Bicalendar 2010-11 Revised

 

 

1st Semester Biology 2010-2011

 

 

AUGUST SEPTEMBER OCTOBER NOVEMBER DECEMBER JANUARY FEBRUARY MARCH APRIL MAY

 

THESE ARE APPROXIMATE DATES FOR ASSIGNMENTS!

 

2010

Monday Tuesday Wednesday Thursday Friday
16 17 18 19 20
 

23 24 25 26 27

 

 

30 31

 

 

TOP

 

2010

Monday Tuesday Wednesday Thursday Friday
1 2 3
TARGET PRE-TEST

6 7 8 9 10

 

13 14 15 16 17

 

20 21 22 23 24

INTERIMS
P-T Conference SHS

 

P-T Conference SJHS

 

27 28 29 30  

 

 

 

TOP

 

2010

Monday Tuesday Wednesday Thursday Friday
1
4 5 6 7 8

 

 

11 12 13 14 15

Professional Development
 

END OF FIRST 9 WEEKS

18 19 20 21 22

25 26 27 28 29
 

 

 

TOP

 

2010

Monday Tuesday Wednesday Thursday Friday
1 2 3 4 5
TARGET TEST #1 MC

8 9 10 11 12

 

15 16 17 18 19
22 23 24 25 26
 
29 30
 

 

TOP

 

2010

Monday Tuesday Wednesday Thursday Friday
1 2 3
SHS SCIENCE FAIR

 

6 7 8 9 10
TARGET TEST #2 EBR

 

13 14 15 16 17

Proficient-Advanced Students Out

TEST REVIEW
Proficient-Advanced Students Out
SEMESTER TEST
Proficient-Advanced Students Out
SEMESTER TEST
Proficient-Advanced Students
Out
 

SEMESTER TEST
Proficient-Advanced Students
Out
END OF 2ND 9 WEEKS

 

Enjoy Your Christmas Vacation December18, 2010 – January 4, 2011!

 

TOP

 

2nd Semester Biology

 

 

2011

Monday Tuesday Wednesday Thursday Friday
 3 4 5 6 7

10 11 12 13 14

17 18 19 20 21

MLK DAY!
24 25 26 27 28
31

 

 

 

TOP

 

2011

Monday Tuesday Wednesday Thursday Friday
1 2 3 4

P-T Conference SJHS

 

P-T Conference SHS

 

7 8 9 10 11
TARGET TEST #3 HE

14 15 16 17 18
President’s Day

 

21 22 23 24 25

  • Protist worksheet DUE
  • Continue Protist & Fungi PowerPoint

28

 

TOP

 

2011

Monday Tuesday Wednesday Thursday Friday
1 2 3 4

 

7 8 9 10 11

LITERACY TEST

LITERACY TEST

 

 

 

14 15 16 17 18

END OF 3RD 9 WEEKS

TARGET TEST #4 CDL

 

21 22 23 24 25
BREAK BREAK BREAK BREAK BREAK
28 29 30 31

 

TOP

 

2011

Monday Tuesday Wednesday Thursday Friday
1
4 5 6 7 8

 

11 12 13 14 15
  • TEST over Vertebrates
  • Handout: Notes on Plants and Plant Worksheet

  • Work on Plant Worksheet

  • Work on Plant Worksheet
  • Work on Plant Worksheet

 

  • Plant Worksheet DUE
18 19 20 21 22

 

EOC GEOMETRY

EOC GEOMETRY

25 26 27 28 29
EOC BIOLOGY
EOC BIOLOGY
  • Start Birds PowerPoint and answer worksheet questions

 

TOP

 

2011

Monday Tuesday Wednesday Thursday Friday
 2 3 4 5 6
  •  Continue Birds PowerPoint and answer worksheet questions

9 10 11 12 13

16 17 18 19 20
  • Teach wing spreading
  • Identify Insects (KEY)
  • Teach Card pointing
  • Identify Insects (KEY)
TARGET POST TEST

 

23 24 25 26 27

 

SEMESTER TEST
REVIEWBOOK RETURN
SEMESTER TEST
REVIEWBOOK RETURN
SEMESTER TEST
30 31
MEMORIAL DAY SEMESTER TEST

 

 

 

  2011

Monday Tuesday Wednesday Thursday Friday
1 2 3
    SEMESTER TEST

 MAKE UP EXAMS

TEACHERS LAST DAY

 

BACK

Semester I Calendar 2001

AP Calendar 2006

 

week topics text chapters/activities supplemental readings/URLs
see you in late August!!

 

week 1

Aug 24-26

  • intro to AP Bio
  • summer assignments   
  • bonding
  • buffers, acids and bases
  • element cycles
  • homeostasis
  • chemical basis of life: carbon, water and life
  • elements for life

 

ch 1, 50-55-due Aug. 24-your first day back!!**check study guide (SG) and question set (QS)

powers of ten see the size of everything!

ch 2

 

 

handouts/AP website

 

scientific method -The Biology Project

size and biology -The Biology Project

 

 

 

dissociation of water

acid-base explained

water movement click on animations

water movement ions, simulations of pH solutions

 

 

 

week topics text chapters/activities supplemental readings/URLs
week 2

Aug 29-Sept 2

  • biomolecules
    • organic compounds
    • carbohydrates, fats, proteins and nucleic acids
ch 3

 

BBC guru condensation and structure of molecules

 

central dogma animation

molecules -animated (rotate)

dehydration synthesis -animation

the biological basis of life for condensation

week 3

Sept 6-9

Sept 5 off-Labor Day

 

  • metabolism
  •  enzyme kinetics t
  • test over ch 2, 3, and 6 on Friday…be ready!!

 

 

ch 6 interactives catalysis, enzyme inhibition

BBC guru this is the enzyme site we used in class

 

week 4

CAMP KERN (but you are responsible for everything!!)

Sept 12-16

  • ecology and geography
  • human ecology
  • population ecology 

 

population clock population estimation

climate graphs/biomes

dissolved O2 lab 

 

 

Nova-The World in Balance  Human Numbers/Time               The Earth in  Peril      Be a Demographer         Population Trend quiz

nitrogen cycle -animation

 

week 5

Sept 19-23

 

  • community ecology
  • ecosystems and the biosphere

 

 

 

 

 

interesting ecology simulations, including:     biomass succession

biomass succession 2

 

 

week topics text chapters/activities supplemental readings/URLs
week 6

Sept 26- 30

 

 

 

  • early evolution of life
 

 

ch 20

burgess shale -resources

early life  -Burgess Shale

geologic time scale

the Origins Game

Aliens in the Milky Way?

week 7

Oct 3-7

  • prokaryotic and eukaryotic cells
  • viruses and bacteria

 

 

 

 

 

ch 4 

 

ch 7 

 

 

Military Medicine through Time
evolution

microbe clock

microscopy  -CellsAlive

eukaryotic vs pro, quizzes on plant and animal cell

show big is? (cell biology) -CellsAlive

microbe world

microbiology

binary fission movie

lysogenic cycle -animation

micro sites

week 8

Oct 10-14

Oct 11 off

-teacher inservice

  • plant vs animal cells
  • subcellular organization
ch 4

these are the links that we used in class-

virtual cell -virtual cell textbook

cell comparison -check yourself

flagella and cilia movement

the cell -ThinkQuest

 

 

 

 

inside the cell -NIH/NIGMS

cell parts -cells.de link of the month

drop/drag organelles

plant and animal cells -structure and function

endoplasmic reticulum

problem sets and tutuorials -The Biology Project, includes: studying cells tutorial  and cytoskeleton  among others-best site

cell similarity -GeoCities-fun stuff

click on animations -HHMI

week 9

Oct 17-21

Oct 22-end of Q1

 

  • biological membranes

 

  • transport across membranes

 

 

ch 5

membrane structure  -simple

pathways in and out of the cell 

antibiotic resistance

BBC guru

drop/drag membrane ZeroBio

membrane structure power point tutorial-advanced

cell signaling Access Excellence

osmosis tutorial Cornell BioG101-104

osmosis simulation Colorado

osmosis simulation2  UMV

osmosis animation Oklahoma State

osmotic pressure sanger

reverse osmosis gearfiltration

ion channels (cell biology, myocyte) CellsAlive

membrane animation HHMI

cell transport animation

Na/K pump Brookscole

membrane tutorial The Biology Project

week 10

Oct 24-28

 

 

  • test over eukaryotic cells and transport
    • aerobic respiration
 

ch 7

 

 

not active….

cell respiration -tutorial

overview  -respiration

electron transport chain and ATP synthesis -movies

glycolysis and fermentation -excellent molecular animations

glucose catabolism U. Alberta-overview

intro to metabolism animation

metabolic process location animation

citric acid cycle animation

oxydative phosphorylation electron transport chain animation

cellular respiration -Kimball’s pages

cellular respiration -showing molecular models

glycolysis explained in animations

oxydative phosphorylation explained inanimations

 

week topics text chapters/ activities supplemental readings/URLs
week 11

Oct 31-Nov 4

 

 

 

  • photosynthesis
  • test over respiration and photosynthesis

 

 

 

ch 8

 

 

 

photosynthesis light rxn -The Biology Project

photosynthesis carbo formation -The Biology Project

photosynthesis sites -Arizona State University

interactives

photosynthesis animation

oxygenic photosynthesis animation U. Alberta

photosynthesis -Maricopa bio181

 

week 12

Nov 7-11

  • The cell cycle

 

  • chromosomes

 

  • mitosis, meiosis/ gametogenesis, cell cycle

 

 

ch 9

Online Onion Tips  -The biology Project

more animations -BSC Courseware

BBC guru

cell cycle  -The Biology Project

cell cycle animation, myocyctes, cytoskeleton, apoptosis (cell biology, cell models, cell gallery)  -CellsAlive

human genome size

mitosis review -Nebraska Wesleyan

interactive mitosis (cell biology)-CellsAlive

mitosis -The Biology Project

modeling mitosis -Cornell BioG101-104  click either random or assignment……

mitosis/meiosis animations -About: homework help

mitosis and meiosis tutorial -Cornell BioG101-104

meiosis vs mitosis -WGBH (PBS)

cancer growth -WGBH (PBS)

metastasis -animation

week 13

Nov 14-18

 

  • principles of heredity
  • inheritance patterns

 

 

ch 10, 15

Dogs Around the World

mendelian genetics  -The Biology Project

human karyotyping  -The Biology Project

genetics -University of Utah

harlequin chromosomes -J. Kimball

drop/drag genetic cross ZeroBio

week 14

Nov 21-22

 

  • RNA and DNA structure and function

 

 

 

Nov 23-25 off Thanksgiving!

 

ch 11

http://ull.chemistry.uakron.

edu/genobc/animations/

 

BBC guru

 

Central Dogma animation

Anatomy of photo 51

DNA movies*

nucleic acids  -The Biology Project

Hershey/Chase -AccessExcellence

DNA basics -The Biology Project

DNA animations*

DNA replication animation

Cracking the code   (PBS)

DNA structure -subunit animation

 

 

 

week topics text chapters/ activities supplemental readings/URLs
week 15

 

 Nov 28-Dec 2

 

 

 

 

  • RNA and DNA structure  and function
  • mutations

ch  12
DNA structure/function

 
transformation movie

translation animation ZeroBio

protein synthesis animation

Beginner’s Guide to Molecular Biology -Molecular Biology Notebook

DNA from the Beginning -DNA from the Beginning

 

week 16

Dec 5-9

 

  • gene regulation
  • molecular genetics
  • genes and development

 

 

 

 

 

 

ch 13, 16

 

gene regulation -Indiana State University

RNA splicing -movie

cloning animation

developement  -The Biology Project

gene expression  -The Biology Project

gene expression in prokaryotes  -The Biology Project

signal transduction animation

week 17

Dec 12-16

 

  • DNA technology
ch 14

build a family tree

 

Sequence for Yourself      Create a DNA fingerprint

Blackett family  -The Biology Project

DNA profiling  -The Biology Project

Western Blot  -The Biology Project

recombinant DNA  -The Biology Project

PCR animation

human genome project -National Human Genome Research Institute

bacterial genetics/recombinant DNA tutorial -Cornell BioG101-104

forensics  -The Biology Project

genetic engineering

 

 

Dec 21

Winter Vacation starts at end of day

 

ch 17 are yours for the break–Happy Holidays!! (you would be bored without this work….

End of First semester

 

 

week topics text chapters/activities supplemental readings/URLs
week 18

Jan 3-7

 

 

  • Darwinian evolution
  • evolutionary changes in populations

 

ch 1718

Evolution

 

The Shape of Life

dating techniques -Geology Labs on-line

Darwin’s life and works -Cal. State University, Dr. C. Urbanowicz

population drift -simulation

genetic drift -simulation

week 19

Jan 10-14

Jan 14-end of Sem I

 

semester I final exam week

 

 

 

AP sample exam

 

 

 

 

 

Ap Genetics Solutions

 

Problem 1
  A rooster with gray feathers is mated with a hen of the same phenotype. Among their offspring, 15 chicks are gray, 6 are black, and 8 are white.

 

What is the simplest explanation for the inheritance of these colors in chickens?

Incomplete or codominance. Feather color is controlled by 2 genes B = black and b = white. The third phenotype is the result of a 50-50 mix of black and white to produce gray.

The 15 gray, 6 black, and 8 white birds represent a 2:1:1 ratio&emdash;the result of mating two heterozygous individuals: (Bb x Bb)

 

1 BB : 2 Bb : 1 bb
 

What offspring would you predict from the mating of a gray rooster and a black hen?

A gray rooster (Bb ) mated to a black hen (BB ) can be represented by the following Punnett square:

 

50% of the offspring should be gray (Bb ) and 50% black (BB )

 

Problem 2
  In some plants, a true-breeding, red-flowered strain gives all pink flowers when crossed with a white-flowered strain: RR (red) x rr (white) —> Rr (pink).

 

If flower position (axial or terminal) is inherited as it is in peas what will be the ratios of genotypes and phenotypes of the generation resulting from the following cross: axial-red (true-breeding) x terminal-white?

Note: Axial (A ) is dominant over terminal (a ).

The genotypes of the parents are AARR and aarr. Therefore the gametes of the parents must be AR and ar so the genotype for all the offspring in the F1 generation will be AaRr, and their phenotype will be axial-pink.

 

What will be the ratios in the F2 generation?

The ratio of genotypes can be determined by examining the Punnett square below:

The ratio of phenotypes will be:

 

6 axial-pink 8 pink
3 axial-red 4 red
3 axial-white 4 white
2 terminal-pink 12 axial
1 terminal-white 4 terminal
1 terminal-red

 

Problem 3

 

Flower position, stem length, and seed shape were three characters that Mendel studied. Each is controlled by an independently assorting gene and has dominant and recessive expression as follows:

 

Character Dominant Recessive
Flower position Axial (A ) Terminal (a )
Stem length Tall (T ) Dwarf (t )
Seed shape Round (R ) Wrinkled (r)

If a plant that is heterozygous for all three characters were allowed to self-fertilize, what proportion of the offspring would be expected to be as follows: (Note – use the rules of probability (and show your work) instead of huge Punnett squares)

a) homozygous for the three dominant traits

AATTRR = 1/4 x 1/4 x 1/4 = 1/64

 

b) homozygous for the three recessive traits

aattrr = 1/4 x 1/4 x 1/4 = 1/64

 

c) heterozygous (assumed for each trait)

AaTtRr = 1/2 x 1/2 x 1/2 = 1/8

 

d) homozygous for axial and tall, heterozygous for seed shape

AATTRr = 1/4 x 1/4 x 1/2 = 1/32

Problem 4
  A black guinea pig crossed with an albino guinea pig produced 12 black offspring. When the albino was crossed with a second one, 7 blacks and 5 albinos were obtained.

 

What is the best explanation for this genetic situation?

Black is dominant over white

 

Write genotypes for the parents, gametes, and offspring.

First cross:

 

Parent’s genotypes = BB (black) x bb (white)
gametes = B b
F1 offspring = all Bb

Second cross:

 

Parent’s genotypes = Bb (black) x bb (white)
gametes = B or b b
F1 offspring = Bb or bb

There should be 50% black to 50% white offspring in this cross.

Problem 5
  In sesame plants, the one-pod condition (P ) is dominant to the three-pod condition (p ), and normal leaf (L ) is dominant to wrinkled leaf (l) . Pod type and leaf type are inherited independently. Determinine the genotypes for the two parents for all possible matings producing the following offspring:

 

a. 318 one-pod normal, 98 one-pod wrinkled

Parental genotypes: PPLl x PPLl or PpLl x PPLl

 

b. 323 three-pod normal, 106 three-pod wrinkled

Parental genotypes: ppLl x ppLl

 

c. 401 one-pod normal

Parental genotypes: PPLL x PpLL or PPLl x PPLL or PPLL x PpLl etc (nine possible genotypes).

 

d. 150 one-pod normal, 147 one-pod wrinkled, 51 three-pod normal, 48 three-pod wrinkled. (a 3: 3: 1: 1 ratio)

Parental genotypes: PpLl x Ppll (see below for details)

3 One-pod normal (PPLl , PpLl , PpLl)

3 One-pod wrinkled (PPll , Ppll , Ppll)

1 Three-pod normal (ppLl)

1 Three-pod wrinked (ppll)

 

e. 223 one-pod normal, 72 one-pod wrinkled, 76 three-pod normal, 27 three-pod wrinkled (a 9: 3: 3: 1 ratio)

Parental genotypes: PpLl x PpLl

Problem 6
  A man with group A blood marries a woman with group B blood. Their child has group O blood. What are the genotypes of these individuals?

 

Father = AO (or IAi)

Mother = BO (or IBi)

First Child = OO (or ii)

 

What other genotypes and in what frequencies, would you expect in offspring from this marriage?

Examine the Punnett square to determine the other genotypes possible.

The other genotypes for children are (according to Campbell’s system): 1/4 IAIB, 1/4 IAi, 1/4 IBi

Problem 7
  Color pattern in a species of duck is determined by one gene with three alleles. Alleles H and I are codominant, and allele i is recessive to both. Note: this situation is similar to the ABO blood system.

 

How many phenotypes are possible in a flock of ducks that contains all the possible combinations of these three alleles?

 

As in the ABO blood system 4 phenotypes are possible in this case:

 

Genotype Phenotype
HH, Hi (H)
II, Ii (I)
HI (HI)
ii (i)

 

Problem 8
  Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her husband are both carriers, what is the probability of each of the following?

 

Under these circumstances assume the following Punnett square to be true.

Where NN or Nn = normal conditions and nn = PKU

 

a. all three of their children will be of normal phenotype

3/4 x 3/4 x 3/4 = 27/64

 

b. one or more of the three children will have the disease (x)

1 – 27/64 = 37/64

 

All three have x 2 out of 3 has x 1 out of 3 has x
+ + =
x x o o o x
3 Combinations x o x o x o
o x x x o o
+ 3(3/4 x 1/4 x 1/4) + 3(3/4 x 3/4 x 1/4) =

Note: the probability of the disease (x) = 1/4 & the probability of being normal (o) = 3/4

 

c. all three children will have the disease

1/4 x 1/4 x 1/4 = 1/64

 

d. at least one child out of three will be phenotypically normal

(Note: Remember that the probabilities of all possible outcomes always add up to 1)

1 – 1/64 = 63/64

Problem 9
  The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring would have the following genotypes?

 

 

a. aabbccdd = x x x = 1/256

b. AaBbCcDd = x x x = 1/16

c. AABBCCDD = x x x = 1/256

d. AaBBccDd = x x x = 1/64

e. AaBBCCdd = x x x = 1/128

Just remember that the probability of a heterozygote (Xx) = 2/4 or 1/2 and the probability of a homozygote XX or xx = 1/4

Problem 10
  In 1981, a stray black cat with unusual rounded curled-back ears was adopted by a family in California. Hundreds of descendants of the cat have since been born, and cat fanciers hope to develop the “curl” cat into a show breed. Suppose you owned the first curl cat and wanted to develop a true breeding variety.

 

How would you determine whether the curl allele is dominant or recessive?

Mate the stray to a non-curl cat. If any offspring have the “curl” trait it is likely to be dominant. If the mutation is recessive, then on ly non-curl offspring will result.

 

How would you select for true-breeding cats?

You know that cats are true-breeding when curl crossed with curl matings produce only curl offspring.

 

How would you know they are true-breeding?

A pure-bred “curl cat” is homozygous.

  1. If the trait is recessive any inividual with the “curl” condition is homozygous recessive.
  2. If the trait is dominant you can determine if the individual in question is true breeding (CC) or heterozygous (Cc) with a test cross (to a homozygous recessive individual).

 

Problem 11
  What is the probability that each of the following pairs of parents will produce the indicated offspring (assume independent assortment of all gene pairs?

 

a. AABBCC x aabbcc —-> AaBbCc

(1)(1)(1) = 1
 

b. AABbCc x AaBbCc —–> AAbbCC

()()() = 1/32
 

c. AaBbCc x AaBbCc —–> AaBbCc

()()() = 1/8
 

d. aaBbCC x AABbcc —-> AaBbCc

(1)()(1) = 1/2
 

Problem 12
  Karen and Steve each have a sibling with sickle-cell disease. Neither Karen, Steve, nor any of their parents has the disease, and none of them has been tested to reveal sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple should have another child, the child will have sickle-cell anemia.

 

In order for Karen and Steve to have siblings with sickle cell anemia their parents must be carriers (Nn). We also know that John and Carol are not homozygous recessive (nn) because they do not have the disease. Therefore the chance that Karen is a carrier is 2/3 (NN, Nn, nN) and the chance that Steve is a carrier is also 2/3. If they have a child and both Karen and Steve are carriers then the child has one chance in 4 of having sickle cell anemia. Since each event is independent of one another the overall probability of the child having sickle cell anemia is:

2/3 x 2/3 x 1/4 = 1/9.

 

Problem 13
  Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the disease and blood group are independently inherited. A normal man with type A blood and a normal woman with type B blood have already had one child with the disease. The woman is now pregnant for a second time. What is the probability that the second child will also have the disease? Assume both parents are heterozygous for the “disease” gene.

 

 

 

Genotype
Father AO
Mother BO

OO expresses the disease

 

The second child’s chance of having the disease is = x = 1/16

 

Problem 14
  In tigers, a recessive allele causes an absence of fur pigmentation (a “white tiger”) and a cross-eyed condition. If two phenotypically normal tigers that are heterozygous at this locus are mated, what percentage of their offspring will be cross-eyed? What percentage will be white?

 

Using the Punnett square below where P = normal pigmenation and p = white

then 25% will be white (pp) and all of the white offspring will also be cross-eyed

 

 

Problem 15
  In corn plants, a dominant allele I inhibits kernel color, while the recessive allele i permits color when homozygous. At a different locus, the dominant gene P causes purple kernel color, while the homozygous recessive genotype pp causes red kernels. If plants heterozygous at both loci are crossed, what will be the phenotypic ratio of the F1 generation?

 

Phenotypic ratios: White (I _ _ _) = 12

Purple (i i P_ ) = 3

Red ( i i p p) = 1

The dominant allele I is epistatic to the p locus, and thus the F1 generation will be:

9 I_P_ : colorless

3 I_pp : colorless

3 i iP_ : purple

1 i i pp: red

 

Problem 16
  The pedigree below traces the inheritance of alkaptonuria, a biochemical disorder. Affected individuals, indicated here by the filled-in circles and squares, are unable to break down a substance called alkapton, which colors the urine and stains body tissues. Does alkaptonuria appear to be caused by a dominant or recessive allele?

 

Recessive

Fill in the genotypes of the individuals whose genotypes you know. What genotypes are possible for each of the other individuals?

If alkaptonuria is recessive George must be a carrier. See below.

 

If alkaptonuria is dominant Carla could not have the disease, as indicated in the pedigree chart, since the parents do not express the trait. See Below.

 

Problem 17
  A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits (5). Extra digits is a dominant trait. What fraction of this couple’s children would be expected to have extra digits?

 

Because the daughter is normal the man’s genotype must be heterozygous for the trait so:

if X = extra digits and x = normal (5) digits then:

 

50% of the offspring will be polydactylic

 

 

Problem 18
  Imagine you are a genetic counselor, and a couple planning to start a family came to you for information. Charles was married once before, and he and his first wife had a child who has cystic fibrosis. The brother of his current wife Elaine died of cystic fibrosis. Cystic Fiborsis is a lethal recessive condition (a person with CF cannot have children).

 

What is the probability that Charles and Elaine will have a baby with cystic fibrosis? (Neither Charles nor Elaine has the disease)

The Probability that Elaine is a carrier is 2/3 (She does not have Cystic Fibrosis which eliminates one of the 4 possibilities. She does have 2 chances out of three of being a carrier. [Nn, nN (carriers) or NN]

The probability that the baby (?) has the disease (if Elaine is a carrier) is 1/4

The total probability is 2/3 x 1/4 or 1/6.

 

Problem 19
  In mice, black color (B ) is dominant to white (b ). At a different locus, a dominant allele (A ) produces a band of yellow just below the tip of each hair in mice with black fur. This gives a frosted appearance known as agouti. Expression of the recessive allele (a ) results in a solid coat color. If mice that are heterozygous at both loci are crossed, what will be the expected phenotypic ratio of their offspring?

 

B = Black – dominant

A = Agouti – dominant

 

BbAa x BbAa =

 

Genotype Phenotype
1
BBAA agouti
2
BbAA agouti
2
BBAa agouti
4
BbAa agouti
1
BBaa black
2
Bbaa black
1
bbAA white
2
bbAa white
1
bbaa white

The phenotypic ratio is:

9 agouti: 4 white: 3 black

 

Problem 20
  The pedigree below traces the inheritance of a vary rare biochemical disorder in humans. Affected individuals are indicated by filled-in circles and squares. Is the allele for this disorder dominant or recessive?

 

The allele is most likely dominant because the #2 individual (see below) with the trait marries a woman with the trait and 50% of their offspring are normal. If the trait were recessive one would expect the following:

 

100% of offspring would have the disease, which is not the case.

 

 

What genotypes are possible for the individuals marked 1, 2, and 3?

 

 

  1. Bb (heterozygous)
  2. Bb ( if 2’s genotype were bb he would not have the disease and if BB all his children would have the condition.)
  3. bb (all normal individuals are homozygous recessive)

 

 

 

Ap Lab 1 Sample 5

 

Osmosis & Diffusion – Lab 1 

Introduction:

All molecules have kinetic energy and are constantly in motion.  This motion causes the molecules to bump into each other and move in different directions.  The result is diffusion.  Diffusion is the random movement of molecules from an area of high concentration to an area of low concentration. This will continue until dynamic equilibrium is reached; no net movement will occur.  Osmosis is a special kind of diffusion.  It is the diffusion of water through a selectively permeable membrane. A selectively permeable membrane means that the membrane will only allow certain molecules through such as water, small solutes, oxygen, carbon dioxide, and glucose, because no additional ATP is required. The membrane will not let ions, nonpolar molecules, or large molecules through because extra ATP is needed for them to travel across the membrane.  Active transport is how molecules (such as ions) move against the concentration gradient.  Additional ATP is required to perform this process.

Water will travel from an area of high water potential to an area of low water potential.  Water potential is the measure of free energy of water in a certain solution.  It is measured by using the Greek letter psi (ψ).  The formula for figuring water potential is:

ψ          =             ψp             +           ψs

Water Potential   =   Pressure Potential   +  Solute Potential

Water potential is affected by 2 different factors.  They are the addition of a solute and the pressure potential.  If a solute is added to the water, then the water potential is lowered.  If more pressure is placed on the water, then the potential is raised. The addition of a solute and water potential are inversely proportional.  Pressure being placed onto the water and the potential of the water are directly proportional.

Solutions can have three relationships with each other; isotonic, hypertonic, or hypotonic.  When the solutions have the same concentration of solutes, they are isotonic.  There is no net change in the amount of water on each side of the membrane.  If the solutions differ in their solute concentrations, the solution that has the most solute is hypertonic to the other solution.  The solution with the smaller amount of solute is hypotonic to the other solution. The net movement of water will be from the hypertonic solution to the hypotonic solution. Net movement will occur until dynamic equilibrium is reached, then there will be no net movement of water.

Hypothesis:

In this lab, osmosis and diffusion will occur between the solutions of different concentration until dynamic equilibrium is reached and there is no net movement of water.

Materials:

Exercise 1A:

The materials used include a 30cm piece of 2.5cm dialysis tubing, string, scissors, 15mL of 15% glucose/1% starch solution, 250mL beaker, distilled water, and 4mL of Lugol’s solution (Iodine Potassium-Iodine or IKI).

Exercise 1B:

This exercise required six 30cm strips of presoaked dialysis tuning, six 250mL cups or beakers, string, scissors, a balance, and 25mL of  these solutions: distilled water, 0.2M sucrose, 0.4M sucrose, 0.6M sucrose, 0.8M sucrose, and 1.0M sucrose.

Exercise 1C:

The materials that were required include 100mL of these solutions: distilled water, 0.2M sucrose, 0.4M sucrose, 0.6M sucrose, 0.8M sucrose, and 1.0M sucrose, six 250mL beakers or cups, a potato, a cork borer, a balance, paper towel, and plastic wrap.

Exercise 1D:

The materials used include a calculator, and a pencil.

Procedure:

Exercise 1A:

Soak the dialysis tubing in water.  Tie off one end of the tubing to form a bag.  Open the bag and place the glucose/starch solution in it.  Tie off the other end of the bag, leaving enough room for expansion of the contents in the bag.  Record the color of the solution in Table 1.1.  Next, test the glucose/starch solution for the presence of glucose.  Record the results in Table 1.1.  Fill a 250mL beaker or cup with 2/3 full with distilled water.  Add 4mL of Lugol’s solution to the distilled water and record the color of the solution in Table 1.1.  Test the solution for glucose and record the results in Table 1.1.  Immerse the bag in the beaker of solution.  Allow the beaker and bag to stand for approximately 30 minutes or until you see a distinct color change in the bag and the beaker.  Record the final color of the solution in the bag, and the solution in the beaker, in Table 1.1.  Test the liquid in the beaker and in the bag for the presence of glucose.  Record the results in Table 1.1.

Exercise 1B:

Obtain the six strips of presoaked dialysis tubing and create a bag out of each one by tying off one end.  Pour 25mL of the 6 solutions into separate bags. Tie off the other end of the 6 bags.  Rinse each bag gently with distilled water and blot dry.  Determine the mass of each bag and record it in Table 1.2.  Immerse each bag in one beaker filled will distilled water and label the beaker to indicate the molarity of the solution in the bag.  Let the setups stand for 30 minutes.  Remove the bags from the water.  Carefully blot them dry and determine their masses.  Record them in Table 1.2.  Obtain the other lab groups data to complete Table 1.3.

Exercise 1C:

Pour 100mL of the solutions into a labeled 250mL beaker.  Use a cork borer to cut potato cylinders.  You need 4 cylinders for each cup.  Determine the mass of the 4 cylinders together and record the amount in Table 1.4.  Place the cylinders into the beaker of sucrose solution.  Cover the beaker with plastic wrap to prevent evaporation.  Let it stand overnight.  Remove the cores from the beaker and blot them gently on a paper towel and determine their total mass.  Record the results in Table 1.4.  Calculate the percentage change.  Do this for the individual and class data.  Graph the class average percentage change in mass.

Exercise 1D:

Determine the solute, pressure, and water potential of the sucrose solution.  Then, graph the information that is given about the zucchini cores.

Results:

Exercise 1A:

 Table 1.1

 

Initial Contents Initial Color Final Color Initial Presence of Glucose Final Presence of Glucose
Bag 15% glucose & 1% starch Cloudy White Purple Yes Yes
Beaker Water & IKI Brown Orange No Yes

 

  1. Which substances are entering the bag and which are leaving the bag? What evidence supports the answer?  Distilled water and IKI are  leaving and entering.  Glucose is able to leave the bag.
  2. Explain the results that were obtained.  Include the concentration differences and membrane pore size in the discussion.  Glucose and small molecules were able to move through the pores.  Water and IKI moved from high to low concentration.
  3. How could this experiment be modified so that quantitative data could be collected to show that water diffused into the dialysis bag?  You could mass the bag before and after it was placed into the solution.
  4. Based on your observations, rank the following by relative size, beginning with the smallest: glucose molecules, water molecules, IKI molecules, membrane pores, and starch molecules.  Water molecules, IKI molecules, Glucose molecules, Membrane pores, and Starch molecules
  5. What results would you expect if the experiment started with a glucose and IKI solution inside the bag and only starch and water outside?  The glucose and IKI would move out of the bag and turn the starch and water solution purple/blue.  The starch couldn’t move inside the bag because its molecules are too big to pass through the membrane of the tubing.

Exercise 1B:

 

Table 1.2: Dialysis Bag Results: Individual Data

 

Contents in dialysis bag Initial mass (g) Final mass (g) Mass difference (g) % Change in mass
Distilled Water 24.7 23.7 1 4.1
0.2M 26.7 27.4 .7 2.62
0.4M 27.4 29 1.6 5.84
0.6M 25.9 29 3.1 12
0.8M 29 32.6 3.6 12.41
1.0M 28 33.7 5.7 20.4

 

Table 1.3: Dialysis Bag Results: Class Data

 

Group 1

Group 2

Group 3

Total Class Average
Distilled Water 4.1% .7% 1.6% 6.4% 2.13%
0.2M 2.62% 6.4% 4.1% 13.12% 4.37%
0.4M 5.84% 9.9% 9.5% 25.24% 8.41%
0.6M 12% 13.4% 9.3% 34.37% 11.57%
0.8M 12.41% 14.6% 15.2% 42.21% 14.07%
1.0M 20.4% 19.7% 15.9% 56% 18.67%

 

  1. Explain the relationship between the change in mass and the molarity of sucrose within the dialysis bags.  The solute is hypertonic and water will move into the bag.  As the molarity increases the water moves into the bag.
  2. Predict what would happen to the mass of each bag in this experiment if all the bags were placed in a 0.4M sucrose solution instead of distilled water.  Explain.  With the 0.2M bag, the water would move out.  With the 0.4M bag, there will be no net movement of water because the solutions reach dynamic equilibrium.  With the 0.6M-1M bags, the water would move into the bag.
  3. Why did you calculate the percent change in mass rather than simply using the change in mass?  This was calculated because each group began with different initial masses and we would have different data.  All the groups needed consistent data.
  4. A dialysis bag is filled with distilled water and then places in a sucrose solution.  The bag’s initial mass is 20g and its final mass is 18g.  Calculate the percent change of mass, showing your calculations.  ((18-20)/20) x 100 = 10%
  5. The sucrose solution in the beaker would have been hypotonic to the distilled water in the bag.

Exercise 1C

 

Table 1.4: Potato Core: Individual Data

 

Contents of Beaker Initial Mass (g) Final Mass (g) Difference in Mass % Change in Mass
Distilled Water 2.8 3.7 .9 32.14
0.2M 2.9 3.1 .2 7
0.4M 2.5 2.2 .3 12
0.6M 2.3 1.9 .4 17.39
0.8M 2.5 1.9 .6 24
1.0M 2.3 1.8 .5 21.74

 

Table 1.5: Potato Core: Class Data

 

Group 1 Group 2 Total Class Average
Distilled Water 32.14% 21.1% 53.24% 26.62%
0.2M 7% 6.7% 13.7% 6.85%
0.4M -12% -6.5% -18.5% -9.25%
0.6M -17.39% -15.2% -32.59% -16.30%
0.8M -24% -20% -44% -22%
1.0M -21.74% -19% -40.74% -20.37%

 

Determine the molar concentration of the potato core.  0.3M

Exercise 1D

 

 

What is the molar concentration of the zucchini cores? .35M

 

  1. If a potato core is allowed to dehydrate by sitting in the open air, would the water potential of the potato cells decrease or increase? Why?  It would decrease because the water would leave the cells and cause the water potential to go down.
  2. If a plant cell has a lower water potential than its surrounding environment and if pressure is equal to zero, is the cell hypertonic or hypotonic to its environment? Will the cell gain water or lose water?  It is hypotonic and it will gain water.
  3. The beaker is open to the atmosphere.  What is the pressure potential of the system?  The pressure potential is zero.
  4. Where is the greatest water potential?  In the dialysis bag.
  5. Water will diffuse out of the bag. Why? It is because the water moves from and area of high water potential to an area of lower water potential.
  6. What effect does adding solute have on the solute potential component of that solution? Why?  It makes is more negative.
  7. Consider what would happen to a red blood cell placed in distilled water: a) Which would have the higher concentration of water molecules?  Distilled Water  b) Which would have the higher water potential?  Distilled Water  c)  What would happen to the red blood cell? Why?  It would lyce, because it would take on too much water.

Error Analysis:

Possible errors that could have affected the results of the lab include incorrectly mixing the solutions, ineffectively tying the dialysis tubing, inaccurately measuring , and inaccurately calculating.

Conclusion:

            During Exercise 1A the data that was collected help determine which molecules can and can not move across a cell membrane. Obviously, because of the color change in the bag, the IKI was able to move across the membrane.  It is small enough to fit through the pores in the selectively permeable membrane, along with water.  Starch was too large to move across the membrane. Glucose, as the Benedict’s test proves, was able to move freely along with the water and IKI solution.

In Exercise 1B, it was proven that water moves faster across the cell membrane than sucrose.  The water moved to help reach dynamic equilibrium between the 2 solutions.  The sucrose molecules are too big to move across the membrane as fast as water can.

The data in Exercise 1C showed that the potatoes contained sucrose.  The sucrose in the potato raised the solute potential, which lowered the water potential.  The beaker of distilled water had a high water potential.  Water moves down the concentration gradient, causing the potato cores to take on water.

Exercise 1D helped better understand the lab with simple algebra equations.  It proved that the data that was collected was correct through mathematics.

 

AP Biology Powerpoints 8th ed

 

 

 

AP Biology PowerPoints

Chapter 1 Introduction Chapter 20 Biotechnology Chapter 39 Plant Responses
Chapter 2 Biochemistry Chapter 21 Genomes Chapter 40 Animal Form & Structure
Chapter 3 Water Chapter 22 Darwin Evolution Chapter 41 Animal Nutrition
Chapter 4 Carbon Chemistry Chapter 23 Population Evolution Chapter 42 Circulation & Respiration
Chapter 5 Macromolecules Chapter 24 Origin of Species Chapter 43 Immune System
Chapter 6 The Cell Chapter 25 Earth History Chapter 44 Osmoregulation
Chapter 7 Cell Membranes Chapter 26 Phylogeny Chapter 45 Endocrine System
Chapter 8 Metabolism Chapter 27 Archea & Bacteria Chapter 46 Animal Reproduction
Chapter 9 Cellular Respiration Chapter 28 Protists Chapter 47 Animal Development
Chapter 10 Photosynthesis Chapter 29 Plant Diversity I Chapter 48 Neurons
Chapter 11 Cell Communication Chapter 30 Plant Diversity II Chapter 49 Nervous System
Chapter 12 Cell Cycle Chapter 31 Fungi Chapter 50 Senses
Chapter 13 Meiosis Chapter 32 Introduction to Animals  Chapter 51 Behavior
Chapter 14 Mendel Chapter 33 Invertebrates Chapter 52 Ecology
Chapter 15 Chromosomes Chapter 34 Vertebrates Chapter 53 Population Ecology
Chapter 16 Molecular Inheritance Chapter 35 Plant Structure Chapter 54 Community Ecology
Chapter 17 From Gene to Protein Chapter 36 Transport in Plants Chapter 55 Ecosystems
Chapter 18 Gene Expression Chapter 37 Plant Nutrition Chapter 56 Conservation
Chapter 19 Viruses Chapter 38 Angiosperms