Genetics II: Terms to know, problems to solve p'299

1
Genetics II Terms to know, problems to solve
(p.299 1-7)

• Drosophila melanogaster, T.H. Morgan
• Chromosome basis for inheritance
• Linked genes
• Sex-linked genes (Drosophila and humans)
• Genetic recombination
• Crossing over
• Genetic maps, map units
• Alterations of chromosome number, nondisjunction,
  aneuploidy, monosomy, trisomy, polyploidy,
• Alterations of chromosome structure, deletion,
  duplication, inversion, translocation
• Downs syndrome, Turner syndrome, Klinefelter
  syndrome

2
Chromosomes tagged to reveal a specific gene
(yellow)
3
The chromosomal basis of Mendels laws p. 281
4
Are the genes for body color and wing size in
fruit flies located on the same chromosome or
different chromosomes?
5
(No Transcript)
6
Chromosomal basis for recombination of linked
genes
7
Constructing a Linkage Map
APPLICATION
A linkage map shows the relative locations of
genes along a chromosome.
TECHNIQUE
A linkage map is
based on the assumption that the probability of a
crossover between two genetic loci is
proportional to the distance separating the loci.
The recombination frequencies used to construct
a linkage map for a particular chromosome are
obtained from experimental crosses. The
distances between genes are expressed as map
units (centimorgans), with one map unit
equivalent to a 1 recombination frequency. Genes
are arranged on the chromosome in the order that
best fits the data. For example, suppose the
observed recombination frequencies between three
Drosophila gene pairs are b?cn 9, cn?vg 9.5,
and b?vg 17. What is the order of these genes
on the chromosome?
8
RESULTS
In this example, the
observed recombination frequencies between three
Drosophila gene pairs (b?cn 9, cn?vg 9.5, and
b?vg 17) best fit a linear order in which cn is
positioned about halfway between the other two
genes
Recombination frequencies
9.5
9
17
vg
b
cn
Chromosome
The b?vg recombination frequency is slightly less
than the sum of the b?cn and cn?vg frequencies
because double crossovers are fairly likely to
occur between b and vg in matings tracking these
two genes. A second crossover would cancel out
the first and thus reduce the observed b?vg
recombination frequency.
9
A partial genetic (linkage) map of a Drosophila
chromosome
10
X and Y Chromosomes
11
Some chromosomal systems of sex determination
12
T. H. Morgans first mutant fly phenotype
13
In a cross between a wild-type female fruit fly
and a mutant white-eyed male, what color eyes
will the F1 and F2 offspring have?
EXPERIMENT
Morgan mated a
wild-type (red-eyed) female with a mutant
white-eyed male. The F1 offspring all had red
eyes.
P Generation
X
F1 Generation
Morgan then bred an F1 red-eyed female to an F1
red-eyed male to Produce the F2 generation.
RESULTS
The F2 generation
showed a typical Mendelian 31 ratio of red eyes
to white eyes. However, no females displayed the
white-eye trait they all had red eyes. Half the
males had white eyes, and half had red eyes.
F2 Generation
14
CONCLUSION
Since all F1
offspring had red eyes, the mutant white-eye
trait (w) must be recessive to the wild-type
red-eye trait (w). Since the recessive
traitwhite eyeswas expressed only in males in
the F2 generation, Morgan hypothesized that the
eye-color gene is located on the X chromosome
and that there is no corresponding locus on the
Y chromosome, as diagrammed here.
W
W
X
X
P Generation
X
X
Y
W
W
Ova (eggs)
Sperm
W
W
F1 Generation
W
W
W
Ova (eggs)
Sperm
F2 Generation
W
W
W
W
W
W
W
W
15
The transmission of sex-linked recessive traits
XAXA
XaY
?
A father with the disorder will transmit the
mutant allele to all daughters but to no sons.
When the mother is a dominant homozygote, the
daughters will have the normal phenotype but will
be carriers of the mutation.
(a)
Sperm
Xa
Y
XAXa
XAY
XA
Ova
XAYa
XAY
XA
XAXa
?
XAY
If a carrier mates with a male of normal
phenotype, there is a 50 chance that each
daughter will be a carrier like her mother, and
a 50 chance that each son will have the
disorder.
(b)
Sperm
XA
Y
XAXA
XAY
Ova
XA
XaY
Xa
XaYA
?
XAXa
XaY
If a carrier mates with a male who has the
disorder, there is a 50 chance that each child
born to them will have the disorder, regardless
of sex. Daughters who do not have the disorder
will be carriers, where as males without the
disorder will be completely free of the recessive
allele.
(c)
Sperm
Xa
Y
Ova
XAY
XAXa
XA
XaY
XaYa
Xa
16
X-inactivation and the tortoiseshell cat
Allele for orange fur
17
Meiotic nondisjunction
Meiosis I
Nondisjunction
Meiosis II
Nondisjunction
Gametes
n 1
n ? 1
n 1
n 1
n 1
n 1
n
n
Number of chromosomes
(a)
(b)
Nondisjunction of homologous chromosomes in
meiosis I
Nondisjunction of sister chromatids in meiosis II
18
Alterations of chromosome structure
19
Down syndrome