Transcript File

Heredity and Genetics
Mr. Erick Santizo
DEFINE the following
 Allele
 Chromosome
 Metaphase
 DNA
 Interphase
Differentiate
Mitosis
Meiosis
Differentiate
Meiosis
Mitosis

Occuris in body cells or somatic cells


# of chromosomes remains the
same in the daughter cells (2n)
diploid
Either occurs in reproductive cells
only or occurs in formation of
gamate only

Daughter cells are identical to
parent cells and each other
# of chromosomes is halved in the
daughter cells (haploid = n)

Daughter cells are genetically
different to parent cell and each
other

Four daughter cells are formed

Exchange of genetic material
between chromosomes


Two daughter cells are formed

No exchange of genetical material
between chromosomes
Variation
 Term
used to show differences between individuals of
the same species.
 Individuals
of the same species varies
 Differences
seen in their physical appearance, such
that no two are alike.
 Even
twins has subtle differences
 Results
from the “GENOTYPE” the genetic make up
Types of Variation

Continuous variation: anything that can be measured:
eg. Human Height, foot length, skin color, leaf size and
pod size in legumes.
 Discontinuous
variation: the differences are separated
and clear cut; they do not merge or grade into each
other. Examples are tongue-rolling in humans and the
presence or absence of horns in cattle.
List as Continuous or discontinuos
 A.
Ear size
 B.
Brown Fur
 C.
Green Eyes
 D.
Wing span
 E.
Eyebrow Thickness
 F.
Widow’s peak
 G.
Elephant’s trunk
Genes
 Pass
on from parents to offspring
 Different
kinds of genes: for eye color, hair color, hair
texture, shape of nose, size of lip, length of finger,
length of arm and the list goes on.
 There
are 23 pairs of chromosomes: 23 that are from
the father or paternal and 23 from the mother which
is maternal. Pairs are called homologous
chromosomes. (the same kind).
 Chromosomes
are made of genes/ units of
inheritance.
*
this is what controls specific characteristics in the
organism.
 Homologous
pair: carries same set of genes; body cells
has two copies of each gene HOWEVER
A
gene can have more than one FORM called an allele.
ALLELES
 So
the cell may have two alleles for a gene that are the
same as each other, or two alleles for a gene that are
different.
 If
the alleles of a gene are the same: Homozygous
 If
the alleles of a gene are different: Heterozygous
Genotype
Phenotype
 BB
: homozygous
 Black
hair
 Bb
: heterozygous
 Black
hair: B is dominant to b
 bb
: homozygous
 Red
 Not
visible; genetic
make up

hair
observable characteristics of
the organism.
 Phenotype
genotype
determined by
Answer the following
2. For each of these genes, state which allele must
be dominant and which must be recessive.
Tt Dominant: ___ Recessive: ___
Gg Dominant: ___ Recessive: ___
Pp Dominant: ___ Recessive: ___
VV Dominant: ___ Recessive: ___
jj Dominant: ___ Recessive: ___
 State
whether each of these genotypes is homozygous
dominant, heterozygous, or homozygous recessive.
 HH
 kk
 Uu
_______________________________
_______________________________
_______________________________
 vv
_______________________________
 Bb
_______________________________
 RR
_______________________________

4. For each of these genes, give the genotype that would be homozygous dominant,
heterozygous, and homozygous recessive.

Y Homozygous dominant: ____ Heterozygous: ___ Homozygous recessive: ___

Q Homozygous dominant: ____ Heterozygous: ___ Homozygous recessive: ___

E Homozygous dominant: ____ Heterozygous: ___ Homozygous recessive: ___

M Homozygous dominant: ____ Heterozygous: ___ Homozygous recessive: ___

F Homozygous dominant: ____ Heterozygous: ___ Homozygous recessive: ___

5. In each of these scenarios, give information about what the
individual’s phenotype will be.

a. With respect to the gene for coat color in mink, if the allele
U produces a yellow coat and the allele u produces a brown
coat, then an individual that is UU will have this coat color:
_______.

b. With respect to the gene for wing shape in house flies, if the
allele I produces rounded wings and the allele i produces
crooked wings, then an individual that is ii will have this wing
shape: ________.

c. With respect to the gene for eye color in Pacific salmon, if
the allele P produces light eyes and the allele p produces dark
eyes, then a salmon with the genotype Pp will have this
phenotype: ___________.

d. With respect to the gene for flipper length in bottlenose
dolphins, if the allele T produces stunted non-functional
flippers and the allele t produces normal flippers, then a
dolphin with the genotype tt will have this phenotype:
________________.

e. Huntington’s Disease in humans is caused by the
manufacture of a damaged version of a protein called
huntingtin. We will use M for the damaged version of the
protein huntingtin and the allele m produces normal
huntingtin. If a person has the genotype Mm, will they have
Huntington’s Disease? ________

6. Given information about these phenotypes, determine which alleles are which.

a. In a group of mice, all individuals that are YY have long whiskers and all
individuals that are yy have short whiskers. What is the allele for long whiskers?
____ What is the allele for short whiskers? ___

b. In a group of fish, all individuals that are OO have thin fins and all individuals that
are oo have thick fins. What is the allele for thin fins? ____ What is the allele for
thick fins? ___

c. In a group of dogs, all individuals that have the genotype Ee have floppy ears.
What is the allele for floppy ears? ___ What is the allele for non-floppy ears? ___

d. We will use the alleles G and g for the gene for coat spots in rabbits. All rabbits
that have one of these alleles have spots. No rabbits without spots have that allele,
they only have the other allele. What is the allele for spots? ____ What is the allele
for no spots? ____

e. We will use the alleles Q and q for the gene for fur smoothness in squirrels. All
squirrels that are heterozygous have smooth fur, whereas only squirrels that are
homozygous for one of those alleles have fluffy fur. Which is the allele for smooth
fur? ___ Which is the allele for fluffy fur? ___

7. In humans, who is heterozygous with respect to their
biological sex? Men or women? Explain.
___________________________________________________
_

8. Suppose that 98% of the ferret population has smooth fur,
and only 2% of the population has fluffy fur. Does this mean
that the allele for smooth fur must be dominant? Explain.
___________________________________________________
Genetic Diagrams

-Shows the cross of two Genotypes. It shows the genotypes and
phenotypes of the parents and the possible genotypes and
phenotypes of the offspring.

Example: a genetic cross between homozygous black hair and a red
hair parent: BB x bb

A.)All offspring would be heterozygous with black hair: Bb

B.) Heterozygous cross with homozygous gives 1Bb:1bb ratio 50%50%

When two heterozygous cross together the ratio is-

1red hair: 3 black phenotypic ratio; 25% of offspring having red hair

Genotypic ratio-1BB: 2Bb:1bb
Question #1

Albinism (absence of pigmentation) in man is caused by a
recessive gene which is transmitted in a normal fashion. A
phenotypically normal (non-albino) couple have four children,
the first three are normal and the fourth is albino.

i. What can you say about the genotype of the parents

ii.What is the possibility that their next child will be albino

iii. One of the normal children marries a normal woman. What
predictions can be made of their First child?

Iv. The albino child marries a normal woman. What predictions
can be made of their first child?
Question #2

A breed of dogs has long hair dominant over short hair. A longhaired BITCH was first mated with a short-haired dog and
produced three long-haired and three short haired puppies.
Her second mating, with a long haired dog, produced a litter
with all the puppies long haired. Use the symbol L to represent
the allele for long hair and “l” to represent for short hair.

I. What was the genotype of the long-haired bitch?

ii. How could it be determined which of the long haired puppies
of the second mating were homozygous?
Incomplete dominance

Shows blending; combination of expression of both alleles in the
heterozygous condition. ( production of a new allele. Eg. Blue and
yellow=green)

Eg. Red (RR) and White (rr) Flowers form pink flowers (Rr)

When crossing a Two pink flowers ( Rw)

Phenotypic ratio: 1red:2pink:1white

Genotypic ratio: 1RR: 1RW: 1WW

2. In some cats the gene for tail length shows incomplete dominance. Cats
with long tails and cats with no tails are homozygous for their respective
alleles. Cats with one long tail allele and one no tail allele have short tails.
For each of the following construct a punnett square and give phenotypic
and genotype ratios of the offspring. a) a long tail cat and a cat with no tail
b) a long tail cat and a short tail cat c) a short tail cat and a cat with no tail
d) two short tail cats.

2. long tails (L), no tails (ll), short tails (Ll). a) a long tail cat and a cat with
no tail LL X ll, gametes for LL (L), gametes for ll (l), resulting phenotype
ratio: 100% short tails, resulting genotype ratio: 100% Ll b) a long tail cat
and a short tail cat LL X Ll, gametes for LL (L), gametes for Ll (L & l),
resulting phenotype ratio: 50% long tails / 50% short tails, resulting
genotype ratio: 50% LL / 50% Ll c) a short tail cat and a cat with no tail Ll
X ll, gametes for Ll (L & l), gametes for ll (l), resulting phenotype ratio: 50%
short tails / 50% no tails, resulting genotype ratio: 50% Ll / 50% ll d) two
short tail cats Ll X Ll, gametes for Ll (L& l), gametes for Ll (L & l), resulting
phenotype ratio: 25% Long Tailed, 50% short tailed, and 25% no tailed,
resulting genotype ratio: 25% LL / 50% Ll / 25% ll
Co-dominance

A condition in which the alleles of a gene pair in a heterozygote are fully
expressed thereby resulting in offspring with a phenotype that is neither
dominant nor recessive. SupplementA typical example showing
codominance is the ABO blood group system. For instance, a person having
A allele and B allele will have a blood type AB because both the A and B
alleles are codominant with each other.

In some cattle the genes for brown hair (B) and for white hair
(W) are co-dominant. Cattle with alleles for both brown and
white hair, have both brown and white hairs. This condition
gives the cattle a reddish color, and is referred to as Roan (BW).
For each of the following construct a punnett square and give
phenotypic and genotype ratios of the offspring.

a) a roan cow and a white bull

b) a brown cow and a roan bull

c) a white cow and a roan bull

d) a roan cow and a roan bull

brown hair (B), white hair (W), roan (BW).

a) a roan cow and a white bull BW X WW, gametes for BW (B & W), gametes
for WW (W), resulting phenotype ratio: 50% roan / 50% white, resulting
genotype ratio: 50% BW / 50% WW

b) a brown cow and a roan bull BB X BW, gametes for BB (B), gametes for
BW (B & W), resulting phenotype ratio: 50% brown / 50% roan, resulting
genotype ratio: 50% BB / 50% BW

c) a white cow and a roan bull WW X BW, gametes for WW (W), gametes for
BW (B & W), resulting phenotype ratio: 50% roan / 50% white, resulting
genotype ratio: 50% WW / 50% BW

d) a roan cow and a roan bull BW X BW, gametes for BW (B & W), gametes
for BW (B & W), resulting phenotype ratio: 25% brown / 50% roan / 25%
white, resulting genotype ratio: 25% BB / 50% BW / 25% WW
Blood groups
Genotype
AA
Phenotype
I
AO
I
Blood group A
BB
Blood group B
I
I BO
I AB
I OO
Blood group A
1.What are possible blood groups of
children Whose parents are blood group
“A” Heterozygous and “B” homozygous?
Blood group B
Blood group AB
Blood group O
b. What if both had heterozygous
genotypes?
Sex determination
Phenotype of parents:
Male
x
Female
Genotype:
XY
x
XX
Gametes:
Offspring genotype:
X
XX
Offspring phenotype: female
Ratio: 1 female: 1 Male
50%-50%
Y
X
x
XX
XY
XY
female
male
male
Sex-linked characteristics

The sex chromosomes carry genes other than those which
determine sex.
“ THE CHARACTERISTICS ARE SAID TO BE SEX-LINKED”
Example: Haemophilia and colourblindness.
Haemophilia or bleeder’s disease: the dominant allele “H” causes
blood to clot NORMALLY. But the recessive allele “h” Causes
HAEMOPHILIA.
Haemophilia
Genotype
H
H
X X
XH X
h
h
XH X
X Y
h
X Y
h
Phenotype
Female, Normal clotting of blood
Female, normal clotting of blood; she is a CARRIER since she
carries the recessive allele but it is not EXPRESSED (no
phenotype)
Female, a haemophiliac
Male, normal clotting of blood
Male, a haemophiliac
Question#1
1. A normal man married a normal woman and all the
female offspring were normal, but about half of the
male offspring were colour-blind and the other half
were normal. How do you account for this (use
knowledge of haemophilia)
Sickle cell anemia

The red blood cell can take a sickle shape instead of the normal
biconcave shape.
Genotype
Hb NN
Hb
SS
HbNS
Phenotype
All red blood cell are normal, the person is
NORMAL
All red blood cells take the sickle shape, the
person suffers from sickle cell anemia
30-40% of the red blood cells are sickle
shaped, the person suffers from the sickle cell
trait.
Question

Karen and Steve each have a sibling with sickle-cell disease.
Neither Karen nor Steve nor any of their parents have the
disease, and none of them have been tested to see if they
have the sickle-cell trait. Based on this incomplete
information, calculate the probability that if this couple has a
child, the child will have sickle-cell disease.
Pedigree charts


shows occurrence of a particular characteristic in a family
tree. The chart can be used to show the possible genotypes of
individuals in the chart, which can be important in genetic
counseling.