Transcript How to complete a Test Cross

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
Ee
Ee
e
ee
ee
E
e
E
EE
Ee
e
Ee
ee
How to complete a
Test Cross
A 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
Answer
Description recognises
possibility
of heterozygous and selecting
for
breeding
B_ × with bb (Test Cross) /
rabbits
with broken coat/ with
homozygous
recessive
Just Test Cross or Back Cross not
sufficient.
Reason for why test cross used
or what possible punnet
squares
Cannot tell by looking at the
dominant trait
whether it is homozygous or
heterozygous, so a cross with a
rabbit
that is homozygous for the
recessive trait (solid) is
carried out.
OR
Explanation may include
diagram.
Discussion includes reasons
how to tell which rabbits are
heterozygous vs homozygous
If any offspring have a solid
coat colour then the rabbit
must be heterozygous. If there
are no offspring with solid coat
colour we can assume the
rabbit must homozygous.
May include idea that suficent
offspring should be produced
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?
Answer
Describe 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