Objective: To review the phases of mitosis with the understanding that in a real tissue or a non-synchronized cell culture, different mitotic phases and cells in interphase will be observed at a certain time. Interphase is the period between mitotic events; all other phases are part of mitosis and their individual probabilities add up to the total probability of seeing cells undergoing mitosis. Pay attention to the individual probabilities for cells in each phase of the cell cycle. Problem N° 10 For a certain type of animal tissue, the probability of a randomly selected cell being in a particular stage of the interphase/mitosis cycle is given by the following table: Stage Probability Interphase 0.5 Prophase 0.1 Metaphase 0.05 Anaphase 0.2 Telophase 0.15 Suppose you examine 100 cells at random from this tissue. A) How many cells do you expect to see in metaphase? _____ cells B) You have selected a high magnification (2000 X) in your microscope and you see only one cell at a time (50 µm in diameter). How frequently you would see a cell undergoing mitosis? _____ % C) What is the probability of seeing no metaphases in 100 cells? _____ %
Objective: To review the phases of mitosis with the understanding that in a real tissue or a non-synchronized cell culture, different mitotic phases and cells in interphase will be observed at a certain time.
Interphase is the period between mitotic events; all other phases are part of mitosis and their individual probabilities add up to the total probability of seeing cells undergoing mitosis.
Pay attention to the individual probabilities for cells in each phase of the cell cycle.
Problem N° 10
For a certain type of animal tissue, the probability of a randomly selected cell being in a particular stage of the interphase/mitosis cycle is given by the following table:
Stage |
Probability |
Interphase |
0.5 |
Prophase |
0.1 |
Metaphase |
0.05 |
Anaphase |
0.2 |
Telophase |
0.15 |
Suppose you examine 100 cells at random from this tissue.
- A) How many cells do you expect to see in metaphase? _____ cells
- B) You have selected a high magnification (2000 X) in your microscope and you see only one cell at a time (50 µm in diameter). How frequently you would see a cell undergoing mitosis? _____ %
- C) What is the probability of seeing no metaphases in 100 cells? _____ %
![Background and/or Instructions
Problem N° 9
Objective: To apply the concept of
codominance in the prediction of horse coat
color.
In codominance, both alleles are expressed;
both gene products are measurable or
visible as an additional phenotype that is
neither dominant nor recessive nor
intermediate. In incomplete (partial)
dominance an intermediate trait is
Cremello:
Palomino:
Chestnut:
produced.
almost white
golden with lighter mane and tail
reddish brown
Coat color of horses is a very complex trait.
The following table gives ratios obtained in matings of the above horse (Equus caballus) varieties:
The colors cremello, chestnut and palomino
can be explained using a codominant model
Cross
Parents
Offspring
of inheritance and are known to be
cremello x cremello
all cremello
determined by alleles at a single locus in
horses.
II
chestnut x chestnut
all chestnut
For this problem, use the genotypes:
cremello x chestnut
all palomino
II
C'C' = cremello
IV
palomino x palomino
% chestnut
C?C? = chestnut
½ palomino
% cremello
C'c? = palomino.
If cross IV (palomino X palomino) is represented by C'C? x c'c², what are the genotypic proportions of
the offspring?
а.
Locus = is the physical position of an allele
on a chromosome (plural loci). As the
position of a gene (or allelic form of one),
the concept of locus becomes important
when mapping chromosomes, i.e., locating
the positions of loci.
Cross III (cremello X chestnut) can be represented by C'C' x C?c?, what is the genotype of the palomino
offspring?
b.
Although mass sequencing projects may
c. Why do crosses I (cremello X cremello) and II (chesnut X chesnut) produce all cremello and all chestnut,
respectively, if cremello X chestnut produces all palomino?
have obviated the need for chromosome
mapping, the association (linkage) between
or among alleles led to important
discoveries and applications in genetics.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F86231d0e-5c74-4bd2-97e4-6f1227ee2b52%2F15597676-805c-47b1-91d7-59eb65a24c04%2Fslwmus_processed.jpeg&w=3840&q=75)
![Background
Problem N° 8
Objective: To understand the role of
probabilities in the transmission of traits in
8.1. An individual of genotype AA Bb Cc DD Ee is test-crossed. Assuming that the loci undergo independent
assortment, what fraction of the progeny are expected to have the genotype Aa Bb Cc Dd Ee?
living organisms.
This problem is about probabilities in
classical geneticsS
You may express your results as fractions,
decimals, or, for large populations, %.
When necessary, always write a zero before
the decimal point. It increases clarity and
8.2. Found on chromosome 4 in humans, two blood group alleles, LM and LN, are equally expressed in
heterozygous individuals. Such heterozygous individuals are said to be in the MN blood group. Assume that a
man and woman, both in the MN blood group, have a child.
precision.
a. What is the probability that this child will be a girl and be in the MN blood group?
For part 8.3., think of the family as a
formula that goes like this:
Family = Parents + Sibship
b. What is the probability that the next two children born to this couple will be in the MM blood group?
.. Sibship = Family - Parents
8.3. Red-green colorblindness in humans is caused by a recessive gene on the X chromosome. A husband and
wife both have normal vision, although both of their fathers are red-green colorblind. What is the probability
that their first child will be:
а.
a son with normal vision?
b. a daughter with normal vision?
с.
a colorblind son?
d.
a colorblind daughter?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F86231d0e-5c74-4bd2-97e4-6f1227ee2b52%2F15597676-805c-47b1-91d7-59eb65a24c04%2Fwe36e6_processed.jpeg&w=3840&q=75)
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