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Multiple Allele and Dihybrid Crosses

Multiple Allele and Dihybrid Crosses

Multiple Allele Crosses

e.g. 3 blood types:

  1. IA = Blood group A (dominant)
  2. IB = Blood group B (dominant)
  3. Io = Blood group O (Recessive)

Dihybrid Crosses

  • The inheritance of two characteristics which are controlled by different genes
  • Each of the two genes will have different alleles
  • Dihybrid crosses can be used to determine the likelihood of offspring inheriting certain combinations of the two characteristics from particular parents
  • A dihybrid cross for two heterozygous parents leads to a 9:3:3:1 phenotypic ration
Monohybrid Inheritance

Monohybrid Inheritance

  • The inheritance of a characteristic controlled by a single gene

Monohybrid Crosses

  • Monohybrid Crosses show the likelihood of the different alleles of that gene (And therefore the different versions of the characteristics) being inherited by offspring of certain parents

e.g. The dominant allele (H) codes for ‘tall’ whereas the recessive allele (h) codes of “short”

Phenotypic Ratios

  • The ratio of different phenotypes in the offspring
  • Genetic diagrams can be used to predict this ration in F1 and F2 offspring
  • Commonly, two heterozygous parents give a 3:1 ratio of dominant: recessive characteristic in the offspring

Monohybrid Inheritance and Codominant Allele

  • In a codominant allele both genes are expressed
  • With two heterozygous parents involving codominant alleles it will commonly give a 1:2:1 ratio in the offspring#





Inheritance of Sex-linked characteristics

  • Sex linkage refers to an allele that code for a characteristic which is located on a sex chromosome
  • As the Y chromosome is smaller than the X, fewer genes are carried on it
  • Therefore, most sex-linked genes are found on the X chromosome (x-linked genes)
  • As males only have one X chromosome they therefore often only have one allele for sex-linked genes, and so there is a higher chance of a recessive gene being expressed
  • Ultimately, males are more likely to express a recessive gene relative to females
  • Genetic disorders caused by a faulty allele located on a sex chromosome include:
    • Colour blindness
    • Haemophilia

e.g. As haemophilia is sex linked, it is therefore on the X chromosome. Therefore, females would require 2 recessive alleles for expression, whereas males would only require one.

As the Y chromosome does not have either gene it is just represented as Y

  XH Xh

If a female carrier and a male non-carrier have offspring there is a 2:1:1 ratio (female without colour blindness : male without colour blindness : male with colour blindness). Overall there is a 50% chance of a male offspring to be colour blind. 

Linkage of autosomal Genes

  • Autosome is any chromosome that is not a sex chromosome
  • Autosomal genes are genes found on the autosomes
  • Genes found on the same autosome are linked (as they stay together during independent segregation of chromosomes in meiosis 1 and the allele will be passed to the daughter cell)
  • The will not occur is crossing over splits them linked genes first
  • The closer the two genes are on the autosome (the closer they are linked due to the decreased chance of crossing over)
Hardy-Weinberg Principle

Hardy-Weinberg Principle

Species: a group of similar organisms that can reproduce to give fertile offspring

Population: A group of organisms of the same species living in a particular area at a particular time

Gene Pool: Complete range of alleles present in a population

Allele Frequency: The frequency of an allele in a population (given as a %)

Hardy-Weinberg Principle

  • Used to predict the frequencies of alleles/genotype/phenotype within a population
  • It can also be test whether or not the Hardy-Weinberg principles applies to a particular allele in a particular population
    • E.g. if frequencies do change between generations in a large population then there’s an influence of some kind
  • The principle will work providing:
    • Large Population size
    • No immigration/Emigration
    • No natural selection
    • Random mating (so all possible genotypes can bread with all others)

Allele Frequency

p + q = 1

p = Frequency of one allele (usually dominant)

q = frequency of one different allele (usually recessive)

  • the frequencies of both p and q must equal to 1 (100%)

Genotype Frequency

p2 + 2pq + q2 = 1

p2 Frequency of homozygous dominant genotype

2pq = Frequency of heterozygous genotype

q2 = Frequency of homozygous recessive genotype

  • Total frequency of all genotypes within a population = 1 (100%)
  • Phenotype can also be determine providing that the relation between phenotype and genotype is known
  • Principle can also be used if two alleles are codominant or if whether the allele is recessive and which is dominant is unknown
    • By making p represent one allele and q represent the other


  • Evolution is the change in an allele frequency over time
  • Natural selection is one explanation of how evolution functions
  • Selection Pressure refers to factors such as food, predators, diseases and competition on a species which results in the frequency of an allele to alter within a gene pool
  • Gene Pool refers to the total number of all alleles within a population at a given time.

Types of Competition:

  1. Intraspecific Competition: Competition between the same species
  2. Interspecific Competition: Competition between different species

Over Production

  • Over production is the result of a high birth rate resulting in intraspecific pressures
  • Those with the allele which is advantageous provide them the limited resources to survive
  • Therefore the allele is passed on


  • The interaction of genes that are not alleles, in particular the suppression of a gene by a different gene.
  • E.g. windows peak:
    • Allele results in a “V shaped” hair
    • Another allele expressed results in baldness
    • Therefore, if carrier of both, as the person is bald the allele for widow’s peak is unclear

Phenotypic ratios for Epistatic Genes

  • Recessive Epistatic Alleles
    • If the epistatic allele is recessive then 2 copies of it will mask the expression of the other gene
    • Homozygous parent + homozygous parent = 9:3:4 phenotypic ratio

(dominant both: dominant epistatic/recessive: recessive epistatic (in F­2))

  • Dominant Epistatic Alleles
    • If the epistatic allele is dominant then 1 copies of it will mask the expression of the other gene
    • Homozygous parent + homozygous parent = 12:3:1 phenotypic ratio

(dominant epistatic: recessive epistatic, dominant other: recessive (in F­2))



  • Where both alleles are equally dominant
  • Therefore, both alleles are expressed in the phenotype

e.g. Snapdragon plant has one allele codes for enzyme which catalyses formation of red pigment and other allele codes for enzyme that lacks this catalytic activity

If alleles showed the usual pattern of 1 dominant and 1 recessive the flower would be either red and white

But as they are co-dominant three colours of flower are found:

Homozygous (1st allele): Both alleles code forthe enzyme and so plant is

Homozygous (2nd allele): No enzyme is produced and so plant is white

Heterozygous: Their single allele for the functional enzyme produces some red and so plant is pink

CR = Red pigment

CW= White pigment

e.g. CRC + CW CW


Phenotypic Ratio: 1

   e.g. CRCW + CRCW


Phenotypic ratio: 2:1:1

 Pink: Red: white

Hund’s Rule

Hund’s Rule

Every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.

Single electrons occupy all empty orbitals before they start to form pairs in orbitals. Two electrons in the same orbital have a repulsion between them due to their negative charge. The more stable configurations is with single electrons in different orbitals.

Evidence for Hund’s Rule

The first ionisation energy for the elements in period 3 has a general increase.

Sulphurs value is below that of phosphorus. The highest energy electrons are both in 3p sub levels this is evidence for Hund’s Rule.

Phosphorus has 3 electrons in its 3p sub level whereas sulphur has 4.

The lowest first ionisation energy for sulphur is because it has a pair of electrons in one of the 3p orbitals. Mutual repulsions between these two electrons make it easier to remove one from the other.

Phosphorus ionisation energy = 1012 KJ mol-1

Sulphur ionisation energy = 1000 KJ mol-1



  • Lipids have a much lower proportion of water than other molecules such as Carbohydrates.
  • insoluble
  • Long carbon chains
  • They are made from two molecules: Lipids can exist as fats, oils and waxes.
    • Fats and oils are very similar in structure (triglycerides).
  • At room temperature, fats are solids and oils are liquids.
    • Fats are of animal origin, while oils tend to be found in plants.
  • Waxes have a different structure (esters of fatty acids with long chain alcohols) and can be found in both animals and plants.
  • They are a diverse group of:
    • Fats
    • Phospholipids
    • Steroids
  • They do not form polymers, they have large molecules (unlike proteins and carbohydrates which are polymers)
  • They have a greater ratio of Oxygen and hydrogen than H2O
  • Lipids perform many functions, such as:
  1. Storage – lipids are non-polar and so are insoluble in water.
  2. High-energy store – they have a high proportion of H atoms relative to O atoms and so yield more energy than the same mass of carbohydrate.
  3. Production of metabolic water – some water is produced as a final result of respiration.
  4. Thermal insulation – fat conducts heat very slowly so having a layer under the skin keeps metabolic heat in.
  5. Electrical insulation – the myelin sheath around axons prevents ion leakage.
  6. Waterproofing – waxy cuticles are useful, for example, to prevent excess evaporation from the surface of a leaf.
  7. Hormone production – steroid hormones. Oestrogen requires lipids for its formation, as do other substances such as plant growth hormones.
  8. Buoyancy – as lipids float on water, they can have a role in maintaining buoyancy in organisms.

Testing for Lipids: Emulsion Test


  • A clean test tube
  • 2cm3 of the sample being tested
  • 5cm3 Ethanol
  • 5cm3 water


  1. Put 2cm3 of the sample being tested into the test tube
  2. Add 5cm3 of ethanol to the test tube
  3. Add 5cm3 of water and shake the mixture gently
  4. For a control, use water as the sample in a new test tube. Repeat instructions.


If the solution turns cloudy white, this indicates lipids are present.





  • Monosaccharides are the monomers which make up carbohydrates
  • Most Simple sugar
  • They have the same number of carbon + oxygen

e.g. Glucose is C6H12O6

  • They have the general formula of (CH2O)n
  • White Crystalline solids
  • Soluble
  • The 3 monosaccharides are Glucose, Fructose and Galactose
  • Glyosidic bonds between each monomer formed under a condensation reaction

Monosaccharide to Disaccharide



Glucose + Glucose


Glucose + Fructose


Glucose + Galactose



  • An Alpha-helix structure
  • Insoluble and is compact
  • No impact of water potential (osmotically inactive)
  • It is the main plant storage of sugar
  • Made of


  • Polymer of β-glucose
  • Each monomer is inverted to the previous monomer
  • Different chains are held with hydrogen bonds
  • These chains run parallel with hydrogen bonds between the chains to form microfibrils.
  • Cellulose is what makes up the structure of a plant cell


  • Similar to amylopectin
  • α-glucose polymer with 1,4 bonds with some 1,6 branches
  • Highly branched structure
  • Can be broken down quickly to glucose when needed
  • Reflects the high metabolic rate demand for energy in animals

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