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BIOL 202 Study Guide - Midterm Guide: Shorthorn, Genotype Frequency, Forward Genetics


Department
Biology (Sci)
Course Code
BIOL 202
Professor
David Hipfner
Study Guide
Midterm

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Midterm __ Biology 202 (2010)
Dr. Schoen’s section (answers in bold—please see notes below each question for
more detailed explanation, when needed):
(1.) (1 point) “Forward genetics” is a term used to refer to the approach:
(a) In which mutations are induced in an organism’s genome and the investigator
attempts to determine how mutational changes in a particular portion of the
sequence influences the organism’s phenotype.
(b) In which the organism’s external environment is modified and the investigator
attempts to determine how this modification influences the organism’s genotype.
(c) In which crosses are conducted between organisms that have mutant and
wild type phenotypes, and the inheritance patterns of progeny are studied,
eventually leading to the identification of the gene(s) responsible for the
mutant phenotype.
(d) In which transcription of a portion of an organism’s DNA sequence is inhibited
and the investigator attempts to determine how this influences the organism’s
phenotype.
(e) In which the organism’s development is modified and the investigator attempts to
determine how this modification influences the organism’s genotype.
(2.) (1 point) In shorthorn cattle, the genotype R/R causes a red coat, R/r causes a roan
coat, and r/r causes a white coat. A breeder has all colors of cattle and makes the
following crosses:
Cross 1: Red x Red
Cross 2: Red x Roan
Cross 3: Red x White
Cross 4: Roan x White
What proportions of progeny are expected in the next generation from each cross?
(a) all Red (Cross 1); all Red (Cross 2); ½ Roan and ½ White (Cross 3);
all Roan (Cross 4)
(b) all Red (Cross 1); ½ Red and ½ Roan (Cross 2); all Roan (Cross 3);
½ Roan and ½ White (Cross 4)
(c) all Red (Cross 1); all Red (Cross 2); all Roan (Cross 3);
½ Roan and ½ White (Cross 4)
(d) all Red (Cross 1); ½ Red and ½ Roan (Cross 2); all Roan (Cross 3);
all Roan (Cross 4)

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(e) None of the above.
(3.) (1 point) Homologous chromosomes pair, synapse, and separate during:
(a) Mitosis
(b) Meiosis II
(c) Meiosis I
(d) Mitosis and Meiosis I
(e) Mitosis and Meiosis II
(4.) (1 points) Consider the pedigree below in which the a allele, responsible for the
expression of a rare trait (filled in symbols), is recessive to the A allele. Which
of the following assignments of genotypes to individuals most likely? (Note: Roman
numeral I refers the parents and II to the progeny. Individuals within generations are
numbered left to right in increasing order).
(a.) I-1: A/a, I-2: a/a and II-1: A/a, II-2: a/a, II-3: A/a, II-4: a/a, II-5: a/a
(b.) I-1: a/a, I-2: A/a and II-1: A/a, II-2: a/a, II-3: A/a, II-4: a/a, II-5: a/a
(c.) I-1: A/A, I-2: A/a and II-1: A/a, II-2: A/A II-3: A/a, II-4: A/A, II-5: A/a
(d.) I-1: XA/Xa, I-2: Xa/Y and II-1: XA/ Xa, II-2: Xa/Y, II-3: XA/ Xa, II-4: Xa/Y, II-5:
Xa/Xa
(e.) None of the above.
Note: Several students wrote that answers (a) and (d) are equally likely. However this
is not the case. Since the trait is rare in the population, then the underlying mutation
must also be rare. For example, suppose that the mutation has a frequency of 0.01 in
the population. Under scenario in answer (a), the probability of sampling an A/a
individual is (from Hardy-Weinberg theory) 2 x 0.99 x 0.01=0.0198, while the
probability of sampling an a/a individual is 0.012 =0.0001. So the probability of
sampling a family in which one parent is A/a and one is a/a is the product of these two
quantities, or 0.00000198.
Under scenario in answer (d), the probability of sampling an XA/Xa mother is
2 x 0.99 x 0.01=0.0198, and the probability of sampling an Xa/Y father is 0.01. Again
using the product rule, the probability of sampling a family with this set of parents is
0.000198, or 100 times more likely than scenario (a).
This question is challenging and tests your ability to put together population genetic
reasoning with pedigree analysis. The idea is that we are searching for the answer that

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best fits what we know about the trait (it is recessive and rare), and using quantitative
reasoning to reach the best solution.
I have accepted answer (a), even though it is not the best answer, because some
students were confused as to ho to minimize the introduction of rare alleles into the
pedigree.
(5.) (1 point) Mitosis and meiosis differ in regard to the presence of:
(a) Chromatids
(b) Mutation
(c) Synapsed homologs
(d) Centromeres
(e) Spindles
(6.) (2 points) Five separate nutritional mutants in Neurospora were independently
isolated. They all require compound F to grow. Intermediate compounds in the
biosynthesis of compound F are known and tested for their ability to support the growth
of each mutant. The results are given in the table below, where (+) indicates growth and
(0) indicates no growth. Assuming a linear pathway, what is the order of the compounds
A through F?
Compounds
Mutants A B C D E F
––––––––––––––––––––––––––––––––––
1 0 0 0 + 0 +
2 0 + 0 + 0 +
3 0 0 0 0 0 +
4 0 + + + 0 +
5 + + + + 0 +
a) EABCDF
b) ECBDAF
c) ECDBAF
d) EACBDF
e) EBACDF
Note: See lecture on Beadle and Tatum’s experiments with Neurospora.
(7.) (2 points) Below is a pedigree for a family with a recessive sex-linked trait. Suppose
we know that the allele for this trait enters the pedigree in the generation of individual A,
and once more in the generation of individual B. What is the probability that the child of
individuals C and D would exhibit the trait?
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