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UNIT AND EXAM REVIEW BOOKLET

BIOLOGY 40S: 2015

NAME: KEY 2015

Biology 40S: UNIT and EXAM REVIEW BOOKLET KEY pg 2

UNIT 1 SCORE:

UNIT 2 SCORE:

UNIT 3 SCORE:

/ 138 MARKS

/110 MARKS

/98 MARKS


UNIT 1: MENDELIAN GENETICS AND HUMAN INHERITANCE

1. FILL IN THE BLANKS: Vocabulary Review ( 5 marks)

Heterozygous Recessive Dominant Test Monohybrid Genotype

1. PUREBRED

Cross pollinating Purebred Cross Homozygous recessive

Phenotype

Before starting an experiment, Mendel allowed each kind of plant to self pollinate to obtain plants.

When Mendel transferred pollen from one pea plant to another, he was the plants.

2. CROSS POLLINATING

3. MONOHYBRM The name of a cross involving only one trait.

4. RECESSIVE A trait that is hidden in the heterozygous condition is said to be a trait.

5. HETEROZYGOUS An organism that has two different alleles for a trait is called .

6. HOMO RECESSIVE A short pea plant contains both genes for a trait that are masked. The trait is considered to be .

7. GENOTYPE The genetic make-up of an organism.

8. PHENOTYPE The physical expression of genes.

9. DOMINANT The rule of dominance states that the trait will hide or mask the expression of other traits.

10. TEST CROSS To determine the unknown genotype of an organism, it can be crossed with an organism that has a recessive genotype and then the offspring can be analyzed. This method of determining genotype is called a .

2. Explain how two organisms can have the same phenotype, but different genotypes. (lmk)

SAME PHENOTYPE: Both have the same dominant physical trait

DIFFERENT GENOTYPE: One organism can be homo dominant BB Other organism can be heterozygous Bb

Biology 40S: UNIT and EXAM REVIEW BOOKLET KEY pg 4 3. a) Explain how gametes from the parents of a dihybrid trait is determined. (1 ink)

FOIL METHOD

b) Determine all the possible gametes that occur, if the parent genotype is AaBb. (1 ink)

AB, Ab, aB, ab

c) Using the product rule, determine the different gametes that can form from a parent with the genotype of TtRRYy. State the different gametes. (lmk)

Product Rule X 1 X % = 4 4 different gametes

Different gametes 4 TRY, TRy, tRY, tRy

4. If two heterozygous organisms for a s • t mate, the genotypic ratio of their offspring should be? Choose one: 1:1 / 1:2 / 3:1 /1:3:1 / 1

your answer. (2 mk)

P1 = Aa X Aa

GR = 1AA: 2Aa:laa

X A a A AA Aa a Aa aa

5. A dihybrid cross b een two heterozygous parents ALWAYS produces a phenotypic ratio of? Choose one: 3:

P1= AaDd X AaDd

PR = 9: 3: 3: 1

:2:1 / 1:6:9) (link)

6. In peas, an autosomal gene for round seeds is dominant over a gene for wrinkled seed. In a cross between a heterozygous round seed with a wrinkled seed, what are the expected phenotypic and genotypic ratios of their offspring? Show your work, including legend, Pl, and Fl. (5mks)

Legend: Round = R Wrinkled = r

P1 = Rr X rr

X R ntf.....,, Rr r

r r

GR = 1 Rr : 1 rr

PR = 1 round: 1 wrinkled


7. In garden peas, the allele for yellow peas is dominant to the allele for green peas. Suppose you have a plant that produces yellow peas, but you don't know whether it is homozygous dominant or heterozygous. What experiment could you do to find out? Explain your answer. With punnett squares. (2mks)

• Preform a test cross

• Cross the unknown genotype with a homo recessive individual

• Analyze the offspring:

• Hetero parent 4 50% dominant offspring and 50% recessive offspring

• Homo dominant parent 4 100% dominant offspring

8. Yellow fruit and dwarf vines are rec- : • atoes. Red fruit and tall vines are dominant traits. Illustrate the cross between t • hybrid plants for • • h traits. What is the probability of obtaining offspring that produce yellow fruit and have tall vines? Show all work. State legend, Pl, Fl, gametes, GR and PR. State the probability of getting yellow tall tomatoes. (7 mks)

Legend: Red = R Yellow = r Tall = T Dwarf = t

P1= RrTt X RrTt

Gametes: RT, Rt, rT, rt

RT Rt rT rt RT RRTT RRTt RrTT RrTt Rt RRTt RRtt RrTt Rrtt rT RrTT RrTt rrTT rrTt rt RrTt Rrtt rrTt rrtt

Genotypic Ratio: 1RRTT : 2RRTt : 2RrTT : 4RrTt: lrrTT : 2rrTt : 2Rrtt: 1RRtt : lrrtt

Phenotypic Ratio: 9 red tall : 3 red short : 3yellow tall: lyellow short Prob 3/16 = 18.75%


9. Match Mendel's THREE laws: Law of Dominance, Law of Segregation, and Law of Independent assortment with the correct definitions. (3mks)

a) LAW OF SEGREGATION

The separation of the pair of parental chromosomes (homologous chromosomes), so that one chromosome from the homologous pair is present in each gamete.

b) LAW OF DOMINANCE The dominant trait will be expressed in the heterozygous form and will mask or hide the recessive form.

c) INDEPENDENT ASSORTMENT Any one pair of characteristics (genes) may combine with either one of another pair.(Seen in dihybrid inheritance)

d) LAW OF SEGREGATION During fertilization, gametes randomly pair to produce four combinations of alleles.

e) INDEPENDENT ASSORTMENT Genes of different traits are inherited separately of each other.

f) LAW OF DOMINANCE Cross a dominant trait with a recessive trait and all offspring show only the dominant trait.

10. Suppose that a child with free earlobes has a mother with attached earlobes. Can a man with attached earlobes be the father? Explain. Free earlobes are dominant to attached earlobes.

Include legend and Pl. (2mks)

Legend: Mother passes her recessive allele (f) to child Free = F Father MUST pass a dominant allele (F) to child, but the father Attached = F only has recessive alleles present

CHILD = Ff Mother = ff Man = ff?????

X f f f ff ff f ff ff

Therefore this man CAN NOT be the father of the child

FEMALE A X MALE A FEMALE B X MALE A

X B b b Bb bb b Bb bb

X B B b Bb Bb b Bb Bb


11. Long second toe and double-jointedness are dominant to short second toe and normal joints Determine the genotype of all family members. The father has a long second toe and normal joints, the mother has long second toe and double-jointedness. Their son has short second toe, and double-jointedness, and their daughter has long second toe and normal joints. Solve this question without the help of a punnett square. (4mks)

Legend: Long = T Short = t

Double = D Normal = d

TOE LENGTH (Tit) JOINTS (Did) FATHER Tt dd MOTHER Tt Dd SON tt Dd DAUGHTER T? dd

12. Two black female guinea pigs are crossed with the same male guinea pig. Based on the information shown in the chart below, answer the following questions. Black is dominant over white.

1'1 GENERATION Fl GENERATION

Female A x Male A 8 black and 6 white Female B x Male A 13 black and 0 white

A) Using the offspring ratios, determine the genotypes of each parent? (3mks)

FEMALE A X MALE A RATIO Black = 8/14 = 57% White = 6/14 = 43% Ratio 1 : 1 (HETERO X HOMO REC)

FEMALE B X MALE A RATIO Black = 13/13 = 100% White = 0/13 = 0% Ratio 1: 0 (HONIO DOM X HOMO REC)

Legend: Black = B White = b

GENOTYPE of FEMALE A x MALE A

GENOTYPE of FEMALE B x MALE A

Bb X bb

BB X bb

B) Show Work using a punnett square for each cross. (2mks)

Legend: Red = C'C' White= CC' Roan = CRCw

X CR CW

CR cRcR cRcW

CW CRC" C 'CW

Cross roan x roan


13. Fill in the blanks: (2 mks)

Polygenic Incomplete dominance Codominance Multiple alleles

A. Incomplete dominance In flowers, pink offspring are produced by crossing a plant bearing white flowers with a plant bearing red flowers. This principle is called .

B. Co-dominance In cattle, red coat color and white coat color are equally expressed, where neither one masks the expression of the other. This principle is called .

C. Polygenic inheritance is the inheritance pattern of a trait that is controlled by two or more genes.

D. Multiple Alleles The ABO blood group is an example of a single gene that has in humans.

14. a) A cross between a purebred animal with red hair and a pure bred animal with white hair produces an animal that has both red and white hair (roan). What type of inheritance is this? Why? (link)

• Inheritance pattern is co-dominant • Both genes equally express in heterozygous form

b) What would a farmer bred, if he wanted three different phenotype to appear in the offspring. (Red, white and roan)? Prove your answer. (2mks)

P1 = CI C' X CRCw

PR = 1 RED : 2 ROAN: 1 WHITE


15. In winter squash, white fruit is dominant over yellow fruit and elongated shaped vegetables are incompletely dominant over sphere shaped vegetables, and bell-shaped vegetables are produced.

A plant which heterozygous for white fruit and sphere shaped is cross-pollinated with one which has yellow fruit colour and an elongated shape. State the legend, Pl, gametes, show the Fl, and give the genotypic and phenotypic ratios of the Fl generation. What is the probability of having white bell shaped squash? (7 mks)

Legend: White = W yellow= w

Elongated = EE Sphere = E'E' Bell = EE'

Pl= Ww E'E' X wwEE

Gametes: WE 2. E wE wE

WE'

wE' wE

WwEE'

wwEE'

Genotypic Ratio: 1 WwEE' : 1 wwEE'

Phenotypic Ratio: 1 white bell shaped: 1 yellow bell shaped

Prob = 50% white bell shaped


16. A man is accused of fathering two children; one with type 0 blood, and another with type A blood. The mother of the children has type B blood. The man has type AB blood. Could he be the father of both children? Explain. (2mks)

Mother = BO

Child 1= 00 Must inherit 0 ALLELE FROM EACH PARENT

Child2= AO Must inherit 0 ALLELE FROM MOTHER and A ALLELE FROM FATHER

Man =AB • DOES NOT HAVE AN 0 ALLELE TO PASS TO CHILD 1(00 BLOOD),

THEREFORE CAN NOT BE THE FATHER OF CHILD 1

• HAS A ALLELE TO PASS TO CHILD 2 (AO BLOOD), THEREFORE COULD BE THE FATHER OF CHILD2

17. A mother with blood type IBi and a father with blood type IA IB have children. Using a punnett square show the possible genotype and phenotypes of their children. (4mks)

X IB i 1A IALB 1A1 113 'BIB pi

GR =1IAIB : F'i :l 1B1

PR =1AB blood group :2B blood group:1A blood group

Legend: BO = IBi AB =

P1 = Ii X IA18

Legend: Woman = BO = Man = AA =

X IB i IA 1113 1Ai

14 IAIB pi

P1 = IAi X I


18. Is it possible for a woman who is blood type B and whose father is blood type 0 and a man with type A blood whose father was type AB and mother was type AA, to have a blood type 0 child? Include a legend, Pl, Fl and the phenotypic and genotypic ratios. (6 mks)

GR = 1 IAIB 1 pp 1 ri

PR = 1 AB blood group: 1 A blood group: 0 0 blood group

Not possible to have 0 blood type child (00) as Man does not have 0 allele to pass to child.


19. In sweet peas, purple flower color (F) is dominant over pink (f), but there is also a control gene such that if the plant has a dominant gene "G", the purple has "permission" to express itself. If the plant is homozygous recessive "gg," the purple "does not have permission" to express itself and the flower will be white flowers.

A plant with homozygous purple, heterozygous controlled flower is crossed with a plant with white flower that has genes for pink expression but contains non-controlled flowers.

Include a legend, P1 (parents genotypes), gametes, and a Punnett square for the Fi generation and calculate the genotype and phenotype ratios. Determine the probability of getting pink flowers.

What type of gene is influencing expression of flower colours? (8 mks)

Legend: GENE 1: Purple = F

Pink= f

GENE 2: Permission to express = G No Permission to express (Epistatic) = g --> leads to White

Pl= FFGg X ffgg

Gametes: FG Fg

FG

Fg fg

FfGg

Ffgg

Genotypic Ratio: 1FfGg : 1 Ffgg

Phenotypic Ratio: 1 purple flower: 1 white flower

Prob = 0% pink plants

Inheritance pattern: Poly genic Inheritance with epistatic gene


20. A man with no signs of the muscular dystrophy, whose father had dystrophy, marries a normal woman. Recall muscular dystrophy is a sex-linked disorder. What is the probability of the couple having a child with the disorder? Show work. (3mks)

Legend: XN = normal X' = muscular dystrophy

= )(NY X XNX?

X XN X? X' XNXN XNX? Y XNY X?Y

PR =0% -25% OF CHILD WITH DISORDER

21. A woman finds out that she has colorblindness. She is pregnant, but does NOT know the gender of her baby. Colorblindness is a sex-linked recessive disorder.

A) Explain how the woman inherited the disease, if it is a recessive disorder? (2 mks)

WOMAN = XeXe

• Woman MUST receive an X chromosome with recessive disorder allele on it from each parent.

• Her mother must have been at least a carrier, and her father would have colorblindness.

B) What advice might a genetic counsellor give the mother-to-be about the chances that her new born baby will also express the disorder? Use a punnett square to support your answer. (2mks)

Genetic counsellor would show the mother a punnett square with the probable outcomes of her future children.

Legend: Xc = normal X' = muscular dystrophy

P1= x X?Y

X X X' X? X?X' X?X' Y XI( XI(

PR = SONS = 100% Colorblindness DAUGHTERS = 100% at least Carriers

Legend: X" = Black X" = Orange X"X" = Calico

X X" X" X" X"X" X"X" Y XBY X"Y


22. In Cats, the sex-linked gene for calico (multi-coloured) is examined. Females are the only one that express this inheritance pattern where if they receive a B gene (representing black color) and an R gene (representing orange colour) we would see black and orange splotches on white coats. Males can only be black or orange never calico.

What would the result be in mating a calico female and a black male? Include a legend, P 1(parents genotypes), punnett square (F1) and determine the phenotypic and genotypic ratios from the Fl generations. What type of inheritance pattern does this question illustrate? (6 mks)

P1 = X"X" x )(BY

: 1 X"Y

PR= MALES = 1 black: 1 orange FEMALES = 1 black: 1 calico KITTENS = 2 black: 1 orange: 1 calico

INHERITANCE PATTERN: Both Sex-linked and Co-dominant

• 0

1 2

II 16-0

1 2

III

3 7

X )1 y\v% x")e%


23. Using the pedigree below, identify the genotype of all the individuals shown below. This is an autosomal dominant pedigree. Use the letter H for dominant and h for recessive. Individual I-1 is Hh. (3 mks)

1 2 3 4 5 6 7 8 9

Pedigree 1. An idealized pedigree of a family vith hypercholesterolemia, an autosomal dominant disease vhere tte heterozygote has a reduced number of functional by density lipoprotein receptors.

1-2 hh

II-1 Hh 11-5 Hh

111-2 hh 111-8 Hh 111-9 hh

24. Draw a pedigree for the following family members studying the trait hemophilia. Include shading for all individuals who express the trait or are carriers of the trait, and genotypes and legend.

A man and woman marry (who both show no signs of the disorder) and they have 4 children; 2 girls and 2 boys respectively. The two youngest boys have hemophilia. Both daughters marry men without hemophilia. The oldest daughter however has one son who is affected with the disorder and one normal girl. The youngest son marries and has three children, two girls and a boy. His first born daughter has haemophilia only and the other children show no signs of disorder. (14 mks)

IL■

gev ro.k Xt"

A) AUTOSOMAL RECESSIVE

Autosomal: male to male transmission

B) AUTOSOMAL DOMINANT

Autosomal: male to male transmission, male and female equally affected Dominant: Affected child with affected parents, does not skips generations

III 1


25. Determine the type of inheritance pattern shown below. State if it is autosomal or sex-linked, dominant or recessive. Explain your answer. (8mks)

A)

B)

C) SEX-LINKED RECESSIVE

Sex-Linked: NO male to male transmission, males are affected more often Recessive: Carriers, Affected child without affected parents, and skips generations

CT!

F-ik 11

D)

Carriu of Trait

•

C)

0:6, 001 •

26. List any 6 facts about meiosis . (6 mks)

I. Cell division that results in sex cell formation

II. Reduction Division (chromosome number is halved)

Biology 40S: UNIT and EXAM REVIEW BOOKLET KEY

III. Produces 4 non-identical haploid gametes

IV. Two divisions (Meiosis I and Meiosis II)

V. Crossing over may occur in prophase 1

VI. Sexual Reproduction and Occurs in the reproductive organs

pg 17

27. How do random assortment and crossing over provide genetic variation among species? Why is variation important to a species? (2 mks)

• Creates new gene combinations not see in parents

• Variation is important as it creates differences among the members of a species and gives them different tolerances, susceptibilities and different characteristics needed for adaptations

28. Explain how crossing over occurs during of Meiosis. Be specific about phase, chromosomes involved and actions occurring. (2mks)

• Occurs during Prophase I • Homologous chromosomes pair up in the process of Synapsis • Homologous chromosomes may become twisted or tangled together. • Pieces of the chromosome may break off and exchange with each other

29. Label the following illustrations of meiosis. (4 mks)

4. METAPHASE I

5. METAPHASE 11

6. TELOPHASE 1

7. PROPHASE 'II

3. PROPHASE 1


2. ANAPHASE II 1. ANAPHASE 1

CYTOKINESIS II

•

11

.4 % lot pi . .

1(

9.

' la


30. What is nondisjunction? Significance of this event occurring? (2 mks)

• FAILURE OF HOMOLOGOUS CHROMOSOMES TO SEPARATE CORRECTLY AS THEY DO IN NORMAL MEIOSIS

• SIGNIFICANCE: LEADS TO INCORRECT CHROMOSOME NUMBER (TRISOMY AND MONOSOMY)

31. Describe the chromosomal arrangement and/or genotype as seen in individuals with the following conditions: (3 mks)

i) Turner's Syndrome

• GENOTYPE XO

• FEMALE

ii) Kleinfelter's Syndrome

• GENOTYPE )0(Y • MALE

iii) Down's Syndrome

• TRISOMY 21 • MALE OR FEMALE

32. Examine the following karyotypes and identify the disorder if present and gender of individual. (2 mks)

A)

; 1 2 3

4

• • 9 9

; r. t ; t ' '' • r

6 7 • 9 10 11 12

1 1

: R ..... .: •

El f-;

13 14 IS 16 17 18

' I i

19 20 21 22 X X

NORMAL FEMALE

B)

TURNER'S FEMALE


33. Discuss the EFFECT (result) of the following conditions (causes) on the offspring. (4mks)

a) A sperm containing an X chromosome fertilizes a female gamete.

X + X = XX - FEMALE

b) If a female has a recessive allele for hemophilia (sex-linked recessive) on both her X chromosomes.

Female =XhX1' X X" Xn X' X'X' VX" Y XnY XnY

100% sons with disorder 100% of daughters with carriers

c) Both parents are normal but carriers for the autosomal recessive disease Tay Sacs.

Pl= Tt x Tt X T t T TT Tt t Tt Tt

25% chance of child with disorder

d) If the X chromosomes fail to separate correctly in the egg and result in n + 1, and are then fertilized by a sperm carrying a Y chromosome.

XX + Y XXY 1CLINEFELTERS SYNDROME, MALE

34. Explain how the prenatal test: Chorionic Villi sampling provide useful information to doctors about

chromosomal abnormities. (1 mk)

• Sample of chorion membrane is removed that contains the same genetic make-up as the fetus

• Karyotype is created from the fetal cells to determine is chromosomal abnormalities are present

K. PEDIGREE


35. FILL IN THE BLANKS: Vocabulary Review (6 mks)

Monosomy Nondisjunction Trisomy Sex-linked Turner X chromosome Karyotype Pedigree Sex chromosomes Amniocentesis Y chromosome Autosomal

C. SEX CHROMOSOMES

D. SEX-LINKED

E. NONDISJUNCTION

F. TRISOMY

The sex of the off spring is determined by the presence of the

There are 22 pairs of matching homologous chromosome that do not contain sex genes are called .

The two chromosomes which determine the gender of an individual.

Traits controlled by genes located on sex chromosomes are called

The name of the process that results in Down's syndrome is called_.

When a gamete with an extra chromosome is fertilized by a normal gamete, a zygote with an extra chromosome is formed. This condition is called .

A condition that results in human females having only one X chromosome instead of the normal two is called .

In Kleinfelter's syndrome, a person has two and a Y chromosome that results in male with underdeveloped sex organs.

A photograph of chromosomes arranged by size and position of centromere.

A technique used to detect genetic disorders in a fetus, by examining amniotic fluid is called .

A graphical representation of genetic inheritance. It shows the presence or absence of a particular trait in each member of each generation.

When nondisjunction caused a gamete to form that has lost a chromosome. (n -1)

A. Y CHROMOSOME

B. AUTOSOMAL

G. TURNER'S

H. )0C CHROMOSOMES

I. KARYOTYPE

J. _AMNIOCENTESIS

L. MONOSOMY


36. Fill in the blanks regarding the correct terms about chromosomal mutations. (5)

A. MUTAGEN An agent that causes a change in the DNA is called , which include such things as radiation, chemicals, and even high temperature.

B. TRANSLOCATION An example of a chromosomal mutation where part of one chromosome breaks off and is added to another chromosome is

called

C. INVERSION Occurs when part of a chromosome is broken off and becomes reinserted backward.

D. DUPLICATION When entire segments of genes are copied.

E. DELETION Part of the chromosome is broken off and lost.

A. DNA

B. URACIL

C. TERMINATOR

F. TRANSCRIPTION

G. MESSENGER (mRNA)

H. DNA REPLICATION

I. tRNA

J. PYRIMIDINES


UNIT 2: MOLECULAR GENETICS 1. FILL IN THE BLANKS: Vocabulary Review (6 MKS)

DNA

DNA replication

hydrogen ribosomes cytosine uracil

terminator translation

transcription messenger

tRNA

pyrimidines

Heredity material that contains the genetic code.

The nitrogen base can not be part of the DNA molecule?

When a codon causes the synthesis of a protein to stop it is called a .

The nitrogen base that would pair with guanine is .

A protein is assembled amino acid by amino acid during the process of .

In the process of , enzymes make an RNA copy of a DNA strand.

The RNA copy that carries information from the DNA in the nucleus into the cytoplasm is RNA.

Two identical strands of DNA are created in the process called

Molecules of bring amino acids to the ribosome for assembly into proteins.

Purines have a double carbon-nitrogen ring, while are a single carbon -nitrogen ring.

Nucleotides form base pairs through a weak bond called a bond.

Translation occurs on the in the cytoplasm.

D. CYTOSINE

E. TRANSLATION

K. HYDROGEN

L. RIBOSOME


2. What is a nucleotide? (lmk)

• SIMPLE BUILDING BLOCKS OF NUCLEIC ACIDS • MONOMERS TO BUILD POLYMERS

3. Name and describe the THREE structures of a basic nucleotide. (3 mks).

i) SUGAR (DEOXYRIBOSE OR RIBOSE)

ii) PHOSPHATE GROUP

NITROGEN BASE (ADENINE, GUANINE, CYTOSINE OR THYMINE)

4. Draw and label the basic structure of a single nucleotide. ( 3mks)

(__Phosphate Group

CH

> Nitrogenous Base (A, T, G, C)

5-C Sugar (Daoxyribose)

5. Describe Chargaffs base pairing rule? How are the base pairs held together? (3 mks)

ADNENINE = THYMINE

CYTOSINE = GUANINE

• BASE PAIRS HELD TOGETHER WITH HYDROGEN BONDS

PHOSPHATE GROUP DEOXYRIBOSE SUGAR

NITROGEN BASE NUCLEOTIDE


6. Label the parts of a DNA nucleotide. (2mks)

7. Explain semi conservation replication. (2mks)

• HALF OF DNA MOLECULE IS ORIGINAL

• HALF OF DNA MOLECULE IS NEWLY SYNTHESIZED

8. Discuss the importance of the following enzymes: (4mks)

a) helicase • Unzips/unwinds DNA molecule

• Forms replication bubbles

b) ligase • Joins together the Okazaki fragments on the lagging strand (discontinuous fragments)

c) DNA polymerase III • POLYMERASE III adds free DNA nucleotides to exposed bases on both leading and

lagging strands

d) RNA polymerase • Unzips/unwinds DNA molecule

• Forms replication bubbles

Biology 40S: UNIT and EXAM REVIEW BOOKLET KEY pg 26 9. In what ways do the chemical structures of DNA and RNA differ? Complete the chart. (7mks)

DNA RNA

Types of Sugar DEOXYIREBOSE RIBOSE

Nitrogen bases ADENINE, THYMINE, GUANINE,CYTOSINE

ADENINE, URACIL, GUANINE,CYTOSINE

Shape DOUBLE HELIX SINGLE HELIX

Location NUCLEUS NUCLEUS AND CYTOPLASM

Size LARGE (BILLIONS OF BASES)

SMALL (MILLIONS OF BASES)

Number of Strands 2 (DOUBLE STRANDED) 1 (SINGLE STRANDED

Function CONTAINS THE MASTER CODE

CARRIES OUT DNA INSTRUCTIONS

10. Discuss the main function of three types of RNA molecules: rRNA, mRNA, and tRNA. (6mks)

mRNA tRNA rRNA STRUCTURE

LINEAR CLOVERLEAF SHAPE

LINEAR INSIDE RIBOSOME

FUNCTION Carries genetic information copied from DNA.

Carries specific amino acids to mRNA to help assemble protein.

Assemble amino acids into proteins.

Biology 40S: UNIT and EXAM REVIEW BOOKLET KEY pg 27 11. Compare a codon and anticodon. Describe the purpose of each. (anks)

Codons are a sequence of 3N-bases found on the mRNA strand. Anticodons are a sequence of 3N-bases found of the tRNA strand.

12. Beginning at the DNA strand, state the mRNA, tRNA, and final protein that would be formed. (4mks)

TAC CCG CAG TTG ATT

ATG GGC GTC AAC TAA

AUG GGC GUC AAC UAA

UAC GGC CAG UUG AUU

First Letter

Second Letter Third Letter U C A G

U

phenylalanine serine tyrosine cysteine U phenylalanine serine tyrosine cysteine C

leucine serine stop stop A

leucine serine stop tryptophan G

C

leucine proline histidine arginine U leucine proline histidine arginine C leucine proline glutamine arginine A

leucine proline glutamine arginine G

A

isoleucine threonine asparagine serine U isoleucine threonine asparagine serine C isoleucine threonine lysine arginine A

methionine threonine lysine arginine G

G

valine alanine aspartate glycine U valine alanine aspartate glycine C valine alanine glutamate glycine A

valine alanine glutamate glycine G

DNA strand Template:

Complementary Strand:

mRNA Strand:

tRNA molecules =

PROTEIN (Amino acids) = Methionine, glycine, valine, asparagine, stop


13. Answer the following questions about base pairings of triplets to codons and anticodons. (5mks)

a) The mRNA that would code for the DNA molecule ACC would be ufiel . b) The complimentary strand to the DNA template ACC would be?

c) The tRNA for the mRNA; UAG would be AOC d) The complimentary DNA strand for the following DNA strand: ATG would be T AC . e) The DNA sequence A CC-. would create a mRNA codon for the amino acid tryptophan.

14. Compare the processes of DNA replication, transcription and translation by completing the following chart. (6mks)

see no-\--e:s DNA Replication Transcription Translation

LOCATION of Process

PURPOSE of Process

ALL TYPES of RNA involved in Process.


15. Read the following statements and determine if the step occurs during the process of DNA REPLICATION, TRANSCRIPTION, OR TRANSLATION. (6 mks)

STEPS Name of PROCESS mRNA is formed from DNA template

-herciriCcii pti 0 n -hia-nidfie n DNA re \\ (Ail on

tRNA binds to specific amino acids, helping form polypeptides

Before a cell divides, the complete DNA is copied

DNA polymerases move along separate strands and nucleotides from the surrounding medium are assembled on parent strand.

bt\1 Pt reph cefh on rRNA directs protein synthesis

Ali o n -ha rig( Introns are removed and exons are spliced together

--IYanWbh on

16. Using the illustration of Protein synthesis, identify the parts shown below. (4mks)

Name the process occurring. P V-Cke 1 r) SrliVV.C.1-C (or - fra r \ & ,,.. Where is this process occurring? CppiCAM 4 ce I I 1. cod on 4. +12-NA 2. en RNA 5. Cit)ti Cocion 3. CitYlinb a Cid

(deçirck) 6. OVY) Ci Ci

Biology 40S: UNIT and EXAM REVIEW BOOKLET KEY pg 30 17. Using the following diagram of protein synthesis, answer these questions:

Explain what is occurring at all the 5 stages of the diagram shown above. (5mks)

* 0 C ir\01-e

21. A -IN( tAis a type of mutation that causes a shift in the reading of codons and changes many amino acids.


18. Complete the following sentences: (2 mks)

A. The ih -hrQn.ic are a segment of DNA that does not code for amino acids of a

protein and is removed by enzymes.

B. The -e_Y 0 r"). become part of the protein. Used in the process of

are a segment of DNA that codes for amino rids that.will

n c p cipri nok)

1-Ydn&leMD ) 19. A -hQ Y) causes a sudden change in the structure or the amount of

genetic material present in an organism's cells.

20. A 11111 01 V* rnvi---41\o y) is a type of mutation in the DNA sequence occurring from h. change in a single base pair and may change only one amino acid.

22. Suppose that a point mutation occurred during transcription from the original DNA; TCG and a nucleotide substitution changed the C to an A and formed a mutated DNA of TAG. Show the codon, anticodon and amino acid that would result. (3mks)

ORIGINAL DNA TCG MUTATED DNA TAG ORIGINAL TRANSCRIPTION (mRNA) Al C

MUTATED TRANSCRIPTION (mRNA) AU C

ORIGINAL TRANSLATION (tRNA)

UCG1 MUTATED TRANSLATION (tRNA)

. um i &O — ,

p 'i i1eN Amino acid chain (protein) MUTATED ??Amino acid chain

(protein)

23. Explanation of effect of the mutation on the protein formation. (1mk)

— or) amify aci - c000 0H-cc -c-bfrivi-6-Div -Rmchon

24. Explain why a frameshift mutation is more damaging to an organism than a point mutation. (2mks)

ct mQ---othr,r) iS == c.,,,,-)av)\---c-ict 6- cef\A—

ptir0-1e6A


25. FILL IN THE BLANKS: Vocabulary Review: (7 mks) -iforsgerrie AeconthirtantN94-- --314ese

---ThrffMtrereiTerrileTIZTEUt ___Cumagatieseligiffinring ---restif.t.i.ue-mokrjqfies 4i - .211,044ste grpy „wine— A. rerOPIAbWgi—

B. N.) .0\r-

pg 32

tor

A section of a mouse DNA is joined to a bacterial plasmid. The DNA formed is called .

When bacterial plasmid transfers a piece of foreign DNA to a host cell, the plasmid serves as a .

J.

c.

D. N E. -1-fan_cyr)i c__

F. bac4 cAj ch

G. Toe_ --tflecap

H. hrrall iueie

I. I .1/ct-e__

c_ On_cji\rNelf tYvAs

K. retc---M cjIe

L. liman C)-ehome4_--

M. dc c.jQ, -encL

N. MAYS

The goal of gene therapy is to insert a into cells to correct a genetic disorder.

The process of gel electrophoresis separates fragments by using an electric field. An organism that contains functional recombinant DNA is called a organism.

cells are the primary source of the plasmids used in DNA technology

Is the insertion of normal genes into human cells to replace or correct genetic disorders.

A distinct unit of hereditary material (segment of DNA) found in chromosomes that controls the protein production and the cell cycle.

The enzyme used to ligate two fragments of DNA together to form recombinant DNA.

or recombinant DNA technology is a method that involves cutting or cleaving the DNA from on organism into small fragments and inserting it into a host organism.

To isolate a DNA fragment, small pieces of DNA must be cut from a larger chromosome using that are bacterial proteins that can cut both ends of a DNA molecule at a specific nucleotide.

International project to completely map and sequence the genes found in humans.

Jagged ends of DNA that can join with matching pieces of DNA from other sources.

Genetically identical copies of two organisms are called .


26. What is recombinant DNA? Importance? (2mks)

no-v-c_c

27. Name and discuss any THREE ways in which genetic engineering are used for human application. Include what has been changed and why it has an impact. (3mks)

+ ce,


28. Name and discuss any THREE ways in which genetic engineering can be harmful or have a negative impact on society. Include what has been changed and why it has an impact. (3mks)

29. List in order (using the numbers 1-6) the correct sequence of events occurring during the process of gel electrophoresis. (3mks)

Loading the DNA into the gel.

View and photograph the gel with the DNA fingerprint under UV light.

Preparing the DNA using restriction enzymes.

Adding electrodes to the gel box and running the gel.

_2_ Preparing and casting the agarose gel to insert the DNA fragments.

3 Remove the gel comb and add TBE buffer to the gel chamber.


30. List and describe two important facts about DNA (charge and fragment size) that we learned from the Gel Electrophoresis Lab. (2mks)

r0A-e_

31. All humans are nearly identical genetically in coding sequences and have many proteins that are identical in structure and function. Nevertheless, each human has a unique DNA fingerprint. Explain this apparent contradiction. Discuss nucleotide base pairing and non-coding regions of DNA called "Satellite DNA. (3mks)


32. DNA fingerprinting can be used to identify a child's parents. Each child inherits one set of chromosomes from each parent. This is why children resemble both of their parents. A child who has a mom with brown hair and blue eyes and a dad with blond hair and brown eyes might end up with brown hair from his mom and brown eyes from his dad. DNA fingerprints are created in the same way, some DNA from the mother and some DNA from the father.

In this example, a family consists of a mom and dad, two daughters and two sons. The parents have one daughter and one son together, one daughter is from the mother's previous marriage, and one son is adopted, sharing no genetic material with either parent. After amplifying the DNA fingerprint from each member of the family, it is cut with a restriction enzyme and run on an agarose gel. Here are the results:

a) From the results who is the adopted child? (1 mk)

b) From the results, who is the daughter from a previous marriage? Da (1 mk) c) From the results, is child D1, more like her mother or her father?

e kber (link)

Bacterium

Strand of DNA from donor cell

Section of donor DNA inserted into bacterial plasmid

Restriction enzyme splits the plasmid open and

removes a section of DNA from donor cell

Recombinant plasmid inserted back into

bacterium


33. Using the diagram below, List in order (using the numbers 1-6) the correct sequence of events occurring during the process of producing recombinant DNA to make bacteria produce human insulin. (3mks)

Plasmid isolated from bacterium

The plasmid containing recombinant DNA is inserted into the host bacterium cell using a gene gun.

2 Same restriction enzymes are used to cut both human DNA and plasmid. Sticky ends are created on human DNA fragments and bacterial DNA (plasmid) fragments.

(:) The host bacterium reproduces producing clones that all contain the human insulin gene.

4 Human DNA fragments and plasmids are ligated together to transfer human DNA into bacterial cell DNA is isolated from two sources; human insulin gene and bacterial plasmid


UNIT 3: EVOLUTIONARY THEORY 1. Define evolution. (link)

• Evolution is the cumulative change in the heritable characteristics of a population over time.

• Any change in the relative frequencies of alleles in a gene pool of a population which changes the genetic composition of that population.

2. Explain the statement" populations evolve not individuals". (link)

• Evolution acts on the phenotype of the individual, but the change must be passed down to future generations over time for it and accumulate within the population

• Therefore evolution ultimately acts on the entire population made up of successful individuals

3. What is the significance of the Galapagos Islands to Darwin's work? (link)

• Galapagos is so important and famous because in 1835 Charles Darwin visited the islands as part of a long expedition to explore South America. He was surprised by how many of the animals and plants he saw were like their "cousins" on mainland South America, but also unique to the islands.

He had many questions. How did the ancestors of these animals get to the islands, so far away and separated by a vast ocean? Why did they look different from their mainland relatives? Was that because they had to change over time to be able to live and reproduce under the harsher conditions of the islands? How did this change happen?

• While Darwin observed the wildlife of the Galapagos, he collected samples to take home with him and he wrote up his notes. Later his ideas and observations would lead him to create his well-known theory of natural selection.

• Another reason why the Galapagos is so important is because it harbors several populations of one of a kind species, and remains to this day, a beautiful, natural laboratory in which to study evolution.


4. Label the following statements with the terms DARWIN, LAMARCK or BOTH for which they apply. (4mks)

a) Both Theory involves heredity

b) Lamarck Acquired characteristics

c) Both Viewed a change over time

d) Darwin Used papers by Malthus and Lye11

e) Darwin Involved competition

0 Lamarck Use it or lose it

g) Lamarck First evolutionist

h) Darwin Natural selection

5. Discuss the contributions of Lyell and Malthus to Darwin's work. (2mks)

• Even: was a geologist who studied the Earth's surface and explained that the Earth's surface was changing over time under its own control (not influenced by god) *)=. Darwin realized that if the Earth (inorganic) could change then living

organisms (organic) should also be able to change.

• Malthus: was an economist who studied population growth and explained that the World population cannot sustain itself with an increase forever. War, famine, disease will reduce the population to keep it in balance.

Darwin realized that overpopulation would lead to competition and cause natural selection to balance the population

6. What is meant by the term "fitness" of an organism? (1 mk)

Fitness is a measure of how well suited an organism is to survive in its habitat and its ability to maximize the number of offspring surviving to reproductive age.


7. List and briefly explain the 6 main points of Darwin's theory. (6mks)

1. • •

OVERPOPULATION All species are capable to producing more offspring than the environment can support. Species populations remain more or less constant as not all individuals can survive.

2. STRUGGLE FOR EXISTENCE (COMPETITION)

• Due to overpopulation, there is a competition for food resources, living space and other necessities of life.

• There is a struggle for existence in which some individuals survive and some die.

3. VARIATION

• Living organisms vary. • Some individuals have characteristics that make them well adapted to their environment and other

individuals have characteristics that make them poorly adapted to their environment. • If the variations are positive and increase the successfulness of the organism, the traits are passed

onto the next generation.

4. SURVIVAL OF THE FITTEST (NATURAL SELECTION)

• Those individuals in a species with traits that give them an advantage are better suited to compete, survive, and reproduce more often than poorly adapted individuals.

• Poorly adapted will die off without leaving offspring. The phase "survival of the fittest" is used to summarize this idea.

• The results of natural selection therefore accumulate. As one generation follows another, the characteristics of the species gradually change- the species evolves.

5. ADAPTATION

• Adaptation is an inherited trait or set of traits that improve the chances of survival and reproduction of organisms. Enables organisms to become better suited to their environment.

6. ORIGIN OF NEW SPECIES (SPECIATION)

• Over numerous generations, new species arise by the accumulation of favorable, inherited variations. Eventually the accumulated changes produces a species that is significantly different from the original, it becomes a new species.

• Speciation is the formation of a new species. New species are formed when a pre-existing species splits.


8. A) Compare how Lamarck and Darwin would each explain the evolution of a cactus plant and its' thorns for protection. (4mks) Choose TWO points from each evolutionist's theories in your answer.

LAMARCK DARWIN 1. UNIVERSAL FORCE/ 1. VARIATION WILLINGNESS TO CHANGE • three neck size variations exist

• Long neck length has an advantage over the short neck length and can reach the high leaves in the trees.

• Giraffes had the ability to produce longer necks to satisfy their needs and become better adjusted to their environment.

2. USE AND DISUSE 2. NATURAL SELECTION

• Giraffes necks that were regularly used became stronger and more highly developed to perform their feeding function

Long neck giraffes have the highest survival rate and the highest reproductive success

3. ACQUIRED CHARACTERISTICS 3.ADAPTATION

• One giraffe acquires a larger, longer neck and will pass this trait from generation to generation to produce longer, larger necked offspring, and eventually these changes make permanent changes to the giraffe species.

Physical/structural adaptations to the giraffe neck length provided advantage to be more successful in the environment

4. COMPETION • Due to overpopulation, there is a

competition for food resources, living space and other necessities of life.

• There is a struggle for existence in which some individuals survive and some die

B) What type of selection (stabilizing, directional, or disruptive) is occurring that favours the thorned cacti over the thornless cacti? Why? (lmk)

DIRECTIONAL: favors one extreme phenotype of thorned


9. Compare the THREE types of natural selection. Give an example of each and provide a graph illustration selection in action. (6mks)

TYPES OF NATURAL SELECTION (with type of phenotype favoured.)

GRAPHS

STABILIZING SELECTION:

/

/

/ ' /

Selection aqaif both extremes

, ,, Population

after selection

Origi nal poPulation

Encourages the formation of average traits

Individuals with the average form of a trait have the highest fitness. The average represents the optimum for most traits.

,/ , ■

\ . .

DIRECTIONAL SELECTION:

so lion et

/

,,/

,

ainstan extreme

,......, , , N / %

/ % Population

Encourages the formation of one extreme trait, such as a very long tongue in anteaters

Individuals that display a more extreme form of a trait have greater fitness than individuals with an average form of the trait.

alter selection

Original

,„ / . , ,

population

DISRUPTIVE SELECTION:

Selects for both extreme traits rather than average traits

Individuals with either extreme variation of a trait have a greater fitness than individuals with the average form of that trait.

Selection against the mean

, • 4. , , , •

t

/ 0

-- , ,

s • Population after selection

Original %

• population

. .,

Biology 40S: UNIT and EXAM REVIEW BOOKLET KEY pg 43 10. Using the terms: STRUCTURAL, BEHAVIOURAL, PHYSIOLOGICAL adaptations determine

which type of adaptation best describes the scenario below. (3mks)

A) PHYSIOLOGICAL Orchids can produce scents that smell like the females of the pollinating insect species. This helps transfer pollen to other plants, as male insects attempt to mate with orchid.

B) STRUCTURAL Monarch butterflies have a bitter taste and are avoided by birds. Viceroy butterflies look and act like Monarch butterfly, but are not bitter tasting, trick birds into thinking they are Monarch butterflies to avoid been eaten.

C) PHYSIOLOGICAL Some strains of bacteria can produce an enzyme that inactivates penicillin, others have cell walls that block the penicillin molecule from entering.

D) BEHAVIOURAL Hibernation of certain mammals during long winter months.

E) BEHAVIOURAL Broken wing behaviour of bird to draw enemy away from nest.

F) STRUCTURAL Orchids are shaped like their insect pollinators and attract male insects as they attempt to mate with the flower.

11. What five conditions that must be met for the Hardy-Weinberg principle to be true? (5 mks)

1. LARGE POPULATION SIZE The population must be very large, and changes in the allele frequencies due to change alone are insignificant and do not alter the gene pool

NO MUTATIONS Allele changes DO NOT occur, NO NEW SOURCE OF GENETIC MATERIAL

3. RANDOM MATING Individuals pair by chance and not according to their genotypes or phenotypes

4. NO GENE FLOW Population is isolated; NO migration of individuals into or out of the population

5. NO NATURAL SELECTION No selective force favors one genotype over another. All individuals have equal reproductive success.


12. What do p, p2, q, q2 and 2pq represent? BE SPECIFIC (5 mks)

p— frequency of dominant allele/gene q = frequency of recessive allele/gene p2 = frequency of homo dominant trait/individual/ phenotype/genotype q2 = frequency of homo recessive trait/individual/ phenotype/genotype 2pq= frequency of heterozygous trait/individual/ phenotype/genotype

13. If the homozygous dominant portion of the population equalled 27%, answer the following questions;

A) Determine the frequency for both the dominant allele and the homozygous dominant phenotype for the trait and the heterozygous trait. (2mks)

GIVEN: Homo dominant phenotype =p2 = 27% = .27

Dominant allele = p = V.27 = .519615242

B) Determine the frequency of the recessive phenotype and recessive allele. (2mks)

Recessive allele = q = 1- p q = 1-.52 q = .48

Recessive phenotype = q2 = (.48)2 q2 .23 (23%)

C) Determine the number of individuals that have the recessive trait if the total population is 500. (Inks)

.23 X 500 = 115 individuals with recessive trait

14. If a population has 135 individuals, and 14 have the homozygous recessive trait,

A) Determine the frequency of the recessive phenotype and recessive allele. (2mks)

GIVEN: Recessive phenotype = q2 = 14/135 q2 = .103703704 (10%)

Recessive allele q = -V.103703704 = .322030594

B) Determine the frequency of both the dominant allele and the homozygous dominant phenotype for the trait. (2mks)

Dominant allele = p = 1 — q p = 1 - .32 p = .68

Homo dominant phenotype = p2 = (.68)2 = .4624 (46%)

C) Determine the number of heterozygous individuals in this population. (3mks)

2pq = 2(.68) (.32) 2pq= .4352 (44%)

.44 X 135 = 59 individuals with heterozygous trait


15. Compare artificial selection to natural selection. Provide an example of when each is used. (3 mks)

NATURAL SELECTION: • is a passive process in natural environment • organisms do not purposefully acquire traits that they may need. • The environment "selects" those traits that will increase within a population. • The kinds of traits that are favorable depend on the demands of the environment. • EXAMPLE: Peppered Moth

ARTIFICIAL SELECTION: • is an active process in an isolated environment that causes a disruption in the

genetic equilibrium. • Humans choose the desirable traits and breed only organisms with those traits. • This process favors specific traits, and determine which organisms get to

reproduce. • This leads to an accumulation of small changes over time. • EXAMPLE: breeders of livestock

16. Explain how the relationship between the geographical location of both; fossils and living species offers compelling evidence for evolution. (2mks)

• The geographical distribution of species suggested that organisms evolve from ancestral species in one location and migrated throughout the world (Pangea: made travel throughout world possible)

• This explains why related/similar species exist throughout the world

• the chronological fossil record gives us the history of life in general and allows us to trace the decent of a particular group from past to present

• studying fossils shows that life of Earth is constantly changing/evolving in response to environmental changes.

17. How do the similarities in a bat wing, whale flipper, horse leg and human hand give evidence for evolution and the notion of common ancestry? (2mks)

• COMPARATIVE STRUCTURES: body structures that are similar in form (homologous structures) and may or may not be used for the same function but are found on dissimilar species.

• Similarity in characteristics t s om common ancestry because different organisms must contain t same genes s form the same internal anatomy as seen in the pentadactyl g whale flipper, horse leg


18. Explain how structures, that were once useful to an organism, could become vestigial. Provide an example of a vestigial organ. (2mks)

• are anatomical features that are fully developed in one group of organisms but reduced and maybe even non-functional in similar groups because of changes in response to environmental pressures to adapt to survive

EXAMPLES: appendix, wisdom teeth, pelvic hones in whales, snakes

19. How can DNA be used to determine the degree of relatedness between organisms? (2mks)

• All organisms have certain biochemical molecules in common. • The degree of similarity between DNA base sequences and amino acid

sequences will indicate the degree of relatedness. • MORE similar molecules are more CLOSELY related • LESS similar molecules are more DISTANTLY related

20. Complete the following paragraph by filling in the appropriate missing vocabulary..

If two species are found in the same habitat and possess very similar physical and behavioural traits,

then one may conclude that they are most likely RELATED/COMMON species. A scientist

may conclude this, because animals of RECENT decent have of similar internal anatomy

called HOMOLOGOUS structures. If this is true, then they must also contain very

similar BIOCHEMICAL (DNA) evidence, which are produced by similar

AMINO ACID chains according to modern studies. By using this data we can construct

diagrams showing relationships among different groups of organisms called phylogenetic trees or

cladograms. (3mks)

21. Some people have argued that humans may no longer be evolving. Give 1 reason why this may be so and 1 reason why this statement may be premature. (2mks)

NOT EVOLVING: 1. we do not see noticeable changes occurring in our lifespan of 100 years 2. environment is stable, no need to evolve

PREMATURE: 1. Rate of evolution is GRADUAL: evolutionary change takes hundreds of thousands to years to accumulate within a population 2. Rate of Evolution is PUNTUATED EQU I LIBIRUM: evolution may occur in response to changes in environment.., in a stable environment there is no driving force or need to cause evolution at this moment in time


22. Explain the concept of the genetic drift, specifically the bottleneck effect, and how this leads to evolution. (3mks)

Genetic Drift is the change in allele frequencies of a population as a result of chance processes. This causes changes in the heritable characterisitcs passed from generation to generation

Bottleneck Effect: • occurs when a population is subjected to near extinction because of a natural

disaster, or human interference. • The disaster acts as a bottleneck preventing the majority of genotypes from

participating in the production of the next generation. This changes allele frequency, therefore leads to evolution

23. Describe how genetic drift affects the amount of genetic variation within very small populations.(2mks)

Genetic drift REDUCES variation within a population because: • only a few individuals with survive after the natural disaster reducing the population

size • only a small gene pool remains and only their genes will be left to pass onto future

generations • some genotypes will be lost from the population • other genotypes will become fixed

24. Define divergent evolution. Explain how geographic isolation can lead to speciation which in turn leads to divergent evolution.(3mks)

DIVERGENT evolution occurs when a group of organisms becomes isolated from the rest of the same species. (common origin) • The isolated species may follow a different line of evolution due to different selective

pressures. • At first the separated population is similar to the ancestral species, but as time goes

on, the separated population each become adapted to their environments and become less and less alike.

• This leads to SPECIATION; the origin of a new species because each of the separated populations will no longer interbreed.


25. List any four factors that tend to increase genetic variation in populations. (2mks)

• Mutations • Gene flow • No natural selection • Large population • Random mating • Large gene pool

26. List any four factors that tend to decrease genetic variation in population (2mks)

• NO Mutations • NO Gene flow • Natural selection • Small population • Selective mating (non-random mating) • Genetic drift • Small gene pool

27. Determine the reproductive isolation mechanism operating in each situation: (3mks)

A) TEMPORAL (SEASONAL) Plants flower at different times of year.

B) STRUCTURAL (MORPHOLOGICAL)Plants have a unique arrangement of their floral parts that stops transfer of pollen from other plants.

C) GAMETE MORTALITY Fertile eggs between two different species fail to develop.

D) HYBRID STERILITY The mule is a sterile offspring between the horse and the donkey.

.E) BEHAVIORAL( ETHOLOGICAL) Birds exhibit courtship displays that are species specific.

F) ECOLOGICAL (HABITAT) Some fish spawn in the shallow water near the shore, while other spawn in the deeper waters in the middle of a lake.

28. Define the two theories on rates of evolution, (2mks)

GRADUALISM is that evolution proceeds very slowly, but large changes can gradually take place over long periods of time.

PUNCTUATED EQUILIBRIUM: periods of sudden change that occasionally occur, may correspond with rapid environmental change followed by long period of stable environment with no changes


29. Draw two separate graphs to represent gradualism and punctuated equilibrium occurring. (2mks)

Graph I Graph 2

30. Explain whether you expect gradualism or punctuated equilibrium to best describe the pace of evolution for the following events: (2mks)

A) A natural disaster such as a flood affects the food supply PUNCTUATED EQUILBRIUM

B) Introduction of a new species to a stable environment PUNCTUATED EQUILBRIUM

C) A pond is expanded to a larger lake area in a wildlife preservation park GRADUALISM

D) A highway is developed through a provincial park. PUNCTUATED EQUILBRIUM


31.FILL IN THE BLANKS: (8 mks) homologous analogous genetic drift punctuated equilibria

A. HOMOLOGOUS Structures that are similar in shape or function because of a close ancestral relationship are called this.

B. ANALOGOUS Structures that are similar in appearance but DO NOT have a close ancestral relationship are called this.

C. PUNCTUATED EQUILBRIUM Pattern of evolution with long stable periods interrupted by brief periods of change.

D. EXTINCTION Represents the end of a species, when the last living individual dies.

E. CO-EVOLUTION Process by which two organisms evolve structures and behaviours in response to changes in each other over time.

F. SPECIATION The formation of a new species from existing species.

G. GENE POOL Common group of genes shared by members of a population.

H. GENETIC DRIFT Random changes in the frequency of a gene.

I. DIVERGENT EVOLUTION Pattern of evolution, in which one species gives rise to many species that appear different externally but are similar internally.

J. ADAPTIVE RADIATION Also known as divergent evolution, Darwin's finches are an example of this.

K. CONVERGENT EVOLUTION Outward pattern of evolution, that produces similarities in appearance and behaviour in organisms but the organisms are not closely related.

L. VESTIGAL Structures that serve no useful purpose or function in an organism

M. GEOGRAPHICAL ISOLATIONA population is prevented from interbreeding with another population of the same species by a natural barrier.

N. REPRODUCTIVE ISOLATION Separation of populations so that they do not interbreed to produce fertile offspring.

0. GRADUALISM

Theory of evolutionary change occurs slowly and gradually over a long period of time.

P. FITNESS

How well an organism can survive and reproduce with its' environment.

co-evolution geographical isolation speciation divergent reproductive isolation extinction convergent gene pool vestigial organs fitness gradualism adaptive radiation

GOOD LUCK ON THE BIOLOGY 40S EXAM 0

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