NCEA 2009

• In cats, the polydactylism gene produces a paw with an extra digit (claw).

• This is a dominant trait (E).• Discuss how a kitten with the normal number of digits /

claws could be produced by two cats that had only produced kittens with extra digits in previous litters. In your answer you should state the genotypes of the parent cats, and explain why you have selected them. Use a Punnett square to support your answer. (Use E for dominant trait.)

Correct statement. (a) Extra digits (E) is dominant over normal digits (e) so (e) may be present in cats, but stay hidden in Ee. Correct genotype combinations (a). Normal digits (ee) could show up in a Ee Ee cross or a Ee x ee cross. OR Correct Punnett square for either / both of above E e E EE Ee e Ee ee

E e e Ee ee e Ee ee

Correct interpretation of punnet phenotypes (a) Eg 25% chance of normal kitten, 75% polydactyl

Correct explanation for choice of parent genotypes. (m) Must use terms allele, gene and recessive correctly. Normal digits (ee) could show up in an Ee Ee cross (or Ee ee). It would be impossible if the parents were Ee and EE. OR At least one parent ie heterozygous / Ee / carries dominant and recessive allele for gene. For kitten to not have extra digit it must inherit the recessive allele from both parents. OR 25% or 50% chance will only be displayed in a large sample, not necessarily a single litter. Idea of lower percentage being less likely is clearly explained

Discusses how a normal digit kitten can be produced AND links the outcome to the statistical probability of it happening. MUST include Punnett square and use gene / allele correctly. (e) At least one of the parents is heterozygous / Ee / carry dominant and recessive allele for gene. For kitten to not have extra digit it must inherit the recessive allele from both parents. AND Statistically every time an Ee Ee cross is carried out there is 25% chance that offspring would be normal (ee) or 50% for ee x Ee The normal trait may not be expressed at all. PLUS Correct Punnett square. OR Explains both possible parent combinations and outcomes with punnett squares. (e) Eg: At least one of the parents is heterozygous, but the other parent may be homozygous recessive or heterozygous. If the other parent is homozygous recessive, statistically there is a 50% chance that the offspring will inherit both recessive alleles, which will be expressed as normal digits. If the other parent is heterozygous, there is a 25% chance that the offspring will inherit both recessive alleles. E e E e e Ee ee E EE Ee e Ee ee e Ee ee

How to complete a Test CrossA test cross is when we breed a known individual who is homozygous recessive with an individual with an unknown genotype.

Fundamentals for test cross• We use a homozygous recessive individual because there

genotypes is known.• Sufficient offspring must be produced to be certain• A pure bred male and female is needed to produce pure

breeding litters.

QUESTION TWO: RABBIT BREEDING (NCEA 2007)

• In some breeds of rabbit, a plain coat colour is called solid; a blotchy coat colour is called broken. Broken coat (B) is dominant to solid coat (b).

• A breeder buys a rabbit with the broken coat trait. • Discuss the process that could be used to identify

whether this rabbit is homozygous or heterozygous for broken coat.

How to answer this question• State what you would need to do….• Explain why you need to do this…• What possible outcomes would there be… and what you

would do• Support with punnet squares

• MAKE SURE YOU INCLUDE THE CORRECT TERMINOLOGY

AnswerDescription recognises Description recognises possibilitypossibility

of heterozygous and selecting of heterozygous and selecting forfor

breedingbreeding

B_ B_ × with bb (Test Cross) / × with bb (Test Cross) / rabbitsrabbits

with broken coat/ with with broken coat/ with homozygoushomozygous

recessiverecessive

Just Test Cross or Back Cross notJust Test Cross or Back Cross not

sufficient.sufficient.

Reason for why test cross used Reason for why test cross used or what possible punnet or what possible punnet squaressquares

Cannot tell by looking at theCannot tell by looking at the

dominant traitdominant trait

whether it is homozygous orwhether it is homozygous or

heterozygous, so a cross with a heterozygous, so a cross with a rabbitrabbit

that is homozygous for thethat is homozygous for the

recessive trait (solid) isrecessive trait (solid) is

carried out.carried out.

OROR

Explanation may include Explanation may include diagram.diagram.

Discussion includes reasons Discussion includes reasons how to tell which rabbits are how to tell which rabbits are heterozygous vs homozygousheterozygous vs homozygous

If any offspring have a solid If any offspring have a solid coat colour then the rabbit coat colour then the rabbit must be heterozygous. If there must be heterozygous. If there are no offspring with solid coat are no offspring with solid coat colour we can assume the colour we can assume the rabbit must homozygous.rabbit must homozygous.

May include idea that suficent May include idea that suficent offspring should be produced offspring should be produced to be certain. to be certain.

Extra• Explain what the breeder must do to establish a pure breeding

colony of Broken coat colour or solid coat colour?

QUESTION ONE: CELL DIVISION AND MUTATION (NCEA 2005)

• Discuss how the processes of meiosis and mutation can contribute to genetic variation.

Brainstorm

• What is a mutation? How does it lead to variation?

• Where must the mutation occur for it to be inherited?

• What processes in meiosis lead to variation?

• How does meiosis and fertilisation lead to recombination of genes?

AnswerDescribe TWO factors that

contribute to genetic variation.

Eg

• mutation change in genetic

makeup of a cell

• meiosis

- independent assortment

- segregation

- recombination

Explain how TWO factors lead

to variation. Must include the

idea of inheritance.

Eg

Mutation – change in the genetic

makeup. If this change occurs in

the somatic cells, it cannot be

passed on, but if in the sex cell, it

may be inherited.

Meiosis – each parent passes on

one member of each pair of

homologous chromosomes selected

at random. At fertilisation the

resulting zygote contains half it’s

chromosome complement from

each parent. Alleles from each

parent are present in new

individual and this gives variation.

(Recombination)

Discussion includes BOTH

mutation and an aspect of

meiosis linked to variation.

The link is made when mutation

in gametes allows it to be

inherited and also includes the

idea of random assortment