BS50 Exam Review 1
Chapter 2: Patterns of Inheritance (lectures 9/21, 9/23, 9/26)
Definitions:
Gene:
Unit of heredity
Allele:
Alternate forms of one type of gene
Phenotype:
Characteristics displayed by an organism
Genotype:
The genetic makeup of an organism (which alleles are present)
Discrete trait:
Traits with two or more distinct phenotypes (ex. pea shape)
Continuous trait:
Traits with a range of variation (ex. height)
Dominant allele (A): Determines phenotype even if heterozygous with another gene
Recessive allele (a): Does not contribute to the phenotype if heterozygous.
Homozygous (AA or aa): True-breeder, all offspring will look parental for the given trait.
Heterozygous (Aa): Carriers, offspring will be a mixture of phenotypes.
Wild type:
Typical genotype or phenotype found in nature.
Autosomal:
Non-sex chromosomes (chromosomes 1-22 in humans)
Sex chromosomes: Chromosomes determining the sex of an organism (X and Y in humans)
Sex linkage:
Genes located on sex chromosomes
Dosage compensation: Mechanisms which allow for equal gene expression in males and females.
X-inactivation:
Dosage compensation mechanism in humans resulting in the inactivation
of one of the two X chromosomes in human females
Barr body:
The inactivated X chromosome which appears as a densely stained mass
Karyotype:
Complete set of chromosomes of an organism
Mendel’s law of segregation (1st law):
Alleles segregate randomly. Offspring have an equal probability of inheriting either allele.
Mendel’s law of independent assortment (2nd law):
Segregation of one gene pair is independent of other gene pairs; traits on non-homologous
chromosomes are inherited independently.
What is the difference between a gene and an allele?
A gene is the region of DNA that encodes a certain trait while the allele is which
variation of the trait is encoded. For example, the gene that encodes eye color can have
different alleles for either brown or blue eyes.
What are the differences between a/a;b/b, ab/ab, a/a * b/b, and ab/ba?
a/a;b/b indicates that the genes are on separate chromosomes. ab/ab designates that the
genes are on the same chromosome and a/a * b/b means that it is not known if the genes
are linked. ab/ba signifies that the order of these genes has been flipped for one of the
chromosomes.
How do you know something is genetic?
Perform crosses and make predictions.
Punnett squares:
Monohybrid Cross (heterozygote for 1 gene)
Dihybrid Cross (heterozygote for 2 genes)
P: parental generation
F1: first filial generation, offspring of the parents
F2: second filial generation, the offspring of F1 siblings
Note: Chance of each box occurring is equal
However, you can take short cuts when determining ratios by applying Mendel’s laws.
From the cross
A/a ; b/b ; C/c ; D/d ; E/e x A/a ; B/b ; C/c ; d/d ; E/e
What is the likelihood that their first offspring will be a/a ; b/b ; c/c ; d/d ; e/e?
For A/a x A/a, 1/4th of the progeny will be a/a.
For b/b x B/b, ½ of the progeny will be b/b. etc …
You can multiply the probabilities to find the final percentage: ¼ x ½ x ¼ x ½ x ¼ =
1/256.
For a diploid organism, when can you not determine the genotype from phenotypic information?
If the phenotype represents the dominant allele then the second allele could be either
dominant or recessive (A/- could be A/A or A/a)
How would you determine the genotype in the above situation?
Perform a test cross with a homozygous recessive organism and observe ratios of the
progeny.
For example:
R/- ; Y/- x
r/r ; y/y
1 R/r ; Y/y : 1 r/r ; Y/y
All progeny will be yellow so you can infer that the parental generation was Y/Y.
Half of the progeny are wrinkled so you can infer that the parental generation was R/r.
Family Pedigrees (for determining human genetics):
Define the following symbols:
Female unaffected
Male affected
Female affected
Male Unaffected
Siblings
Parents
Sex unknown
Consanguineous mating
I
II
III
IV
A
What can you say about the inheritance of this trait (dominant/recessive, sex-linked/autosomal)?
Autosomal recessive. Recessive because there is an affected offspring without an
affected parent. It is autosomal because an X-linked trait cannot be passed to a daughter
unless the father is affected (remember, a father cannot be a carrier for an X-linked trait
since he has only one X) and it is not Y-linked since a female is affected.
What are the genotypes of A’s parents?
Since they have an affected offspring, they both must be heterozygous.
What is the probability that A will have an affected sibling?
We know that each parent is a carrier and has a ½ chance of passing on the affected allele
so ½ * ½ = ¼. Remember that the genotype of each offspring is an independent event
and so, even if 5 siblings are affected, the chance of an affected 6th sibling is still 1/4th.
This is a pedigree for a rare trait:
What can you say about the inheritance of this trait?
Sex-linked (on the X) recessive. Recessive because there is an affected offspring without
an affected parent. The gene cannot be on the Y since there are affected males without an
affected father. The key to this problem is that this is a rare trait so you can assume that
people marrying into the family are not carriers. Since males marrying into the family
are wild type homozygous, the females must be carriers. The only way that the offspring
would be affected would be if they inherited one affected allele from the mother and
nothing from the father (i.e. an X from mom and a Y from dad).
What are the most likely genotypes of:
A: XA/XA, this is a rare mutation
B: Xa/Y, affected
C: XA/Xa, she has an affected offspring but is not affected herself (she is a carrier). Also, her
father must have passed Xa to her so she couldn’t be homozygous dominant.
D: Xa/Y, affected
What is one way to determine sex linkage?
Different results based on whether mom or Dad has the trait (reciprocal crosses).
Vs.
Affected gene is recessive on the X
b. For their child to have PKU, both A and B must be carriers and both must donate the recessive allele. The
probability that individual A has the PKU allele is derived from individual II-2. II-2 must be P/p since her
father must be p/p. Therefore, the probability that II-2 passed the PKU allele to individual III-2 is 1/2. If
III-2 received the allele, the probability that he passed it to individual IV-1 (A) is 1/2. Therefore, the
probability that A is a carrier is 1/2 x 1/2 = 1/4.
The probability that individual B has the allele goes back to the mating of II-3 and II-4, both of whom are
heterozygous. Their child, III-3, has a 2/3 probability of having received the PKU allele and a probability
of 1/2 of passing it to IV-2 (B). Therefore, the probability that B has the PKU allele is 2/3 x 1/2 = 1/3.
If both parents are heterozygous, they have a 1/4 chance of both passing the p allele to their child.
p(child has PKU) = p(A is P/p) x p(B is P/p) x p(both parents donate p)
1/4
1/4
x 1/3
x
= 1/48
c.
If the first child is normal, no additional information has been gained and the probability that the second child
will have PKU is the same as the probability that the first child will have PKU, or 1/48.
d.
If the first child has PKU, both parents are heterozygous. The probability of having an affected child is now
1/4, and the probability of having an unaffected child is 3/4.
Chapter 3: The Chromosomal Basis of Inheritance (lectures 9/28,
10/3)
Definitions:
Haploid:
one set of genetic information (1n)
Diploid:
two sets of genetic information (2n)
Centromere: region on the chromosome where spindle fibers attach
Non-disjunction: Chromosomes fail to separate in meiosis. Results in a monosomic (2n-1) cell.
Aneuploid:
Cell with different chromosome number than normal (can be more or less).
Chromatin: material that makes up chromosomes, composed of protein and DNA; can be
either euchromatin (loosely packed, contains most of the active genes) or
heterochromatin (tightly packed)
Sister Chromatids: Pair of identical chromosomes resulting from DNA replication and held
together with one centromere until anaphase
Homologous chromosomes: Pair of similar chromosomes, one from each parent
Synapsis:
Pairing of homologous chromosomes, held together by a synaptonemal complex.
Results in a tetrad (4 chromatids) and can also be known as a bivalent (2
chromosomes)
Chiasma:
The point of connection between non-sister chromatids as a result of crossing
over.
Mitosis
1 division
2n
2n
Produces two identical daughter cells
No recombination
Body cells (asexual)
Meiosis
2 divisions
2n
1n
Produces four haploid non-identical cells
Recombination
Gametes
Note: Number of chromosomes determined by number of centromeres
Key events to remember:
Reductional division (diploid to haploid): Anaphase I
Synapsis/recombination: Prophase I
Mendel’s 1st law: Anaphase I
Mendel’s 2nd law: Metaphase I
Number of chromosomes doubles: Anaphase Mitosis, Anaphase II
Chromosomes replicate: Interphase
Note: the majority of a cell’s life is spent in interphase
40. (from textbook)
The plant Haplopappus gracilis is diploid and 2n=4. There are one long pair and one short pair
of chromosomes. The accompanying diagrams represent anaphases of individual cells in meiosis
or mitosis in a plant that is genetically a dihybrid (A/a ; B/b) for genes on different chromosomes
or chromatids, and the points of the V’s represent centromeres. In each case, indicate if the
diagram represents a cell in meiosis I, meiosis II, or mitosis. If a diagram shows an impossible
situation, say so.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
Impossible: the alleles of the same genes
are on nonhomologous chromosomes
Meiosis II
Meiosis II
Meiosis II
Mitosis
Impossible: appears to be mitotic anaphase
but alleles of sister chromatids are not
identical
Impossible: too many chromosomes
Impossible: too many chromosomes
Impossible: too many chromosomes
Meiosis I
Impossible: appears to be meiosis of
homozygous
Impossible: the alleles of the same genes
are on nonhomologous chromosomes