Download SBI 3U – Genetic Continuity

SBI 3U – Genetic Continuity
Name: _______________________
Test Review
Test – Thursday, April 23rd
Topics Covered:
 Cell cycle
 Cancer
 Cell division: Mitosis, Meiosis
 Formation of gametes
 Non-disjunction disorders
 Mendel and his Pea Plants
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Dominant and Recessive traits
Punnett squares: mono- and dihybrid crosses
Multiple alleles
Incomplete dominance, Codominance
Terms you should know: (You should know these terms)
 Mitosis
 Polar bodies
 Meiosis
 Chromosomes
 Interphase
 Chromatin
 Prophase
 Sister chromatids
 Metaphase
 Homologous chromosomes
 Anaphase
 Tetrad
 Telophase
 Centromere
 Cytokinesis
 Spindle fibers
 Centriole (centrosome)
 Trisomy
 Crossing over
 Down syndrome
 Somatic cell
 Trait
 Gametes
 Allele
 Diploid
 Genotype
 Haploid
 Fertilization
 Zygote
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Phenotype
P generation
F1 generation
F2 generation
Dominant allele
Recessive allele
Incomplete dominance
Codominance
Pure-bred
Hybrid
1. Indicate which process the following cells use to reproduce:
a) Brain cells
Mitosis
b) Fat cells
Mitosis
c) cells of a zygote
Mitosis
d) sperm-producing cells of the testes
Meiosis
2. The figure below shows a cell with 36 chromosomes undergoing meiosis.
a) How many chromosomes would be in each
cell during stage B?
18
b) How many chromosomes would be in each
cell during stage c?
9
c) In which stage(s) would you find a cell with
a diploid chromosome number?
B
d) In which stage(s) would you find a cell with
a haploid chromosome number?
C
3. The figure below shows plant and animal cells during cell division. Identify each cell as either
a plant or an animal cell. Identify the phases of cell division.
a) Plant , Metaphase
b) Plant , Telophase
c) Animal , Anaphase
d) Animal , Cytokinesis
4. Suggest reasons why skin cells, blood cells and the cells that line the digestive tract
reproduce more often than other types of cells such as muscle cells.
The more specialized a cell is the less able it is to undergo cell division. Skin cells and the cells
that line the digestive tract are not as specialized as muscle cells.
5. Provide a hypothesis that explains why the skin is so susceptible to cancer.
The skin is exposed to sun’s radiation. Radiation can cause damage to the cell’s DNA.
6. Compare and contrast the stage of metaphase for mitosis and metaphase I and II. Draw
diagrams to illustrate your answer.
In metaphase for mitosis sister chromatids line up at the equator. In metaphase I tetrads (2
pairs of homologous chromosomes) line up at the equator. In metaphase II homologous pairs
line up along the equator.
7. A microscopic water animal called daphnia can be reproduced because male gametes are not
required. Indicate the sex of the offspring produced. Explain your answer.
The daphnia has to be female because if there are no male gametes, then there would be no Y
chromosome.
8. For Labrador retrievers, black fur (B) colour is dominant over yellow (b).
a) What would be the genotype of a homozygous black dog? A heterozygous black dog?
b) Could the heterozygous black dog have the same genotype as a dog with yellow fur?
Explain.
a) BB (homolozygous black dog) and Bb (heterozygous black dog)
b) No because to have a recessive trait the dog would have to be homozygous (bb)
9. For each of the following situations, create a Punnett square using the Mendel Pea Plant
handout or look on page 138 of your text. Determine the following information:
- parent phenotype
- parent genotype
- parent gametes
- F1 generation phenotypes
- F1 generation genotypes
a) Two heterozygous tall parents are crossed.
Parent: Tt (Tall) x Tt (Tall)
F1 : TT (Tall) Tt (Tall) tt (short)
b) A heterozygous tall plan is crossed with a dwarf plant.
Parent: Tt (Tall) x tt (short)
F1: Tt (Tall) tt (short)
c) Two plants that are heterozygous for purple flowers are crossed.
Parent: Pp (Purple) x Pp (Purple)
F1: PP (Purple) Pp (Purple) pp (yellow)
d) A plant that is homozygous for green pods is crossed with a plant that has yellow pods.
Parent: GG (Green) x gg (yellow)
F1: Gg (Green)
e) A plant that is homozygous for round seeds is crossed with a plant that is heterozygous
for round seeds.
Parent: RR (Round) x Rr (Round)
F1: RR (Round) Rr (Round)
10. In guinea pigs, the allele for a black coat (B) is dominant over the allele for a white coat (b),
and short hair length (H) is dominant over long (h). A black guinea pig was crossed with a
white guinea pig. All F1 offspring have black coats.
a) Describe how you can determine whether or not the black parent is homozygous or
heterozygous for the condition. Indicate the letter you will use to represent an allele.
B – black
b – white
If the offspring were ALL black, then the black guinea pig had to be homozygous to give
a B to all offspring. If it were heterozygous then at least one would be white.
b) If 10 offspring were produced, indicate how many you would expect to have black coat
colour, if the black parent were heterozygous.
If they had 4 offspring then 3 of them would be black (either BB or Bb) so if they had 10
I would expect at least 7 to have black coats.
Indicated the genotypes and phenotypes that would result from the following crosses:
c) A guinea pig that is homozygous for black and heterozygous for short hair is crossed with
a white, long-haired guinea pig.
Parents: BBSs(black and short hair) x bbss(white and long hair)
F1: BbSs (Black short hair)
Bbss (Black long hair)
d) A guinea pig that is heterozygous for black and for short hair is crossed with a white,
long-haired guinea pig
Parents: BbSs (Black and short hair) x bbss (white and long hair)
F1: BbSs (black short hair) Bbss (black long hair) bbSs (white short hair) bbss (white long hair)
e) A guinea pig that is homozygous for black and for long hair is crossed with a guinea pig
that is heterozygous for black and for short hair.
Parents: BBss (Black long hair) x BbSs (Black short hair)
F1: BBSs (Black short hair) BBss (Black long hair) BbSs (Black short hair) Bbss (Black long hair)
11. In fruit flies, the Red allele (E1) for eye colour is dominant to all others. The Apricot allele
(E2) is recessive to E1 and dominant to the Honey (E3) and White (E4) alleles. Honey is
recessive to E1 and E2 and dominant to the white allele (E4), which is recessive to all others.
Two fruit flies are crossed, and they have 137 apricot-eyed, 65 honey-eyed and 72 whiteeyed offspring. Determine the phenotypes and genotypes of the parents.
Possible genotypes of offspring
E2
E4
Apricot (E2E2 or E2E3 or E2E4)
E2E3 E2E4 E3E4
E4E4
3
2 3
3 4
E
E E E E
Honey (E3E3 E3E4)
Apricot
Honey White
4
2 4
4 4
E
E
E
E
E
4 4
White (E E )
12. Suppose you have two rose plants, both with pink flowers. You cross the two plants and are
surprised to find that, while most of the offspring are pink, some are red and some are white.
You decide that you like the red flowers and would like to make more. What cross would you
perform to produce the most red-flowered plants?
Must be incomplete dominance because pink is a blending of red and white.
Cross either the red and red offspring or the red and pink offspring
13. Explain how it is possible to produce a trisomic XXX female.
During anaphase of meiosis the sex chromosomes don’t separate cause one daughter cell to have
2 X chromosomes and the other to have none. When that egg is fertilized with a sperm
containing an X chromosomes the three X chromosomes pair up to form a tri.