 1 white because the heterozygote has its own recognizable phenotype.

Segregation has not changed: each heterozygote still makes the two gamete types equally. The different phenotypic ratio comes from how the heterozygous genotype is expressed.

homologous chromosomes carry two alleles in a diploid parent → meiosis separates one allele into each gamete → fertilization restores a two-allele genotype → allele relationship and environment shape phenotype → offspring data test the prediction

For any genetic cross:

define allele symbols and their relationship
write parental genotypes, not only phenotypes
derive gametes before drawing the grid
combine gametes, then translate genotype to phenotype
state probabilities and the assumptions behind them

If the simple 3:1 expectation fails	First question to ask
heterozygote has a third phenotype	incomplete dominance?
both products appear	codominance?
phenotype changes with conditions	environmental or plastic response?
more than two alleles occur in the population	multiple-allele locus such as ABO?

Gamete contribution	X-bearing sperm	Y-bearing sperm
X-bearing egg	XX zygote	XY zygote
expected probability	about 1/2	about 1/2

Y chromosome with functional SRY → testis-determining factor (TDF) is produced → embryonic gonads develop as testes → testicular hormones direct typical male reproductive development

Without a functional SRY signal, the undifferentiated gonads normally follow ovarian development and typical female reproductive structures develop.

XX/XY is the core human chromosomal model, but chromosome number, SRY location or function, hormone synthesis and tissue response can vary. Chromosomal pattern alone does not describe every aspect of sex or gender.

Most genes in the non-homologous region of X have no matching allele on Y. An XY individual is therefore hemizygous for those X-linked genes: the single allele on X is expressed even if it is recessive.

Genotype	Typical status for an X-linked recessive allele h
XᴴXᴴ	unaffected XX
XᴴXʰ	unaffected carrier XX
XʰXʰ	affected XX
XᴴY	unaffected XY
XʰY	affected XY

a son receives his X chromosome from his mother and Y from his father
a father passes his X chromosome to every daughter and to no son
an affected father cannot pass his X-linked allele directly to a son

Write the allele as a superscript on X, such as Xᴴ or Xʰ. Do not place the X-linked allele on Y when the gene is absent from the relevant Y region.

XᴴXʰ × XᴴY	Xᴴ sperm	Y sperm
Xᴴ egg	XᴴXᴴ unaffected daughter	XᴴY unaffected son
Xʰ egg	XᴴXʰ carrier daughter	XʰY affected son

Among daughters: 1/2 carriers and 1/2 non-carriers; none affected in this cross. Among sons: 1/2 affected and 1/2 unaffected. Across all offspring, each grid outcome has probability 1/4.

Haemophilia A and B result from deficient clotting factors. The inheritance pattern reflects an X-linked recessive allele; treatment supplies or supports the missing clotting function but does not alter the inherited allele.

A carrier female XH Xh crossed with an unaffected male XH Y produces unaffected daughters, carrier daughters, unaffected sons and sons with haemophilia.

Pedigree mark	Meaning
square / circle	male / female in the standard convention
filled symbol	individual shows the tracked phenotype
horizontal partner line	mating or reproductive partnership
vertical descent line	offspring connection
shared sibship line	siblings from the same parents
I, II, III …	generations
1, 2, 3 …	individuals within a generation

identify the phenotype being tracked and whether filled means affected
trace parents, siblings and generations accurately
mark genotypes that are forced by phenotype or parent–child transmission
keep uncertain alleles as alternatives until more evidence removes them

A pedigree records observed family evidence, not every conception or every relative. Small families and missing records can make more than one inheritance model possible.

Observation	Strong inference
two unaffected parents have an affected child	supports recessive inheritance; contradicts simple complete-dominant inheritance
affected person has an affected parent in every generation	supports dominance, but does not prove it alone
affected father has an affected son	contradicts X-linked transmission from father to son
affected sons arise from unaffected mothers	consistent with X-linked recessive carrier mothers
males and females affected similarly	supports autosomal inheritance

Start with forced genotypes. For an autosomal recessive trait, affected people are aa; each unaffected parent of an affected child must carry a. For an X-linked recessive trait, an affected son XʰY proves that his mother supplied Xʰ.

Treat each inheritance mode as a hypothesis. Predict transmissions that must or cannot occur, compare them with the pedigree, and reject a model only when a relationship contradicts it.

Skipping a generation suggests recessiveness but is not a proof. Chance, incomplete records, late onset and variable expression can obscure the apparent pattern.

Feature	Continuous variation	Discrete variation
possible values	many intermediate measurements	separate categories
common cause	polygenic effects plus environment	often one or few major loci
examples	height, body mass, skin pigmentation	ABO group, biological sex chromosome category
useful display	histogram or box plot	bar chart or counts table

Ask whether the variable is measured on a scale or counted in named categories. Height remains continuous even if a researcher later groups measurements into artificial height bands.

Continuous does not mean every imaginable value must occur in one sample, and discrete does not mean the categories are controlled by only one allele.

In polygenic inheritance, several loci contribute to one characteristic. Different combinations of small allele effects generate many genotypic values rather than a few Mendelian phenotype classes.

many contributing genotypes + nutrition, health, temperature or other environmental effects + genotype-specific responses to those conditions → overlapping phenotypes across a continuous range

Trait	Genetic contribution	Environmental contribution
height	many growth-related loci	nutrition, disease and developmental conditions
body mass	metabolism and appetite loci	diet, activity and health
skin pigmentation	several melanin-related loci	ultraviolet exposure changes melanin production

Do not confuse multiple genes contributing to one trait with multiple alleles at one gene locus. These are different levels of genetic variation.

order the measurements from lowest to highest
find the median
find Q1 as the median of the lower half and Q3 as the median of the upper half
calculate IQR = Q3 − Q1
draw the box from Q1 to Q3 and the median line inside
extend whiskers to the most extreme non-outlier values and plot outliers separately

A common rule marks values below Q1 − 1.5 × IQR or above Q3 + 1.5 × IQR as outliers. Check the convention used before treating the raw minimum and maximum as whisker ends.

A horizontal box-and-whisker plot labels minimum, lower quartile Q1, median, upper quartile Q3, maximum, interquartile range and an outlier.

Feature	What to compare
median line	typical central value
box length, IQR	spread of the middle 50%
whiskers	spread beyond the quartiles among non-outliers
isolated points	potential outliers requiring biological or measurement review
overlap between boxes	descriptive overlap, not automatically a significance test

If group A has a higher median but strongly overlapping boxes with group B, report both observations. Do not erase the overlap or claim a significant difference without an appropriate inferential test.

A box plot does not show sample size, separate peaks or every raw value unless these are added. Two very different distribution shapes can share the same median and quartiles.

X and Y contributions establish the usual chromosomal starting point → unequal X-linked gene dosage changes inheritance risk → pedigrees reveal transmission across families → polygenes and environment create continuous population variation → box plots summarize centre and spread

Evidence source	Reliable question
X-linked cross	which parent supplies the allele to sons or daughters?
pedigree	which inheritance hypotheses are contradicted by family relationships?
continuous measurements	where is the median and how wide is the middle 50%?
apparent outlier	biological extreme, data error or different process?

Match the claim to the evidence scale. One Punnett grid predicts a fertilization probability, a pedigree constrains family genotypes, and a box plot describes a sample distribution; none alone proves a universal biological rule.

Before meiosis, an Aa diploid cell carries A on one homologous chromosome and a at the same locus on the other homolog.

At anaphase I, homologous chromosomes move to opposite poles. A and a therefore enter different daughter cells; meiosis II separates sister chromatids without reuniting the alleles.

If meiotic products survive equally, an Aa parent makes approximately 1/2 A gametes and 1/2 a gametes. Random fertilization then restores two alleles in the zygote.

Mendel's law of segregation is therefore a chromosome-mechanics statement: paired alleles separate because their homologous chromosomes separate during meiosis.

Each bivalent can face either spindle pole at metaphase I. The orientation of the A/a pair does not determine the orientation of a B/b pair on a different chromosome, so maternal and paternal homologs enter daughter cells in different combinations.

Gene positions	Assortment expectation
on different chromosomes	independent
far apart on the same chromosome with recombination approaching 50%	behaves approximately independent
close together on the same chromosome	linked; parental combinations exceed recombinants

Independent assortment is not the random separation of sister chromatids. It concerns the relative orientation and separation of different homologous chromosome pairs in meiosis I.

Two unlinked homologous chromosome pairs orient in two equally likely ways at meiosis I, producing AB and ab gametes in one orientation or Ab and aB gametes in the other.

Every gamete from AaBb must contain one allele from the A/a locus and one from the B/b locus. For unlinked loci, multiply the independent probabilities.

A-locus choice	B-locus choice	Gamete	Probability
A, 1/2	B, 1/2	AB	1/4
A, 1/2	b, 1/2	Ab	1/4
a, 1/2	B, 1/2	aB	1/4
a, 1/2	b, 1/2	ab	1/4

A valid gamete list has one allele per locus, no duplicate types and probabilities summing to 1. Writing A, a, B and b as four separate gametes loses one gene from each gamete.

The same multiplication rule scales: an individual heterozygous at n independently assorting loci can make 2ⁿ gamete types, before considering linkage or unequal viability.

Assume both genes are autosomal and unlinked, complete dominance acts at each locus, both parents are AaBb, the four gamete types are equally frequent, fertilization is random and offspring classes survive equally.

Phenotype class	Probability calculation	Expected fraction
A_B_	3/4 × 3/4	9/16
A_bb	3/4 × 1/4	3/16
aaB_	1/4 × 3/4	3/16
aabb	1/4 × 1/4	1/16

The ratio is two independent 3:1 monohybrid probabilities multiplied together. A 4 × 4 grid lists all 16 genotype combinations; the product rule reaches the same phenotype expectation more efficiently.

An AaBb by AaBb dihybrid Punnett grid groups offspring into four phenotype classes in a 9:3:3:1 ratio and contrasts an AaBb by aabb test cross with a 1:1:1:1 ratio.

Cross the double heterozygote AaBb with aabb. The tester can contribute only ab, so each offspring genotype reveals which gamete came from the heterozygote.

AaBb gamete	Tester gamete	Offspring	Phenotype class
AB	ab	AaBb	both dominant
Ab	ab	Aabb	A dominant only
aB	ab	aaBb	B dominant only
ab	ab	aabb	both recessive

If A/a and B/b assort independently, AB, Ab, aB and ab gametes each occur with probability 1/4, so the test-cross phenotype expectation is 1:1:1:1.

A strong excess of two reciprocal classes and a shortage of the other two suggests linkage: parental chromosome combinations are being transmitted more often than recombinant combinations.

search an authoritative gene database with the gene name or symbol
confirm the organism is Homo sapiens and select the correct record
record chromosome, cytogenetic band or genomic coordinates from genomic context
record the named RNA and polypeptide product and its supported function
compare a second gene's chromosome and coordinates before calling the pair linked

Database field	Question it answers
gene symbol and stable identifier	which gene record?
chromosome and genomic coordinates	where is the locus?
transcript or product annotation	what is produced?
curated functional summary	what evidence supports its role?

Genes on different chromosomes are unlinked. Genes on the same chromosome are physically linked, but how often crossing over separates them depends on the distance between their loci.

A nearby genomic feature is not automatically the same gene or product. Record the database version, species, coordinates and identifier so the evidence can be checked again.

homologs separate alleles → independently oriented bivalents combine alleles from unlinked genes → AaBb makes AB, Ab, aB and ab gametes equally → random fertilization produces 9:3:3:1 in an F2 or 1:1:1:1 in a test cross

Prediction	Required conditions
1:1 allele segregation	normal meiosis and equal gamete contribution
four equal AaBb gametes	loci unlinked or recombination effectively 50%
9:3:3:1 F2	complete dominance, random fertilization and equal survival
1:1:1:1 test cross	double heterozygote × double recessive tester

A database locates the genes; the cross tests their behaviour. If observed offspring depart systematically from the unlinked expectation, inspect chromosome location, parental phase, crossing over and viability before blaming random sampling alone.

Same AaBb genotype	Alleles on one homolog	Alleles on the other	Name
AB/ab	A with B	a with b	coupling phase
Ab/aB	A with b	a with B	repulsion phase

Writing only AaBb records which alleles are present but loses which combinations share a chromosome. For linked genes, parental phase determines which gametes are common and which require crossing over.

Autosomal linkage means two loci share a non-sex chromosome. Sex linkage is different: it refers to a locus on a sex chromosome, usually X.

A diagram contrasts linked A and B loci on the same chromosome, unlinked loci on different chromosomes and crossing over that creates Ab and aB recombinant chromatids from AB and ab parental chromatids.

Start with a heterozygote in coupling phase: parental homologs carry AB and ab. After DNA replication, each homolog has two sister chromatids.

A chiasma between the A and B loci exchanges matching segments between non-sister chromatids. Two chromatids remain parental, while two become Ab and aB.

Chromatid or gamete class	Allele combination	Origin
parental	AB and ab	no exchange between the loci
recombinant	Ab and aB	crossover between the loci

A crossover elsewhere on the chromosome does not recombine these two loci. The exchange must occur between their positions to separate their parental allele combination.

Cross AB/ab with ab/ab. Because the tester contributes only ab, each offspring's phenotype or genotype identifies the gamete produced by the double heterozygote.

Heterozygote gamete	Offspring	Classification for AB/ab parent
AB	AaBb	parental
ab	aabb	parental
Ab	Aabb	recombinant
aB	aaBb	recombinant

two largest reciprocal classes identify the parental haplotypes
two smaller reciprocal classes identify recombinant gametes
an exact 1:1:1:1 pattern supports independent assortment, not strong linkage

Classify by comparison with the heterozygous parent's chromosome combinations, not by whether an offspring looks common or unusual in the population.

$r=\frac{R}{N}\times 100\%$

Here, R is the number of recombinant offspring and N is the total. If a test cross gives 420 AB, 400 ab, 90 Ab and 90 aB offspring, r = (90 + 90) / 1000 × 100% = 18%.

Observation	Inference
recombinants much below 50%	loci are linked
lower recombinant frequency	loci are usually closer, so fewer chiasmata fall between them
about 50% recombinants	loci behave as unlinked; they may be on different chromosomes or far apart

Recombinant frequency does not exceed 50% in an ordinary two-locus test. Multiple crossovers can restore parental combinations, so frequency underestimates large physical distances.

Statement	Meaning in a dihybrid goodness-of-fit test
null hypothesis, H₀	observed departures from the expected genetic ratio are due to chance sampling
alternative hypothesis, H₁	the departures are too large for that chance model; another explanation is needed

Convert the ratio to expected counts before calculating χ². For a total of 556 offspring and a 9:3:3:1 model: multiply 556 by 9/16, 3/16, 3/16 and 1/16 to obtain 312.75, 104.25, 104.25 and 34.75.

Categories must be mutually exclusive, observations should be independent and expected counts should be large enough for the approximation. The expected ratio must come from a biological model chosen before inspecting the deviations.

Chi-squared tests whether counts fit a specified expectation; it does not identify the true alternative mechanism when they do not fit.

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

Category	O	E	(O − E)² / E
both dominant	315	312.75	0.016
A dominant only	108	104.25	0.135
B dominant only	101	104.25	0.101
both recessive	32	34.75	0.218
sum	556	556	χ² = 0.470

For four mutually exclusive categories with one fixed total and no parameters estimated from the data, degrees of freedom = categories − 1 = 3.

Use counts, not percentages, inside the formula. Keep expected values unrounded during calculation, and include every category contribution before comparing with a critical value.

A chi-squared table compares observed and expected Drosophila phenotype counts, sums category contributions to 0.47 and compares the result with the p equals 0.05 critical value for three degrees of freedom.

At p = 0.05	Statistical decision	Genetic conclusion
χ² ≤ critical value	fail to reject H₀	deviations are not significant; data are consistent with the expected ratio
χ² > critical value	reject H₀	deviations are significant; the stated ratio does not adequately explain the data

For df = 3, the p = 0.05 critical value is 7.815. Because 0.470 < 7.815, fail to reject H₀: the observed Drosophila counts are consistent with a 9:3:3:1 expectation.

Do not say the test proves H₀ or proves independent assortment. A non-significant result means the observed departure is small enough to be explained by sampling under the model.

A significant result directs further investigation. Possible causes include linkage, differential survival, non-random fertilization, scoring error or an incorrect dominance model; chi-squared alone does not choose among them.

meiosis predicts gamete combinations → chromosome loci determine whether genes assort independently or travel together → crossing over creates reciprocal recombinants → test-cross counts reveal parental phase and recombinant frequency → chi-squared tests whether departures from a chosen ratio exceed chance expectation

Offspring pattern	First model to examine
9:3:3:1 F2	two unlinked genes with complete dominance
1:1:1:1 test cross	four equally frequent gametes from an unlinked double heterozygote
two large and two small reciprocal test-cross classes	linked loci with crossing over
significant χ² departure	assumptions, linkage, viability, fertilization or scoring require review

Keep prediction and evidence separate: define the genetic model first, calculate expected counts from it, compare observations with the expectation, and change the model only when chromosome location or statistical evidence requires it.

1 mark

For what reason do gametes contain only one allele of each gene?

3 marks

L. purpureus can have purple or white flowers. Two pure-breeding varieties were crossed: HA 4 with white flowers and GL 424 with purple flowers. All of the $F_{1}$ plants had purple flowers. The $F_{1}$ plants were self-pollinated to produce an $F_{2}$ generation. There were 97 plants with purple flowers and 38 plants with white flowers in the $F_{2}$ generation.

Using a Punnett grid, explain the results of this cross.

1 mark

Define the term genotype.

2 marks

Identify the phenotypes of each part of the phenotypic ratio.

Ratio	Phenotypes
9
3
3
1

4 marks

Many genetic diseases are due to recessive alleles of autosomal genes that code for an enzyme. Using a Punnett grid, explain how parents who do not show signs of such a disease can produce a child with the disease.

1 mark

Scientists incubated larvae of the moth Utetheisa ornatrix at either $15^{\circ} \mathrm{C}$ or $22^{\circ} \mathrm{C}$ until they hatched. They found the hatched moths had different wing colour patterns due to phenotypic plasticity.

Moth from larvae incubated at \(15^{\circ

Moth from larvae incubated at \(22^{\circ

Which of the following explains the observed differences in wing colour?

4 marks

Discuss the causes and treatments of phenylketonuria.

1 mark

Which statement defines alleles?

9 marks

Describe the inheritance of ABO blood groups.

1 mark

A Mirabilis jalapa plant with red flowers was crossed with one with white flowers. All plants in the F1 generation had pink flowers. What phenotype ratio would be expected in the F2 generation?

4 marks

Distinguish between autosomes and sex chromosomes in humans.

8 marks

Explain how males inherit hemophilia and how females can become carriers for the condition.

2 marks

Explain how the pedigree chart shows that the dominant allele causing PKD is not on the X chromosome.

7 marks

Explain the reasons for variation in human height.

3 marks

Using the data, deduce whether the incidence of CHF or the incidence of anemia has a greater effect on the blood hepcidin concentration.

6 marks

Outline the relationship between Mendel's law of independent assortment and meiosis.

3 marks

The expected ratio of phenotypes in the offspring of a cross between a plant with narrow, yellow leaves and a plant heterozygous for the genes for leaf width and colour is 1: 1: 1: 1.

Justify this expected ratio using a Punnett grid or other diagram.

3 marks

Outline how it can be shown that the genes for shell base colour (Cc) and presence or absence of bands (Bb) are linked.

1 mark

An individual is heterozygous for two linked genes $\frac{\mathrm{AB}}{\overline{\mathrm{ab}}}$ .

To investigate the frequency of crossing over, a test cross is carried out between the individual and another that is homozygous recessive for both genes. What are the possible recombinants in the offspring of this cross?

2 marks

The chi-squared value was calculated as shown. Deduce, with reasons, whether the observed ratio differed significantly from the expected Mendelian ratio.

c^{2}=\Sigma \frac{(\text { Observed }- \text { Expected })^{2}}{\text { Expected }}=1002.6

	Probability
Degrees of freedom	0.995	0.975	0.20	0.10	0.05	0.025	0.02	0.01	0.005	0.002	0.001
1	0.00004	0.001	1.642	2.706	3.841	5.024	5.412	6.635	7.879	9.550	10.828
2	0.010	0.051	3.219	4.605	5.991	7.378	7.824	9.210	10.597	12.429	13.816
3	0.072	0.216	4.642	6.251	7.815	9.348	9.837	11.345	12.838	14.796	16.266
4	0.207	0.484	5.989	7.779	9.488	11.143	11.668	13.277	14.860	16.924	18.467
5	0.412	0.831	7.289	9.236	11.070	12.833	13.388	15.086	16.750	18.907	20.515
6	0.676	1.237	8.558	10.645	12.592	14.449	15.033	16.812	18.548	20.791	22.458
7	0.989	1.690	9.803	12.017	14.067	16.013	16.622	18.475	20.278	22.601	24.322

Next Topic

—D3.3 HomeostasisRead

Haploid gametes + fusion = diploid

Genetic crosses in flowering plants

Genotype exam focus

Phenotype exam focus

Dominant and recessive alleles

Phenotypic plasticity

Phenylketonuria (PKU)

SNPs and multiple alleles

ABO blood groups

Incomplete dominance and codominance

Sex determination

Haemophilia exam focus

Pedigree charts

Continuous variation

Box-and-whisker plots

Segregation and independent assortment

Dihybrid crosses

Autosomal gene linkage

Recombinants exam focus

Chi-squared test