Bio 10 (revision Version)

Question 1

101 Define crossing
over and its effect.

Answer 1

Crossing over is the exchange of DNA material
between non-sister homologous chromatids.
Crossing over produces new combinations of
alleles on the chromosomes of the haploid cells.

Question 2

10.1 Describe the
formation of
chiasmata.

Answer 2

pairing of homologous chromosomes/
synapsis;chromatids break (at same point); (do not
accept chromatids overlap)non-sister chromatids
join up/swap/exchange alleles/parts;X-shaped
structure formed / chiasmata are X-shaped
structures; chiasma formed at position where
crossing over occurred;
chiasmata formation between non-sister
chromatids can result in an exchange of alleles.

Question 3

10.1 What event
occurs prior to meiosis?

Answer 3

Chromosomes replicate in interphase before
meiosis.

Question 4

10.1 Distinguish
between metaphase 1 and 2

Answer 4

Homologous chromosomes separate in meiosis I,
while sister chromatids separate in meiosis II

Question 5

10.1 Explain
Mendel’s second
law of
independence.

Answer 5

Law of independent assortment:
Independent assortment of genes is due to the
random orientation of pairs of homologous
chromosomes in meiosis I.

*independent assortment only occurs with
unlinked gene, because linked genes that are
present on the same chromosome would be
inherited together.

Question 6

10.1 Skill: Drawing
diagrams to show
chiasmata formed
by crossing over.

Answer 6

**diagrams of chiasmata should show sister
chromatids still closely aligned, except at the
point where crossing over occurred and a
chiasma was formed.

Question 7

10.2 Distinguish
between linked and
unlinked genes.

Answer 7

Gene loci are said to be linked if on the same
chromosome.
Unlinked genes are on different chromosomes,
and segregate independently as a result of
meiosis.

Question 8

10.2 Explain the two
types of variation,
and provide
specific examples
of it.

Answer 8

Variation can be discrete or continuous.
Discrete:
e.g. flower color (it can only be red or purple)
Continuous:
e.g. human height
The phenotypes of polygenic characteristics tend to show continuous variation.
Polygenic traits such as human height may also be
influenced by environmental factors.

Question 9

10.2 Application:
Explain Morgan’s
discovery of non-
Mendelian ratios in
Drosophila

Answer 9

His breeding experiments involving fruit flies
clearly demonstrated that linked genes were not
independently assorted.
Sex Linked:
-he discovered a clear sex bias in phenotvoic
distribution in white-eyed mutants
Gene Linkage:
-certain phenotypic combinations occurred in
much lower frequencies than was to be expected
-gene linkage occurred and hence did not
independently assort
-linked alleles could be uncoupled via
recombination (crossing over) to create
alternative phenotypic combinations
**these new phenotypes would occur at a much lower frequency
Cross over Frequency:
-genes with a higher crossover frequency are
further apart
-genes with a lower crossover frequency are
closer together
-developed the first gene linkage maps that
showed the relative positions of genes on a
chromosome

Question 10

10.2 Explain how
polygenic traits can
be influenced by
environmental
factors.

Answer 10

g. human height may also be influenced by
environmental factors.
-the added effect of environmental pressures
functions to increase the variation seen for a
particular trait
-human height is controlled by multiple genes
(polygenic), resulting in a bell-shaped spectrum
of potential phenotypes
-environmental factors such as diet and health
(disease) can further influence an individual
human’s height
e.g. human skin color
-skin colour is controlled by multiple melanin
producing genes, but is also affected by factors
such as sun exposure

Question 11

10.2 Application:
Completion and
analysis of Punnett
squares for dihybrid
traits.
Skill: Calculation of
the predicted
genotypic and
phenotypic ratio of
offspring of
dihybrid crosses
involving unlinked
autosomal genes.

Answer 11

A Punnett square is a useful tool for predicting the outcomes of genetic crosses. When considering a dihybrid cross, which involves two different traits, we can use a 4x4 Punnett square to calculate the possible genotypes and phenotypes of the offspring.

To complete a Punnett square for a dihybrid cross, we first write out the genotypes of the two parents for both traits. For example, if we are considering a cross between two pea plants, one with yellow round seeds (YYRR) and one with green wrinkled seeds (yyrr), we would write:

Parent 1: YYRR
Parent 2: yyrr

Next, we write out the possible gametes that each parent can produce. For each trait, the parent can produce one of two possible alleles. Therefore, Parent 1 can produce the gametes YR, Yr, yR, and yr, while Parent 2 can produce the gametes yR, yr, Yr, and Yr.

We then fill in the Punnett square by combining the possible gametes from each parent. This gives us a grid of 16 squares, each representing a possible genotype for the offspring.

YYRR YYRr YYrR YYrr
YyRR YyRr YyrR Yyrr
YyRR YyRr YyrR Yyrr
yyRR yyRr yyrR yyrr

We can then count the number of squares that contain each genotype to calculate the predicted genotypic ratio of the offspring. In this example, there are 9 squares with the genotype YyRr, 3 squares with YYRR, 3 squares with YYRr, and 1 square each with YYrr, YyRR, YyrR, yyRR, yyrR, and yyrr. Therefore, the predicted genotypic ratio of the offspring is:

YyRr: 9/16
YYRR: 3/16
YYRr: 3/16
YYrr: 1/16
YyRR: 1/16
YyrR: 1/16
yyRR: 1/16
yyrR: 1/16
yyrr: 1/16

We can also calculate the predicted phenotypic ratio of the offspring by considering the expression of each trait. In this example, both yellow color and round shape are dominant traits. Therefore, any offspring with at least one Y allele will have yellow seeds, and any offspring with at least one R allele will have round seeds. Based on the predicted genotypic ratio, we can calculate the predicted phenotypic ratio as:

Yellow round: 9/16 + 3/16 + 3/16 + 1/16 = 16/16 = 1
Yellow wrinkled: 1/16
Green round: 1/16
Green wrinkled: 1/16

Therefore, the predicted phenotypic ratio of the offspring is:

Yellow round: 1
Yellow wrinkled: 1/16
Green round: 1/16
Green wrinkled: 1/16

By using a Punnett square to calculate the predicted genotypic and phenotypic ratios of the offspring of a dihybrid cross, we can gain a better understanding of how traits are inherited and expressed in subsequent generations

Question 12

10.2 Skill:
Identification of
recombinants in
crosses involving
two linked genes.

Answer 12

*Recombinant phenotypes can be identified by
performing a test cross
-linked genes become separated by a chiasma,
there will be an exchange of alleles between the
non-sister chromatids
creates new allele combinations that are
different to those of the parent
*the frequency of recombinant phenotypes within
a population will typically be lower than that of
non-recombinant phenotypes
-crossing over is a random process and chiasmata
do not form at the same locations with every meiotic division
-relative frequency of recombinant phenotypes
will be dependent on the distance between linked
genes
-greater when the genes are further apart on the
chromosome

Question 13

10.2 Skill: Use of a
chi-squared test on
data from dihybrid
crosses.

Answer 13

A chi-squared test is a statistical method used to determine whether the observed data fits a theoretical model or expected values. In the context of dihybrid crosses, a chi-squared test can be used to determine whether the observed ratios of offspring genotypes and phenotypes match the predicted ratios.

To perform a chi-squared test on data from a dihybrid cross, we first need to calculate the expected ratios of genotypes and phenotypes based on the known genetics of the traits being studied. This can be done using a Punnett square, as described in the previous application.

Once we have the expected ratios, we can compare them to the observed ratios by collecting and counting the actual number of offspring with each genotype or phenotype. We can then calculate the chi-squared value using the formula:

χ² = Σ ((observed - expected)² / expected)

where Σ represents the sum of the values for each category being tested. The expected values are based on the predicted ratios from the Punnett square, while the observed values are the actual counts from the experiment.

The resulting chi-squared value can then be compared to a critical value from a chi-squared distribution table, which takes into account the degrees of freedom (df) in the data. In the case of a dihybrid cross, the df is 3, since there are four possible outcomes for each trait and two traits being studied.

If the calculated chi-squared value is less than the critical value, then we can conclude that the observed data fits the expected ratios and the hypothesis of independent assortment is supported. However, if the calculated chi-squared value is greater than the critical value, then we can reject the hypothesis of independent assortment and conclude that the observed data does not fit the expected ratios.

By using a chi-squared test on data from dihybrid crosses, we can determine whether the observed ratios of offspring genotypes and phenotypes are consistent with the predicted ratios based on known genetics. This can provide important insights into how traits are inherited and expressed, and can also help to identify any deviations from expected patterns that may be due to other factors, such as genetic linkage or environmental influences

Question 14

10.2 Show how
linked genes are
represented

Answer 14

Alleles are usually shown side by side in dihybrid
crosses, for example, TtBb. In representing
crosses involving linkage, it is more common to
show them as vertical pairs.

Question 15

10.3 Define gene
pool.

Answer 15

A gene pool consists of all the genes and their
different alleles, present in an interbreeding
population.
large gene pool = high amounts of genetic
diversity
small gene pool = low amounts of genetic
diversity
Gene pools can be used to determine allele
frequency

Question 16

10.3 Explain how
evolution occurs
with reference to
alleles.

Answer 16

Evolution requires that allele frequencies change
with time in populations in ways like:
-mutations
-gene flow: the movement of alleles into, or out
of, a population as a result of immigration or
emigration
-sexual reproduction
-natural selection

Question 17

10.3 Explain what
reproductive
isolations are
provide examples

Answer 17

reproductive isolation is when barriers prevent
two populations from interbreeding (keeping their
gene pools separate)
Temporal Isolation (sympatric)
-occurs when two populations differ in their
periods of activity or reproductive cycles
-e.g. Leopard frogs and wood frogs reach sexual
maturity at different times in the spring
Behavioral Isolation (sympatric)
-occurs when two populations exhibit different
specific courtship patterns
-e.g. certain populations of crickets may be
morphologically identical but only respond to
specific mating songs
Geographic Isolation (allotropic)
-occurs when two populations occupy different
habitats or separate niches within a common
region
e.g. ions and tigers occupy different habitats and
do not interbreed (usually)

Question 18

10.3 List out and
explain the different
paces of speciation.

Answer 18

Speciation due to divergence of isolated
populations can be gradual
-allopatric and sympatric speciation results in
gradualism
-speciation is seen as a smooth and continuous
process (big changes result from many cumulative
small changes)
-supported by the fossil record of the horse, with
many intermediate forms connecting the ancestral
species to the modern equivalent
Speciation can occur abruptly.
-punctuated equilibrium suggests that species
remain stable for long periods before undergoing
abrupt and rapid change

Question 19

10.3 Application:
Identify examples
of directional,
stabilizing and
disruptive selection.

Answer 19

Stabilising
-an intermediate phenotype is favoured at the
expense of both phenotypic extremes
-e.g. human birth weights (too large = birthing
complications; too small = risk of infant mortality)
Directional
-one phenotypic extreme is selected at the cost
of the other phenotypic extreme
-e.g. development of antibiotic resistance in
bacterial populations
Disruptive
-both phenotypic extremes are favoured at the
expense of the intermediate phenotypic ranges
-e.q. the proliferation of black or white moths in
regions of sharply contrasting color extremes

Question 20

10.3 Application:
Explain speciation in
the genus Allium by
polyploidy.

Answer 20

Polyploidy is when offspring have additional sets
of chromosomes
Sympatric speciation will result if the polyploid
offspring are viable and fertile but cannot
interbreed with the original parent population.
Examples:
Triploid (3n) = - 24 chromosomes (e.g. Allium
carinatum - keeled garlic)
Tetraploid (4n) = ~ 32 chromosomes (e.g. Allium
tuberosum - chinese chives)

Question 21

10.3 Skill:
Comparison of
allele frequencies
of geographically
isolated
populations.

Answer 21

To compare the allele frequencies of geographically isolated populations, we first need to obtain genetic data from each population. This can be done by collecting samples of DNA from individuals in each population and analyzing it using techniques such as polymerase chain reaction (PCR) and gel electrophoresis.

Once we have the genetic data, we can calculate the allele frequencies for each population by counting the number of times a particular allele appears in the sample and dividing by the total number of alleles. We can then compare the allele frequencies between populations to determine whether they are significantly different.

One common method for comparing allele frequencies is the chi-squared test. This involves calculating a chi-squared value to determine whether the observed differences in allele frequencies between populations are statistically significant. The chi-squared test can be used to test the null hypothesis that the allele frequencies are the same in both populations.

Question 22

10.3 Name an
example of a
polyploid species
and explain how it
leads to speciation.

Answer 22

Allium carinatum - Triploid (3n) = ~ 24
chromosomes
non-disjunction/failure of chromosome pairs to
separate during meiosis V
diploid gamete «can lead to polyploidy»

fusion of diploid and haploid gamete produces
triploid cells V
unable to produce viable gametes during meiosis

Question 23

10.1 Define
chiasmata.

Answer 23

The connection points between non-sister
chromatids where crossing over has occurred.

Question 24

10.1 Outline the
process of crossing
over

Answer 24

-homologous chromosomes are paired up to
form bivalents in a process called synapsis.
-sister chromatids are tightly associated.
Homologous chromosomes are tightly associated.
-a breakage occurs in the same lace of adjacent
non sister chromatids.
-non-sister chromatids join each other at the
break points.
-this results in a section of the non-sister
chromatids being exchanged (producing
recombinant chromatids).
-late in prophase I, sister chromatids are still
tightly associated.
-homologous pairs start to move apart and
chiasmata become visible.
-chiasmata represent the locations where crossing
over occurred.
-non-sister chromatids remain joined at these
points and appear as cross shaped structures.
-chiasmata are lost when the homologous pairs
are separated during anaphase I

Question 25

10.1 State two
consequences of
chiasmata formation
between non-sister
chromatids.

Answer 25

Increased stability of bivalents at chiasmata
An increase genetic variability (if crossing over
occurs)

Question 26

10.1 Explain how
crossing over
between linked
genes can lead to
genetic
recombinants.

Answer 26

• (the process of crossing over results in the
  exchange of DNA between the maternal and
  paternal chromosomes.)
  -crossing over can decouple linked combinations
  of alleles and therefore lead to independent
  assortment.
  -furthermore, crossing over can occur multiple
  times and between different chromatids within the
  same homologous pair.

Question 27

10.1 Contrast meiosis
I with meiosis Il.

Answer 27

There are a number of ways in which meiosis I
differs from mitosis and meiosis I1):
-in meiosis I sister chromatids remain associated
with each other
-the homologous chromosomes behave in a
coordinated fashion in prophase
-homologous chromosomes exchange DNA
leading to genetic recombination and meiosis I is
a reduction division in that it reduces the
chromosome number by half.
-(The processes that result in the creation of
genetic variety of gametes are initiated in meiosis l.
-the segregation of homologous chromosomes
occurs during anaphase I resulting in two haploid
cells, each with only one copy of each
homologous pair.

Question 28

10.1 Describe the
experiment of
Bateson and
Punnett that lead to
results that did not
support Mendel’s
law of independent
assortment.

Answer 28

-William Bateson and Reginald Punnet conducted
crosses with sweet peas.
-one of the parent cells had long pollen (LL) and
purple flowers (PP).
-the other had round pollen (Il) and red flowers
(pp).
-the surprising result came in the F(2) generation
of a dihybrid cross.
-instead of the expected ratio of 9:3:3:1, there
were far more of the individuals with the paternal
phenotypes seen in the P generation
-and much smaller numbers of the non-parental
phenotypes, known as recombinants.

Question 29

10.1 Describe the
trends and
discrepancies that
led Morgan to
propose the idea of
linked genes.

Answer 29

Morgan observed similar discrepancies in fruit
flies as Bateson and Punnett. Morgan’s discovery
of sex linkage led him to develop a theory of
gene linkage that accounted for the high than
expected number of paternal phenotypes and the
notion of crossing over to explain the presence of
the recombinants.

Question 30

10.1 State the
difference between
independent
assortment of genes
and segregation of
alleles.

Answer 30

Segregation is the separation of the two alleles of
every gene that occurs during meiosis.
Independent assortment is the observation that
the alleles of one gene segregate independently
of the alleles of other genes.

Question 31

10.1 Describe what
makes genes
“linked.”

Answer 31

A linkage group is a group of genes whose loci
are on the same chromosome and hence don’t
independently assort
• Linked genes will tend to be inherited together
and hence don’t follow normal Mendelian
inheritance for a dihybrid cross
• Instead the phenotypic ratio will be more closely
aligned to a monohybrid cross as the two genes
are inherited as a single unit
• Linked genes may become separated via
recombination (due to crossing over during
synapsis in meiosis I)
• The alleles for these traits were located on a
shared chromosome (gene linkage) and hence did
not independently assort
• Linked alleles could be uncoupled via
recombination (crossing over) to create
alternative phenotypic combinations, but these
new phenotypes would occur at a much lower
frequency

Question 32

10.1 Contrast
discrete and
continuous variation

Answer 32

Variation of characteristics can either be discrete
or continuous.
-for discrete variation the characteristics is in
categories.
-there is no intermediate phenotype.
-with discrete variation the characteristic is
determined by a single gene.
-for continuous variation the characteristic has a
continuous range, usually a normal distribution.
-for continuous variation the characteristic
involves genes and the environment.

Question 33

10.1 Explain
polygenetic
inheritance using an
example of a two
gene cross with
codominant alleles.

Answer 33

-a polygenic characteristic is determined for by 2
or more genes.
-the genes have an additive effect.
-polygenic characteristics show continuous
variation.
-Mendel discovered an example of this in beans,
where a cross between a purple-flowered plant
and a white-flowered plant gave purple-flowered
plants in the F(1) generation
-but when these were self pollinated, the
expected 3:1 ratio did not occur and instead a
range of flower colours was seen.
-this can be explained if there are two unlinked
genes, with co-dominant alleles.
-self-pollination of the F(I) should give five
different shades of flower colour in a ratio of
1:4:6:4:1.
-if the number of unlinked genes, with co-
dominant alleles, were larger then there would be
more phenotypic variants.

Question 34

10.1 Outline the use
of Pascal’s triangle
to determine
phenotype
frequencies that
results from
polygenic crosses.

Answer 34

The number and frequency of variants can be
predicted using alternate rows of Pascal’s
triangle. As the number og genes increase, the
distribution becomes increasingly close to the
normal distribution.

Question 35

10.3 State the
formula for
calculating allele
frequencies.

Answer 35

of copies of a particular allele/total # of alleles

for that population

Question 36

10.3 State the
formula for
calculating
genotype
frequencies for
genes in a gene
pool.

Answer 36

p^2 = frequency of homozygous dominant, where
p= the allele frequency of dominant allele
q^2 = frequency of homozygous recessive, where
q = the allele frequency of recessive allele
2pq = frequency of heterozygous
***p^2 + q^2 + 2pq = 1