1. DIHYBRID CROSSES & GENE LINKAGE
AHL Topic 10.2
IB Biology
Miss Werba

2. DIHYBRID CROSSES & GENE LINKAGE POLYGENIC INHERITANCE
AHL TOPIC 10 – GENETICS
10.1 MEIOSIS
10.2
DIHYBRID CROSSES & GENE LINKAGE
10.3
POLYGENIC INHERITANCE
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3. THINGS TO COVER Genotype & phenotype ratios
Autosomes v Sex chromosomes
Crossing over = exchange of alleles
Linkage group
Dihybrid cross between two linked genes
Identification of recombinants
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4. DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
10.2.1
DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
Linked genes:
Genes that occur on the same chromosome
Likely to be inherited together
Unlinked Genes:
Genes that occur on separate chromosomes
Follow Mendel’s Law of Independent Assortment - inheritance of alleles is random
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5. DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
10.2.1
DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
The basic principles applied to solving monohybrid problems is also used for dihybrid problems.
The only difference is that 2 traits are considered at the same time, rather than one.
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6. DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
10.2.1
DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
Steps:
Assign letter symbols for alleles
Determine the parental genotypes
Work out the possible allele combinations for each parent
Draw up the Punnett square & complete.
Determine genotype ratio – by counting the different genotypes
Determine the phenotype ratio – by determining the possible phenotypes from the genotypes
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7. DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
10.2.1
DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
Example:
Two traits will be considered in Drosphilia flies:
Eye colour
Wing length
These traits are coded for by unlinked genes.
Assume that:
Red eyes is dominant over pink
Long wings are dominant over short
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8. DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
10.2.1
DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT Q:
A fly with pink eyes and short wings was crossed with a pure breeding one that had red eyes and long wings.
Determine the genotype and phenotype ratios in the F1 and F2 generations.
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9. DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
10.2.1
DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
Example:
Let:
R = red eyes, r = pink eyes
L = long wing, l = short wings
Parents are:
pink eyes and short wings :
possible gametes:
red eyes and long wings:
rrll rl
RRLL RL
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10. DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
10.2.1
DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
Example:
F1 Genotype ratio:
F1 Phenotype ratio: rl RL
RrLl
100% RrLl
100% red eyes, long wings
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11. DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
10.2.1
DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
Example:
If F1 generation are left to interbreed:
Possible gametes formed are: RL Rl rL rl
RrLl x RrLl
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12. DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
10.2.1
DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
Example:
F2 Genotype ratio: RRLL; RRll; rrLL; rrll
RrLl; rrLl; Rrll; RrLL; RRLl
F2 Phenotype ratio: 9 red eyes, long wings
3 red eyes, short wings
3 pink eyes, long wings
1 pink eyes, short wings RL Rl rL rl
RRLL
RRLl
RrLL
RrLl
RRll
Rrll
rrLL
rrLl
rrll
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13. DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
10.2.1
DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
Example:
In this cross the parents were:
red eyes, long wings and
pink eyes, short wings.
The recombinants are:
red eyes, short wings
pink eyes, long wings
The recombinant ratios observed are:
9 red eyes, long wings (same as parent genotype)
3 red eyes, short wings (recombinant)
3 pink eyes, long wings (recombinant)
1 pink eyes, short wings (same as parent genotype)
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14. DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
10.2.1
DIHYBRID CROSSES (UNLINKED GENES) Command term = CALCULATE and PREDICT
When two pure breeding individuals are crossed, the F1 will all have the same genotype and phenotype.
They will all be heterozygous.
If allowed to interbreed, the F2 will have a ratio of:
9 : 3 : 3 : 1
This is the same ratio that Gregor Mendel found over 300 years ago in his studies of pea plants.
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16. CHROMOSOMES Command term = DISTINGUISH
10.2.2
CHROMOSOMES Command term = DISTINGUISH
Chromosome:
molecule of DNA found in cell nuclei
Homologous chromosomes:
two chromosomes which have the same sequence of genes
Autosome:
a chromosome which determines the characteristics of an organism, except for its sex
Sex chromosomes:
determine gender in humans (XX= ♀; XY=♂)
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17. CHROMOSOMES Command term = DISTINGUISH
10.2.2
CHROMOSOMES Command term = DISTINGUISH
Human body cells have 46 chromosomes.
Exist as 23 homologous pairs (½ from each parent)
One pair are the sex chromosomes (allosomes):
Determine gender and some non-sexual traits
Females XX
Males XY
22 pairs are autosomes (determine non-sexual traits)
You get 22 autosomes and 1 sex chromosome from each parent
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18. CROSSING OVER Command term = EXPLAIN
10.2.3
CROSSING OVER Command term = EXPLAIN
Crossing over and recombination:
when chromosomes exchange segments during meiosis, giving rise to variation
Crossing over and recombination occurs during...?
Prophase I of meiosis
Crossing over and recombination occurs at the...?
chiasma
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19. CROSSING OVER Command term = EXPLAIN
10.2.3
CROSSING OVER Command term = EXPLAIN
When homologous chromosomes line up during prophase I they form a...?
synapsis
The pair of homologous chromosomes is referred to as a...?
bivalent
Portions of the adjacent chromosomes may swap, resulting in the formation of ...?
recombinants or recombinant chromosomes
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20. CROSSING OVER Command term = EXPLAIN
10.2.3
CROSSING OVER Command term = EXPLAIN
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Ref: Biology Key Ideas 20

21. LINKAGE GROUP Command term = DEFINE
10.2.4
LINKAGE GROUP Command term = DEFINE
Linked genes are pairs or groups of genes which are carried on the same chromosome and are inherited together.
Mendel’s law of independent assortment is not always true for genes on the same chromosome.
It is still possible for recombination to occur between linkage groups, but it is far less likely.
Gene linkage can occur on the autosomes and sex chromosomes.
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22. GENE LINKAGE Command term = EXPLAIN
10.2.5
GENE LINKAGE Command term = EXPLAIN
The distance between two gene loci is measured as a recombination fraction (θ).
It is measured in CentiMorgans (cM) – equivalent to a 1% chance of recombination.
If two loci are not linked, then θ = 0.5
indicating that the genes separate together in 50% of all meioses
If two loci are linked, then θ = 0.1
indicating that the genes separate together in 90% of all meioses and crossing over rarely only (only 10%)
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23. GENE LINKAGE Command term = EXPLAIN
10.2.5
GENE LINKAGE Command term = EXPLAIN
Unlinked genes:
are found on different chromosomes
homologous pairs line up independently of each other during metaphase I of meiosis
Independent Assortment allows for recombination T t B b
T and B not linked
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24. GENE LINKAGE Command term = EXPLAIN
10.2.5
GENE LINKAGE Command term = EXPLAIN
Linked genes:
are found on the same chromosome
can only be broken apart by crossing over and gene recombination during prophase I of meiosis T t B b
T and B are linked
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25. GENE LINKAGE Command term = EXPLAIN
10.2.5
GENE LINKAGE Command term = EXPLAIN T t B b
T and B not linked T t B b
T and B are linked
Alleles are normally shown side-by-side in dihybrid crosses
TtBb
Alleles in linkages groups are shown as vertical pairs:
This means that T is linked to B (on same chromosome) and t is linked to b (on same chromosome).
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26. GENE LINKAGE EXAMPLES Command term = EXPLAIN and IDENTIFY
10.2.5
10.2.6
GENE LINKAGE EXAMPLES Command term = EXPLAIN and IDENTIFY
A test cross was conducted on a fly heterozygous for eye colour and body colour
The results are shown below:
151 wild-type (normal colour eyes & normal body colour)
8 purple eyes and normal body colour
10 normal eyes and black body colour
131 purple eyes and black body colour
What is a test cross?
Explain how these F1 offspring support the idea that genes P and B are linked.
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27. GENE LINKAGE EXAMPLES Command term = EXPLAIN and IDENTIFY
10.2.5
10.2.6
GENE LINKAGE EXAMPLES Command term = EXPLAIN and IDENTIFY
What is a test cross?
a cross with a homozygous recessive genotype to allow the genotype of unknowns to be inferred
Explain how these F1 offspring support the idea that genes P and B are linked.
the genes did not move independently of one another; only crossing over & recombination will allow you to get the other 2 phenotypes in smaller numbers
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28. GENE LINKAGE EXAMPLES Command term = EXPLAIN and IDENTIFY
10.2.5
10.2.6
GENE LINKAGE EXAMPLES Command term = EXPLAIN and IDENTIFY
Draw a chromosome picture for the heterozygous parent.
Which F1 phenotypes represent:
parental-type offspring?
recombinant offspring?
Calculate the percentage crossing over or recombination fraction (θ).
If a recombination fraction of 1% is equivalent to one map unit (used to map chromosomes), calculate the map distance between genes P and B.
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29. GENE LINKAGE EXAMPLES Command term = EXPLAIN and IDENTIFY
10.2.5
10.2.6
GENE LINKAGE EXAMPLES Command term = EXPLAIN and IDENTIFY
Draw a chromosome picture for the heterozygous parent.
Which F1 phenotypes represent:
parental-type offspring? (normal eyes & normal body and purple eyes & black body)
recombinant offspring? (purple eyes & normal body and normal eyes & black body) P B p b
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30. GENE LINKAGE EXAMPLES Command term = EXPLAIN and IDENTIFY
10.2.5
10.2.6
GENE LINKAGE EXAMPLES Command term = EXPLAIN and IDENTIFY
Calculate the percentage crossing over or recombination fraction (θ).
If a recombination fraction of 1% is equivalent to one map unit (used to map chromosomes), calculate the map distance between genes P and B.
6%  6 map units apart
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31. GENE LINKAGE EXAMPLES Command term = EXPLAIN and IDENTIFY
10.2.5
10.2.6
GENE LINKAGE EXAMPLES Command term = EXPLAIN and IDENTIFY
Predict the genotypes of the offspring (and their expected frequency) for the following mating:
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32. GENE LINKAGE EXAMPLES Command term = EXPLAIN and IDENTIFY
10.2.5
10.2.6
GENE LINKAGE EXAMPLES Command term = EXPLAIN and IDENTIFY
20 map units apart
 20% recombination fraction (θ= 0.2)
Parents: x
Gametes: AB, ab x ab
F1 genotype ratio:
parental-type offspring – ;
recombinant offspring – ;
 must add up to 80% (ie. 40% each)
 add up to 20% (ie. 10% each)
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33. Sample questions
Q1.
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34. Sample questions
Q2. In some maize plants the seed is enclosed in a green sheath called a tunica. The allele (T) for this is dominant to the allele (t) for normal, unenclosed seeds. The endosperm of the seed can be starchy (allele E) or sugary (allele e). The genes for these two characteristics are linked. The table below shows the outcome of crosses between a plant heterozygous for both characteristics and one that is homozygous recessive for both characteristics.
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35. Sample questions
Q2.
State the genotype of the heterozygous parent using the correct notation. [1]
Identify which individuals are recombinants in this cross. [1]
Explain what has occurred to cause these results. [2]
Maize belongs to the group of plants known as angiospermophyta. Distinguish between angiospermophytes and bryophytes. [2]
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36. Sample questions
A1. C
A2.
unenclosed seeds, starchy and tunica present, sugary
and (both needed)
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37. Sample questions
A2.
crossing over; between non-sister chromatids (in prophase I); results in exchange of alleles / change in linkage groups; so some gametes are T e or t E; test cross expect ratio of two phenotypes / correct Punnett Square showing test cross; but instead get four phenotypes with smaller percentage of recombinants; Above points can be shown in diagrams.
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