linked genes, recombinant and parental types

1. Ch. 12
Bi - two
Centro - the center
Chroma - colored
Cyclo - a circle
Gamet - a wife or
husband
Gen - to create,
produce
Inter - in between
Mal - bad, evil
Meio - less
Mere - a part
Meta - between
Mito - thread
Pro - before
Soma - body
Telos - an end

2. Plans for the week
Tuesday – linked genes
Wednesday – mutations to know and
Thursday – DNA structure &
replication
Friday – videos, discussion, diagrams

3. Challenging Genetics
Practice Problems
11 problems
You and a partner or partners will
work them together and report
the answer and show the
solution on the screen

4. Ch. 12
Linkage Groups & Chromosome Maps
When Mendel crossed his F1 generation: PpRr x PpRr, he got
a 9:3:3:1 ratio. He would have not seen this pattern if the
alleles had been located on the same chromosome and
inherited together.

5. Objectives
1. Define and explain how we get parental and recombinant
types
2. Create a chromosome map based on crossing
over/recombinant types data

6. Chromosome Theory of Inheritance
1. Genes have specific
locations on chromosomes
2. Genes on homologous
chromosomes segregate
away from each other
during meiosis
3. Each gene pair segregates
independently of other gene
pairs

7. Thomas Hunt Morgan studied fruit flies and
found that in some crosses, expected
outcomes weren't happening. Further
experiments confirmed that alleles located
on the same chromosome are often
inherited together. (i.e. they DON’T follow the
law of independent assortment)
A common cross used to demonstrate linkage groups is the
cross of a heterozygote wild type vestigial wings/ black body with
a recessive mutant.

8. What is up with all those letters?
or just use the old fashioned way
A a B b x a a b b

9. Because these alleles are found on the same chromosome, a
Punnett square would show them inherited together. There are
two possible arrangements for the heterozygote (AaBb) in the
above cross.
If the dominant alleles AB
are on the same
chromosome, it is called a
CIS arrangement
If the dominant alleles
are on different
chromosomes (Ab) then
it is called TRANS

10. Consider a cross with a parent that has the CIS arrangement to
one that is a mutant for both traits (aabb)
a
b
a
b
A
B
a
b X
50% wild
50%
mutants

11. Thomas Hunt did not observe a perfect 1:1 (or 50/50) ratio.
Instead, his results looked like this….
Expected Observed
Wild Type 50 40
Mutant 50 40
Vestigial wings, Wild 0 10
Black body, Wild 0 10
Question: How would you explain these results?

12. Answer: The two offspring that did not look like either parent are
called recombinants. They are a result of a CROSS-OVER that
occurred during meiosis
The alleles switched position.

13. Parental and Recombinant Types
Parental Types – offspring
have the same phenotype as
one of the two parents
Recombinant Types –
offspring with a new
combination of phenotypes

14. If recombinant phenotypes are seen 50% of the time, the genes
are not linked (they are on separate chromosomes)

15. If some other pattern appears (less than 50%), the
genes are linked
How often you see recombinants (the result of crossing over
thus “unlinking” the genes) allows you to calculate the distance
between any 2 genes on a chromosome

16. 1. What is a parental type?
2. What is a recombinant type?
3. How do we get recombinant types?
4. How do we get parental types?

17. Practice Questions (assume no crossing over occurs)
1. A dumpy winged (dd) fruit fly with long aristae (AA) is crossed
with a long winged (Dd) short aristae (aa). Show the cross and
the phenotypic proportions.

18. 2. A fruit fly with short legs (ll) and vestigial wings (ww) is crossed
with one that is heterozygous for both traits. Assuming the
dominant alleles are on separate chromosomes, show the cross
and the expected phenotypic proportions.

19. 3. In fruit flies, red eyes is a dominant allele located on the X
chromosome. The recessive condition results in white eyes. The
tan body trait is also X-linked and is dominant to yellow bodies. A
female who is heterozygous both traits with the dominant alleles
located on the same chromosome is crossed with a white eyed,
yellow bodied male. Show the cross and the phenotypic proportions
(Don't forget these traits are X-linked!)

20. 4. Chromosome Map Problem
In pea plants, flower color and pollen shape are located on the same chromosome. A plant
with purple flowers and long pollen (AaBb) is crossed with one that is recessive for both traits
(aabb).
The results are as follows:
Results
Purple, long 47
Red, round 47
Purple, round 3
Red, Long 3
a) Are the chromosomes of the AaBb parent in the cis or trans position? Sketch the punnett square showing the
expected offspring.
b) How far apart are the two alleles?

21. 116
5. In minions, the phenotype for number of eyes is
governed by two alleles – one eye = (E) and two
eyes = (e). Two heterozygous minions mate to
produce 465 offspring.
Calculate how many of these offspring are expected
to have two eyes. Round your response to the
nearest whole number.

22. 6.

23. In corn, purple kernels (R) are dominant to
yellow kernels (r). Cobs from the offspring of a
cross between a purple plant and yellow plant
were used in a lab. A student counts 329
purple and 299 yellow kernels on one cob.
Calculate the chi-squared value for the null
hypothesis that the purple parent was
heterozygous for purple kernels. Give your
answer to the nearest tenth.
7.

24. A plant geneticist is investigating the inheritance of genes for
bitter taste (S) and explosive rind (e) in watermelon. Explosive
rind is recessive and causes watermelons to burst when cut.
Non-bitter watermelons are a result of the recessive genotype
(ss). The geneticist wishes to determine if the genes assort
independently. She performs a testcross between a
bitter/nonexplosive hybrid and a plant homozygous recessive
for both traits. The following offspring are produced:
bitter/non-explosive – 88
bitter/explosive – 68
non-bitter/non-explosive – 62
non-bitter/explosive – 82
Calculate the chi-squared value for the null hypothesis that the
two genes assort independently. Give your answer to the
nearest tenth.
Are these genes linked? How do you know?
8.

26. In DNA, A pairs with ___ and G pairs with ___
So…if the genome of a fruit fly contains 2378
thymine nucleotides, how many adenine, guanine,
and cytosine nucleotides would you expect if there
are 4000 base pairs?
2738 adenine
1622 cytosine
1622 guanine
9.

27. Are these genes linked?
How do you know?
10.

28. Is the student correct?
11.