Download File

Read and answer questions (in
your brain) from pages 596-597.
Pick up page 1 of the “Falcon
Family Murder”. Fill in the
pedigree symbols on the back of
this sheet from the legend on page
605 of your text.
What follows is an account of the
famous Falcon Family Murder. The
case remained unsolved for over 20
years but, finally, there was enough
evidence to convict a killer. As you
go through each stage of the
investigation you will learn the
science you need to narrow down
the suspects and find the killer.
OBJECTIVES: STUDENTS WILL…
• CONSTRUCT AN ACCURATE PEDIGREE CHART FOR
THE FALCON FAMILY
• LEARN ABOUT GENE INTERACTIONS
• DETERMINE ANY GENOTYPES POSSIBLE FOR THE
TRAITS OF BLOOD TYPE AND FOOT SHAPE.
• USE THE EVIDENCE FROM THE MURDER TO MAKE A
PRELIMINARY LIST OF SUSPECTS.
• PRODUCE A KARYOTYPE FROM THE EVIDENCE TO
ELIMINATE SUSPECTS.
• CONSTRUCT A DNA FINGERPRINT FOR THE
EVIDENCE AND CERTAIN FAMILY MEMBERS THEN
INTERPRET IT TO NARROW DOWN POTENTIAL
SUSPECTS.
• TRANSCRIBE AND TRANSLATE A GENE TO FIND THE
KILLER’S PHENOTYPE AND ELIMINATE ALL BUT ONE
SUSPECT.
Grampa Jim Falcon and Gramma Trudy Falcon had five kids. A
daughter, Jayni who married Scott and had three brown eyed
sons: Jason, Greg, and Dave. Jayni had a kid sister, Kerri-Ann,
who married Myles. These two had two sons, Ray and Roy, a
daughter, Cynthia, and another son, Dean (all four had blue
eyes). Dean married Audra at a very young age and had a
precocious daughter, Bonnie. Jim and Trudy also had a pair of
identical twin boys, Bobby and Denny. Denny married Nina and
they had four children: blue eyed Lori, blue eyed Carol (who
was born with Down Syndrome), brown eyed Jo-Anne, and
blue eyed Larry. Jim and Trudy’s youngest was a son, Virgil,
who married Jody and were blessed with three charming, blue
eyed daughters: Lynne, Lynda, and Karissa.
CONSTRUCT THE PEDIGREE CHART FOR THIS FAMILY.
A. CONSTRUCT A PEDIGREE CHART
A pedigree chart is like a family tree. It
shows the generations of parents and
offspring, matings, and sexes of
individuals. It can be used to predict
genetic possibilities for certain traits. In
biological pedigrees, certain symbols are
used to designate certain things...
WHAT is a PEDIGREE??
PEDIGREE PRACTICE
•Do the pedigree chart case
study (WITHOUT
GENETICS!!) (JUST
circles, squares, lines, and
numbers.)
•Check with teacher.
•Do Falcon Family
Pedigree.
PEDIGREE CASE STUDY
I
LINE A
LINE B
1
2
II
1
2
3
4
5
6
7
III
1
2
3
4
5
6
7
IV
1
2
3
4
5
B. FILL IN GENOTYPES
Grampa Jim and Gramma Trudy were set in their
ways and did not believe in doctors or modern
medicine. Consequently, they were never tested
to see what type of blood they had. Before having
their kids, Jayni was tested, however, and was
found to be AB+. Kerri-Ann was tested and found
to be A-. Virgil was B- and Bobby was also B-.
Denny was as ornery as his old man and always
refused to be tested. Scott was found to be O+, as
was Jody, and Nina. Ray was O+, his sister-in-law
Audra was A-, and her daughter, precious little
Bonnie, was AB+. WHAT CAN WE TELL ABOUT
THE GENETICS OF THE FAMILY??
GENETICS
HEREDITY - passing of traits from parents to offspring
GENES - pieces of DNA in chromosomes which instruct
the cell how to produce certain proteins.
Gregor Mendel (1822-1884)
•
didn’t know about genes but worked with pea plants.
• Peas are good to work with because:
1) easy (fast, cheap) to grow
2) many offspring from one cross
3) easy to cross (self or cross pollinate)
4) they have many different characteristics with only
two combinations (ex. green vs. yellow, tall vs. short)
• Mendel crossed pea plants and carefully studied what
happened to the various traits through the generations.
When crosses are done, the generations are given
the following designations:
• P = Parental; F1 = First Filial generation; F2 = 2nd Filial gen. etc.
When Mendel crossed Tall plants with Tall plants, the F1
were all TALL
When he crossed short plants
with short plants, the F1 were
all SHORT
When he crossed Tall plants with short plants, the F1
were ALL TALL
Mendel determined that traits were controlled by
factors (genes) and some factors were
DOMINANT to others (the effects were seen for
one even in the presence of another). Traits not
expressed are RECESSIVE.
TALL is DOMINANT to short because it is seen
in the F1 and the short trait “disappears”
NOTE: DOMINANT does NOT mean “BETTER”
or even “MORE COMMON”. In some organisms,
short may even be dominant to tall. Type O
blood is recessive to type AB blood but is more
common.
The dominant gene is sometimes named by using a
CAPITAL letter and the recessive gene uses the
same letter in lower case (ex. Tall=T, short=t)
Different forms of the gene are called ALLELES.
Mendel then crossed the Tall F1 plants with one another (or
self pollinated them). What do you think he saw in the F2?
75% Tall : 25% short
In a plant, there are two copies of every gene for each trait.
To show the Dominant trait, the plant only needed to have
ONE copy of the dominant allele (T) but to show the
recessive trait BOTH copies had to be recessive (tt).
The “purebred” tall plants of the P generation would have
been TT. All the short plants would be tt. When both
alleles are the same, they are HOMOZYGOUS.
When a tall (TT) was crossed with a short (tt) the resulting
F1 would ALL have been Tt and were ALL tall. They were
HETEROZYGOUS (also called HYBRIDS).
When an organism makes sex cells, the two genes for a
trait separate or SEGREGATE (during meiosis I).
Because of this, a hybrid tall plant (Tt) could pass on either
a T allele OR a t allele to its offspring.
The gene compliment an organism has is called its
GENOTYPE (TT, Tt, or tt)
The physical expression of the genes (how it “looks”) is
called its PHENOTYPE (Tall, short)
GENOTYPIC OR PHENOTYPIC RATIOS – How many of each
genotype and phenotype there are.
MONOHYBRID CROSS - crossing one gene pair of
contrasting traits (tall x short; green x yellow)
On page 600, Answer #1-3 verbally to a partner then
answer #4-5 in your notebook.
On page 600,
Answer #1-3
verbally to a
partner then
answer #4-5 in
your notebook.
QUESTIONS – PG 600, #4-5
4 a). What is the difference in genotypes between a
homozygous black dog and a heterozygous black dog?
Homozygous black = BB
Heterozygous black = Bb
4 b). COULD THE HETEROZYGOUS BLACK DOG
HAVE THE SAME GENOTYPE AS THE YELLOW
DOG?
NO. THE YELLOW DOG (RECESSIVE
PHENOTYPE) MUST HAVE A GENOTYPE OF
bb WHERAS THE HETEROZYGOUS DOG HAS
A GENOTYPE OF Bb.
QUESTIONS – PG 600, #4-5
5. Heterozygous ROUND crossed with WRINKLED
a) GENOTYPE OF PARENTS:
ROUND: Rr
WRINKLED: rr
b) GAMETES PRODUCED from round:
R AND r
c) GAMETES PRODUCED from wrinkled:
r AND r
d) Possible genotypes and phenotypes of F1:
Rr (Round) and rr (wrinkled)
Genetics problems often involve determining probabilities of
certain crosses producing various results. These predictions
can be done
1) Mathematically
2) Using a Punnett Square
1) Mathematically – Genetics involves determining the
mathematical probability (P) of something happening.
P = number of ways that a given outcome can occur
total number of possible outcomes
Ex) flipping heads on a coin = ONE head/TWO possibilities
P = 1/2
PRODUCT RULE: the probability of more than one event
occuring simultaneously is equal to the probability of each
event multiplied together.
ex) The probability of rolling snake eyes on two dice is…
1/6 x 1/6 = 1/36
The probability of a boy picking a pair of kings on Tuesday…
½ x 1/13 x 1/17 x 1/7 = 1/3094
DO THE PROBABILITY
ACTIVITY WITH A
PARTNER.
YOU NEED:
- ONE SHEET PER PAIR
- TWO COINS PER
PAIR
Genetics problems often involve determining probabilities of
certain crosses producing various results. These predictions
can be done
1) Mathematically
2) Using a Punnett Square
2) PUNNETT SQUARE - a geneticist’s chart to map crosses.
Made by putting the possible gametes from one parent
across the top and the possible gametes from the other
parent down the left side and charting the resulting, possible
zygotes.
Sample exercise 1, pg. 601
A breeder crosses a wrinkled (r) seed plant with a round (R)
seed plant. The round seed is heterozygous. Determine
genotypic and phenotypic ratios for this cross.
What are the genotypes of the parents?
Gametes from
wrinkled on top…
wrinkled
round
Gametes from
round on left…
Fill in the resulting
genotypes…
Rr
rr
RATIOS:
Rr
rr
GENOTYPES: 1:1
Rr:rr
PHENOTYPES: 1:1
ROUND:WRINKLED
PUNNETT SQUARES
• Mendel crossed pure breeding tall plants with
short plants (pure breeding = ____________ )
– DRAW THE PUNNETT SQUARE FOR THIS
CROSS.
• He then crossed two of the F1.
– DRAW THE PUNNETT SQUARE FOR THIS
CROSS.
TEST CROSS
In sheep, white wool is much nicer than black (black sheep
of the family = bad) and is dominant.
WHITE = W; BLACK = w
Imagine you were given a white
ewe and wanted to go into sheep
farming.
How could you test to see if a
white ewe was pure bred
(homozygous dominant) or
hybrid (heterozygous)?
TEST CROSS
ww
What is the genotype of a black sheep?
What is the genotype of the white ewe?
WW OR Ww
What would the F1 look like in either cross if the white was
crossed with a black?
WW x ww
W
W
Ww x ww
W
w
w
Ww
Ww
w
Ww
ww
w
Ww
Ww
w
Ww
ww
ALL WHITE
½ WHITE: ½ BLACK
So the only way an offspring can be black is if the ewe was
HETEROZYGOUS.
TEST CROSS
This is called a TEST CROSS.
When an individual showing the dominant phenotype but
unknown genotype is crossed with an individual who shows
the recessive phenotype (thus the homozygous recessive
genotype).
If any offspring show the recessive trait, the unknown is
HETEROZYGOUS. If all the offspring show the dominant
trait (and sufficient numbers of offspring are born to be
reliable) the unknown is probably HOMOZYGOUS
dominant.
A horticulture worker has seeds from a particular cross, but has no
information about the genotype or the phenotype of the parents. He
plants and grows the offspring, and records the traits of each
offspring (Table 1). What was the genotype and phenotype of the
parent plants?
POSSIBLE
GENOTYPES?
RR or Rr
rr
POSSIBLE PARENT GENOTYPES?
Could either be RR? NO – all F1 round
Could both be rr?
NO – all F1 wrinkle
Rr x rr
1:1
Rr x Rr
3:1
READ QUESTIONS
3 AND 4 ON PAGE
604 (TOP). BE
PREPARED TO
ANSWER THEM IN
5 MINUTES.
DO QUESTIONS 1
AND 2 AT THE
BOTTOM OF PAGE
604 IN YOUR
BOOKS.
REVIEW – PG 604, #1a-b
1. DALMATIONS: SPOTTED (S); NON-SPOT (s)
a) CROSS OF TWO HETEROZYGOTES
S
S
s
SS
Ss
s
Ss
ss
GENOTYPES: 1:2:1 SS:Ss:ss
PHENOTYPES: 3:1
SPOT:NON-SPOT
b) SPOTTED (S __ ) x UNKNOWN ( __ __ ) =>
F1 = 3 SPOT (S __ ); 3 NON-SPOT (ss)
PHENOTYPE OF UNKNOWN?
FATHER DEFINITELY GAVE s (TO NON-SPOTS)(AS DID
MOTHER.
IF DAD Ss, F1 = 3:1 SPOT:NON-SPOT.
IF DAD ss, F1 = 2:2 SPOT:NON-SPOT.
REVIEW – PG 604, #2
2. HAIRLESS (H) AND HAIRY (h).
F1 = 6 HAIRLESS; 2 HAIRY . WHAT ARE PARENTS?
EACH OF THE HAIRY PUPS HAVE A GENOTYPE OF hh.
THERFORE, EACH PARENT DONATED AN h ALLELE.
THE HAIRLESS PUPS HAVE TO HAVE AT LEAST ONE H
ALLELE SO AT LEAST ONE PARENT HAS TO HAVE
GIVEN AN H. ONE PARENT IS Hh.
IF THE OTHER PARENT IS hh, WE WOULD EXPECT THE
RATIO OF HAIRLESS:HAIRY IN F1 TO BE 1:1.
IF THE OTHER PARENT IS Hh, WE WOULD EXPECT
THE RATIO OF HAIRLESS:HAIRY IN F1 TO BE 3:1
MULTIPLE ALLELES
Mendel was lucky to study peas which had only two alleles
for each trait (T or t; Y or y; R or r)
Most traits have more than two possible alleles.
Ex. Fruit flies (drosophila melanogaster) have many
different possible eye colours. The wild type is red but
there are also apricot, honey, and white eyes.
At any time, one fly can only have 2 eye colour genes. Red
is dominant to apricot, dominant to honey, dominant to
white.
MULTIPLE ALLELES
CAPITAL LETTERS AND LOWER CASE DON’T WORK NOW.
INSTEAD, A LETTER REPRESENTING “EYE COLOUR” IS
USED WITH A SUPERSCRIPT NUMBER FOR EACH ALLELE:
E1 = WILD (RED) EYE ALLELE
DOMINANT TO
E2 = APRICOT EYE ALLELE
DOMINANT TO
E3 = HONEY EYE ALLELE
DOMINANT TO
E4 = WHITE EYE ALLELE
MULTIPLE ALLELES
WHAT GENOTYPES COULD A WILD TYPE HAVE?
E1E1, E1E2, E1E3, E1E4
WHAT GENOTYPES COULD A APRICOT TYPE HAVE?
E2E2, E2E3, E2E4
WHAT GENOTYPES COULD A HONEY TYPE HAVE?
E3E3, E3E4
WHAT GENOTYPES COULD A WHITE TYPE HAVE?
E4E4
MULTIPLE ALLELES
(P. 609, Sample 1)
A CROSS IS MADE OF A E1E4 AND A E2E3.
WHAT ARE THE PHENOTYPES OF THE PARENTS?
WILD TYPE AND APRICOT
WHAT WOULD BE THE GENOTYPES AND PHENOTYPES OF
THE F1?
RATIOS:
E1
E4
GENOTYPES:
1:1:1:1
E1E2
E2E4
E2
PHENOTYPES: 2:1:1
E3
E1E3
E3E4
WILD: APRICOT: HONEY
DO PRACTICE #1,
PG. 609
ANSWER:
E1E2 x E2E4
E1
E2
E2
E1E2
E2E2
E4
E1E4
E2E4
RATIOS:
GENOTYPES:
1:1:1:1
PHENOTYPES:
1:1
WILD: APRICOT
MULTIPLE ALLELES –
blood type as an example
There are 4 possible
phenotypes for the ABO
blood system: type A, B, ,
AB, and O
There are 3 possible alleles
for blood type
IA
IB
i
... IA and IB are codominant
and i is recessive to them
both.
Phenotype
Genotypes
A
IAIA or IAi
B
IBIB or IBi
AB
O
IAIB
ii
MULTIPLE ALLELES –
blood type as an example
• In 1944 Charlie Chaplin, the legendary
comedian was involved a legal battle over
the paternity of a child born to Joan Barry,
an upcoming Hollywood star. Chaplin had
described Miss Barry as “a big handsome
woman of 22, well built and made alluring by her
low decollete dress.”
The baby was blood type B,
Joan Barry was type A, and
Chaplin was type O.
From what you know of blood type inheritance
could Chaplin possibly have been the father??
Show the baby’s and parents possible genotypes
to help explain your answer!
Guess what – blood type data were not admissible
evidence in California at the time of the trial, and since it
was the ONLY evidence the Charlie was not the father
the court deemed it had no alternative but to declare him
responsible for the child support!
INCOMPLETE DOMINANCE
Mendel also never had a blending of traits.
Ie. TALL x SHORT did not give medium. But this “blending” happens
often in nature.
When two alleles are equally dominant, they can interact to produce a
new phenotype. This is called INCOMPLETE DOMINANCE.
EX. RED (CRCR) SNAPDRAGONS x WHITE (CWCW)= PINK (CRCW)
WHAT WOULD PINK x PINK GIVE? CRCW x CRCW
RATIOS:
CR
CW
GENOTYPES:
1:2:1
CRCR CRCW
CR
PHENOTYPES: 1:2:1
CW
CRCW
CWCW
RED: PINK: WHITE
CODOMINANCE
(P. 610)
This is a type of incomplete dominance where both traits are seen at the
same time.
EX. SHORTHORN CATTLE.
RED
BULL (HrHr) x WHITE COW (HwHw) = ROAN CALF (HrHw) A
ROAN HAS BOTH RED AND WHITE HAIRS
X
WHAT WOULD ROAN x ROAN GIVE? HrHw x HrHw
Hr
Hw
RATIOS:
Hr
Hw
HrHr
HrHw
HrHw
HwHw
GENOTYPES:
1:2:1
PHENOTYPES:
1:2:1
RED: ROAN: WHITE
- MURDER MYSTERY PG 611
- GENETICS PROBLEMS
SHEET
- HOODED MURDER
- MORE GENETICS
PROBLEMS SHEET
- Do Page 612 #1-3
Pg. 612, #1a
1. C __ =GRAY > CchCch =CHINCHILLA (SILVER) >
CchCh, CchCa=LIGHT GREY > ChCh, ChCa = HIMALAYA >
CaCa = ALBINO
a) P = HETERO. HIM. x ALBINO
ChCa
:
CaCa
Ch
Ca
Ca
ChCa
ChCa
GENOTYPES?
Ca
CaCa
CaCa
GENOTYPES:
1:1
PHENOTYPES:
1:1
HIMALAYA: ALBINO
Pg. 612, #1b
1. C __ =GRAY > CchCch =CHINCHILLA (SILVER) >
CchCh, CchCa=LIGHT GREY > ChCh, ChCa = HIMALAYA >
CaCa = ALBINO
b) P =
FULL
C C__
x
LIGHT
:
CchC__
F1 = 2 FULL: 1 LIGHT: 1 ALBINO
C C__ : CchC__ : CaCa
GENOTYPES?
GENOTYPES?
WHAT ARE PARENT GENOTYPES?
P = C Ca
:
CchCa
Pg. 612, #1c
1. C __ =GRAY > CchCch =CHINCHILLA (SILVER) >
CchCh, CchCa=LIGHT GREY > ChCh, ChCa = HIMALAYA >
CaCa = ALBINO
c) P = CHINCHILLA
Cch Cch
x
LIGHT
: CchC__
GENOTYPES?
LIGHT ONE’S MOM WAS ALBINO (CaCa) SO…
WHAT ARE RESULTS IN F1?
Cch
Cch
GENOTYPES:
1:1
chCch
chCch
ch
C
C
C
PHENOTYPES: 1:1
CchCa
CchCa
Ca
CHINCHILLA: LIGHT GREY
Pg. 612, #1d
1. C __ =GRAY > CchCch =CHINCHILLA (SILVER) >
CchCh, CchCa=LIGHT GREY > ChCh, ChCa = HIMALAYA >
CaCa = ALBINO
d) TEST CROSS = UNKNOWN DOM. x
HOMOZYGOUS RECESSIVE
P = TEST CROSS OF LIGHT GREY GENOTYPES?
Ca Ca
: CchC__
F1 = 5 HIMALAYAN: 5 LIGHT GREY GENOTYPES?
Ch Ca
: CchCa
… Ca GIVEN TO ALL BABIES BY ALBINO. Cch GIVEN TO
HALF THE BABIES (BY GREY) SO Ch MUST COME FROM
GREY… GENOTYPE?
CchCh
Pg. 612, #2
2. CrCr = CHESTNUT; CmCm = CREMELLO; CrCm = PALOMINO
PALOMINO x CREMELLO (CrCm x CmCm)
Cr
Cm
Cm
CrCm
CrCm
Cm
CmCm
CmCm
GENOTYPES:
1:1
PHENOTYPES:
1:1
PALOMINO: CREMELLO
Pg. 612, #3
• The environment can also affect phenotype.
Nutrient amounts, sunlight availability, and
even consumers (herbivores, predators) can
determine what results will be.
• Probability can only tell you what to
EXPECT. Especially in small numbers,
probabilities might not match expected
results.
CRAZY HYBRIDS
CROSS A MOUSE AND A KIWI?
CRAZY HYBRIDS
CROSS A SNAKE AND A BANANA?
CRAZY HYBRIDS
CROSS A ROOSTER AND A LILY?
CRAZY HYBRIDS
CROSS A BEETLE AND A WALNUT?
CRAZY HYBRIDS
CROSS A SHEEP AND
CAULIFLOWER?
CRAZY HYBRIDS
CROSS A FROG AND AN ORANGE?
CRAZY HYBRIDS
CROSS A TURTLE AND A
LADY BUG?
CRAZY HYBRIDS
CROSS AN ELEPHANT AND A
RHINO?
(B) DETERMINE ANY GENOTYPES
POSSIBLE FOR THE TRAITS OF
BLOOD TYPE AND FOOT SHAPE.
Blood has 3 possible genes:
the ‘A’ gene, the ‘B’ gene and the ‘O’ gene.
The symbols for these genes are:
A = IA
B = IB
O=i
If you have:
IAIA you’ll have ‘A’ blood but also if you are IAi
IBIB you’ll have ‘B’ blood but also if you are IBi
If you are IAIB you will have type ‘AB’
If you are ii you will have type ‘O’.
(B) DETERMINE ANY GENOTYPES
POSSIBLE FOR THE TRAITS OF
BLOOD TYPE AND FOOT SHAPE.
Rh is a different gene altogether. Rh positive is
dominant to Rh negative.
If you are Rh+ your genotype could be:
RR or Rr
If you are Rh- your genotype could only be:
rr
(B) DETERMINE ANY GENOTYPES
POSSIBLE FOR THE TRAITS OF
BLOOD TYPE AND FOOT SHAPE.
Under each person on your pedigree chart, draw
four blanks like this: ___ ___ ___ ___
IN PENCIL (to correct mistakes) fill in what you
KNOW about the genotypes of each individual. If
you don’t know, don’t fill it in.
Example: If someone is type ‘O+’ you KNOW they
are: i i
R ___  the last blank stays blank
because you don’t know if it
is a R or r.
(B) DETERMINE ANY GENOTYPES
POSSIBLE FOR THE TRAITS OF
BLOOD TYPE AND FOOT SHAPE.
By looking at blood types of either parents OR
children, it is sometimes possible to KNOW
otherwise unknown genes.
Examples: if Mom is type ‘O’ then she is i i. She can
ONLY give an i gene to each child.
- if a child is ‘AB’ it MUST have got an IA from one
parent and an IB from the other. You may not know
which parent is which but you KNOW neither parent
has type ‘O’ blood (genotype = i i ). You also KNOW
the next generation can NOT have type ‘O’ blood.
(B) FILL IN GENOTYPES FOR
BLOOD TYPE ON YOUR
PEDIGREE CHART. (use pencil)
-PUT: ___ ___ ___ ___ UNDER EACH PERSON
- READ GIVEN INFO TO FILL IN POSSIBLE GENES
- USE CLUES TO FILL IN POSSIBLE GENES
- HINTS: Type O IS i i. This gives you a gene from Mom
and Dad (__ i ) AND a gene to all children.
-Type Rh- IS rr. This gives you a gene from Mom and Dad
(__ r ) AND a gene to all children.
- Type AB IS IAIB and means none of Mom, Dad, or kids
are type O.
- etc.
B. FILL IN GENOTYPES
Grampa Jim and Gramma Trudy were set in their
ways and did not believe in doctors or modern
medicine. Consequently, they were never tested
to see what type of blood they had. Before having
their kids, Jayni was tested, however, and was
found to be AB+. Kerri-Ann was tested and found
to be A-. Virgil was B- and Bobby was also B-.
Denny was as ornery as his old man and always
refused to be tested. Scott was found to be O+, as
was Jody, and Nina. Ray was O+, his sister-in-law
Audra was A-, and her daughter, precious little
Bonnie, was AB+. WHAT CAN WE TELL ABOUT
THE GENETICS OF THE FAMILY??