Download Flip Folder 5 KEY - Madison County Schools

Flip Folder #5: Unit 5: Meiosis and Genetics (Note: Sample PS = Punnett square) KEY
1. Meiosis
a. Purpose**
To create haploid gametes for sexual reproduction. The purpose of this is to INCREASE GENETIC
VARIATION.
b. Sexual Life Cycle – diploid, haploid, gametes, zygote
Diploid – both member of homologous chromosome pairs (double the haploid number)
Haploid – only one member of homologous chromosome pairs (half of total chromosome number)
Gametes – haploid sex cells
Zygote – 1 diploid cell baby; Result of fertilization of haploid sperm and egg
c. Result
4 haploid cells (4 sperm or 1 egg/3 polar bodies)
d. Law of Segregation
The law of segregation is that our diploid homologous chromosomes separate into haploid gametes at the end
of metaphase (anaphase). This means heterozygous people can have homozygous kids. This is the reason why
recessive traits can be “hidden” but then show back up.
e. Law of Independent Assortment
We have 23 pairs of chromosomes THAT LINE UP INDEPENDENTLY of each other. The information on
each chromosome is inherited independent of the other (like hair color and eye color).
2. Stages of Meiosis I
a. PMAT – Note: Terms tetrads or homologous chromosomes signify it’s meiosis I instead of mitosis/meiosis II
i. Prophase I (explanation must include crossing over, tetrad, and synapse)
 (Like mitosis) - Nucleus break down, spindle fibers form, chromatin goes to chromosomes
 (Different from mitosis) - Homologous chromosomes synapse forming a tetrad. Crossing over
between homologous chromosomes occurs here.
ii. Metaphase I (explanation must include homologous chromosomes and the law of
independent assortment)
 Homologous chromosomes line up at the metaphase plate (middle) of the cell. The law of independent
assortment means that each homologous pair lines up independently of the others. (For example,
they could line up as follows: 12 34 or 12 43)
 Difference from mitosis: Mitosis = Centromeres, Sister Chromatids, or Chromosomes in middle
Meiosis = Tetrads, Homologous Chromosomes in the middle
iii. Anaphase I (explanation must include the law of segregation)
 Homologous chromosomes separate. (Anaphase, apart, and away). The law of segregation just says
that alleles separate from one another. This is true because they are on the homologous chromosomes
that are separated.
 Difference from mitosis: Mitosis = Centromeres, Sister Chromatids, or Chromosomes pulled apart
and away Meiosis = Tetrads, Homologous Chromosomes pulled apart and away
iv. Telophase I
 Nuclei reform – You now have 2 haploid cells.
b. Major difference from mitosis - Prophase 1: Tetrads, Homologs, Synapsis, Crossing over
3. Stages of Meiosis II
a. PMAT 2 – Essentially same as mitosis but only half of chromosomes
PMAT II is basically just like mitosis. The only difference is that the 2 cells are now haploid (23 chromosomes) and
will divide into a total of 4 haploid cells containing 23 chromosomes still (because we have separated the sister
chromatids).
b. Spermatogenesis, oogenesis
Spermatogenesis = production of 4 fully functional, equally important haploid sperm cells. This is because the
evolutionary drive for males is to have as many offspring as possible. This is best accomplished through producing
as many sperm as possible. As long as the sperm fertilizes the egg it has done its job (so lots of nutrients don’t have
to be put into each individual sperm).
Oogenesis = the production of 1 haploid egg cell and 3 polar bodies that eventually degrade. This is because women
must actually carry the young if the egg is fertilized; therefore, more nutrients must be put into the egg to ensure the
survival of the zygote.
c. Contrast meiosis vs. mitosis**
1
1
2
46
1
2
4
23
Centromeres
Homologous Chromosomes
Identical
Different
Grow, Repair
Replace
Make Gametes
4. Probability
a. Law of Multiplication – Give example**
This says that the likelihood of 2 independent events BOTH OCCURING TOGTHER can be found by
multiplying their individual probabilities. For example, the chance of a flipped coin landing on heads is ½. The
chances of a 2nd flipped coin landing on heads is also ½. The chances of them BOTH LANDING ON HEADS is ¼
(1/2 * 1/2).
** Note: The law of independent assortment says the traits found on separate chromosomes are inherited
INDEPENDENTLY of each other. THIS IS WHY YOU CAN MULTIPLY 2 SEPARATE PUNNETT
SQUARES IN A DIHYBRID CROSS TO FIND THE LIKELIHOOD THAT THEY’LL BOTH OCCUR
TOGETHER IN THE OFFSPRING.
b. Law of Addition – Give example**
Chance that AN EVENT CAN OCCUR IN 2 OR MORE DIFFERENT WAYS. Sum of separate probabilities.
Ex.1/4 Pp +1/4 Pp 1/2 Pp (OR in other words ¼ Pp and ¼ pP  ½ Pp)
5. The Spectrum of Dominance
a. Complete dominance
Recessive means that the DNA sequence for that allele does not code for a functioning protein
Dominant means that the DNA sequence for that allele does code for a functioning protein. This is why
dominant alleles “mask” recessive alleles.
Note: Dominant does not mean better or more common.
b. Codominance
Codominance = both alleles dominate. Heterozygous phenotype shows both homozygous phenotypes. (black +
white = black & white)
c. Incomplete dominance
Incomplete dominance = Neither allele dominates the other. Heterozygous phenotype is a blend of the 2
homozygotes. (Black + white = grey)  Don’t think they just mix together though. Grey + grey = ¼ black, ½
grey, ¼ white.
6. Multiple Alleles
a. Define**
Multiple alleles = There are more than 2 alleles for a single gene at a single locus. (Ex. Blood type has I A, IB,
and i alleles)
b. ABO
 Can have A allele (IA), B allele (IB), or O allele (i). The “I” represents immunoglobin (a type of red blood
cell that that the glycoprotein would be found on.
 Blood type is named for the type of glycoprotein coded for on the red blood cell. The glycoprotein can also
be thought of as an antigen (so blood type A would have glycoprotein/antigen A). This is because if the A
blood is placed into someone with B or O blood then they would produce antibody A to fight the blood
(hence it is called antigen A).
o A blood type could make antibody B (because it wouldn’t fight itself)
o B blood type could make antibody A
o O blood type could make both antibody A and B




o AB blood type would make neither antibody
IA = A glycoprotein, IB = B glycoprotein, i = no glycoprotein coded for
IA and IB are both dominant to O (because A and B alleles code for a protein while O doesn’t). They are
codominant to each other.
O is universal donor. AB is universal recipient.
**Note: A+ or A- refers to Rhesus factor (another glycoprotein that can be found on blood cells). + =
dominant because make protein and - = recessive because it isn’t made.
c. Sample PS
If a woman with type A blood had a child with a man that had type B blood, what are all the possible blood types
that the child could have? Show the Punnett square. (There are multiple Punnett squares that could be made, but one
particular Punnett square would show all the possible offspring).
IA
i
IB
i
A, B, AB, or O
IA IB
IBi
IAi
ii
7. Pleiotropy
a. Define**
This is where one gene affects multiple phenotypes. ( “Pleio” means “multiple”)
Sickle Cell Disease is a great example. This gene affects the red blood cells shape, Oxygen carrying ability, Malaria
resistance, etc.
b. Example
8. Epistasis
a. Define**
A gene at one locus affects a gene at a second locus.
b. Sample PS
This usually has a 9:3:4 ratio; not the normal 9:3:3:1 ratio as seen with most dihybrids.
EPISTASIS
NORMAL
c. Example
Hair is a great example. Several genes are interacting to “create” hair’s phenotypes – These are: color, shape,
thickness, texture. (“epi” means “source”; “stasis” means location) The source is affecting another location.
9. Polygenic Inheritance
a. Define**
 This is where there exist many different degrees of phenotypic outcomes.
 This is due to Quantitative Characters. (Quantity -how many alleles did you inherit from your parents.)
 “poly” means “many”; “ genie” refers to “genes”; “inheritance” from your parents. MANY GENES AT
MANY LOCI (PLURAL FOR LOCUS) AFFECTING 1 TRAIT
 Note: It would look like a dihybrid/trihybrid problem (AaBbCC) but it is not BECAUSE ALL OF
THE GENES LISTED AFFECT ONE SINGLE TRAIT.
b. Example
Skin Color is a great example. We have many different degrees of skin pigmentation ranging from Albino  Black,
Black. It depends on how many copies of the same gene for making the skin pigment melanin you inherited from
your parents in the sperm and egg.
c. How it differs from simple dominance patterns?
 THERE IS A CONTINUOUS RANGE OF PHENOTYPES (HUNDREDS OF POSSIBILITIES) 
THINK POLYGENIC = MANY GENES AND MANY PHENOTYPES
 Norm of the Reaction – This refers to where the majority of organisms fall on the bell curve for that trait.
 Evolution? The norm can tell you about the type of environment organisms live in. The norm is
because some trait is beneficial in that environment.
10. Punnett Squares & Ratios
a. Monohybrid
1 Punnett square
Phenotype ratio = Dominant : recessive
Genotype ratio = Homozygous dominant : heterozygous : homozygous recessive
b. Dihybrid
2 Punnett Squares  If asks for one specific phenotype or genotype, multiply probability of each together (LAW
OF MULTIPLICATION)
Phenotype ratio = Dominant/Dominant : Dominant/recessive : recessive/Dominant : recessive/recessive
Wouldn’t ask for genotype ratio
c. Testcross – THIS TOPIC IS COVERED ON ANOTHER CARD. DIDN’T MEAN TO PUT IT TWICE
11. Pedigrees
a. Recessive alleles – how do you identify?
I.
Someone who has the disorder must be homozygous recessive
II.
If a child has a recessive disorder but neither of the parents have the disorder, then the parents must be
carriers
III.
Males and Females will acquire the disorder in roughly the same percentages.
IV.
If both parents have the disorder, then the child must also have the disorder
i. Example
I.
II.
b. Dominant alleles – how do you identify?
Someone who has the disorder can be homozygous dominant or heterozygous
Two parents who have the trait can have a child who does not (if they are both heterozygous)
i. Example
I.
II.
III.
IV.
V.
VI.
c. Sex-linked inheritance – how do you identify?
It will appear to mainly affect males (as they only have one X chromosome). This is because if the
inherited X chromosome has a recessive gene on it; it will not be covered up by a dominant one on another
X chromosome (as is the case in most females).
A. Females tend to be carriers.
If a son has the trait, they MUST have gotten it from their mother (because that is where their X
chromosome comes from).
For a daughter to have the trait, they must have gotten it from both their mother and father.
If a father has the trait, his daughters MUST be carriers but his sons will not be affected.
If a mother is a carrier, 50% of her sons will have the trait and 50% of her daughters will be carriers.
If a mother has the trait, 100% of her daughters will be carriers and 100% of her sons will have the trait.
i. Example
12. Recessive Diseases
a. Examples with descriptions**
1. Cystic Fibrosis (Also referred to as “CF’.)
a. This is the most common lethal genetic disease among people of European descent.
b. This disorder affects 1 in 2,500 births.
c. In Caucasians, 1 in 25 people is a carrier for the disorder.
d. The disorder creates a faulty Chloride ion (Cl-) protein carrier on cell membranes in the lungs. This
causes fluid (water) to build up in the lung tissues.
i.
People drown in their own fluid.
ii.
They are also prone to get multiple infections in the lungs.
e. Treatment? Since it is genetic there is no cure. Patients have to get the fluid drained from the lungs
periodically for their entire life. There are medicines to help reduce the number of times this has to occur.
2. Sickle-cell Disease
a. This disorder is the most common genetic disorder among people of African descent.
b. It affects 1 in 400 births.
c. The 6th Amino Acid in the hemoglobin molecule is changed (Glutein  Valine) in the primary sequence
needed to make red blood cells. (The easy way to remember this is: 666 is the number of the beast. 6 is
the amino acid that changed to create this horrible disease. It went from good [glutein] to very bad
[valine].)
d. Sickle- cell trait (“trait” is used to refer to individuals that are carriers.)
i.
These individuals have resistance to Malaria because of the one recessive allele they
possess but mainly have normal red blood cells for carrying oxygen.
ii.
This is referred to as the Heterozygote Advantage. They have an advantage over
individuals that are homozygous dominant or homozygous recessive. Homozygous
dominant are not resistant to Malaria. Homozygous recessive are also resistant to Malaria;
BUT they have the disease to contend with.
e. These sickle shaped cells have reduced oxygen carrying ability. They also are painful when the points of
the cell jab into the walls of the blood vessels.
f. Treatment? There is no cure as it is genetic. Some medicines help with the pain or low oxygen levels.
13. Dominant Diseases
a. Examples with descriptions**
1. Achondroplasia (This is referred to as Genetic Dwarfism.)
a. This disorder affects 1 in 10,000 births.
b. Most people are homozygous recessive and there for much taller than these individuals.
2.
Huntingdon’s Disease
a. This disorder affects 1 in 10,000 births.
b. It has a late life onset – usually in the 40-50 age range. (Usually after children are born.)
c. The dominant gene has a locus on tip of Autosome 4.
d. Family history is important in diagnosis of this disorder. (Pedigree can help.)
e. It is a slow degenerative disorder affecting the brain that is almost always fatal.
14. Linked Genes and Linkage Maps
a. What does it mean for genes to be linked?
I.
These are usually inherited as a linked unit because they are found on the same chromosome.
II.
This term usually is associated with genes on Autosomes (1-22).
III.
In the picture below, why would gene 1 only be slightly linked to gene 3?
b. How is crossing over involved?
The closer 2 genes are together on a chromosome, the less likely they are to have crossover occur between them;
therefore, the more linked they are.
c. How do you map genes from recombination frequencies?
15. Sex-linked Inheritance patterns
Males will inherit the traits at a higher frequency (because they must only get 1 X).
If a dad has the trait, his mom AND daughter MUST be AT LEAST CARRIERS.
For a daughter to have the trait, dad must have it and mom must be at least a carrier.
a. Sample PS for hemophilia**
If a female is a carrier for hemophilia mates with a man who has hemophilia, what would be the expected offspring?
XA
Xa
Xa
X AX a
XaXa
Y
X AY
XaY
b. Sample PS for colorblindness**
If a female that is colorblind (a sex-linked recessive trait) and is crossed with a normal male, what genotypes and
phenotypes would be possible in the offspring?
Xb
Xb
XB
XBXb
XBXb
Y
XbY
XbY
16. X-inactivation
a. What is it?
In female mammal embryos, one of the two X chromosomes in each cell does not uncoil into chromatin (called
X-inactivation). Different X chromosomes in each embryonic cell are inactivated so heterozygous female still
expresses both X chromosomes.
Colors of calico cat are an example (if X chromosome with orange allele is deactivated then patch is black and vice
versa). White color is controlled by another gene.
b. Barr Bodies
The chromosome that remains coiled (dark, compact) is inactive and is called a Barr body.
17. Polyploidy
a. Polyploidy - Many extra sets of chromosomes. (so more than diploid  di = 2 sets of each homologous
chromosome). This refers to ALL OF THE PAIRS OF CHROMOSOMES.
i. 3n (triploid) - Three “halves” are in this cell.
ii. 4n (tetraploid) - Four “halves” are in this cell.
a. Aneuploidy
Chromosomal Number (Aneuploidy means “Abnormal number of chromosomes”)
Deadly in most animals; Plants not really affected because they are simple organisms.
This refers to ALL OF THE PAIRS OF CHROMOSOMES.
b. Trisomy
Trisomic - Three of ONLY 1 CHROMOSOME PAIR.
Ex = Trisomy 21 = Down’s Syndrome
c. Monosomy
Monosomic - Missing ONE MEMBER OF A SINGLE HOMOLOGOUS PAIR, the other half of the pair. (It is
located in the Trisomic gamete.)
d. Nondisjunction
Failure to separate” during Anaphase.
Different impact if it happens in Anaphase 1 or Anaphase II.
18. Chromosome Alteration
Individual Chromosome Structure - These occur because of faulty crossover
a. Duplication
A chromosome segment was “copied” twice. Two genes on one chromosome. It is “missing” from the other
homologous chromosome.
b. Inversion
A chromosomal segment is “backwards”. It was inverted (“backwards”) during crossover.
c. Translocation
A chromosomal segment is attached to a different autosome. It accidentally broke loose and ended up on another
chromosome.
d. Deletion
Chromosome segment is “missing”. It got stuck on the other homologous chromosome during crossover.
e. Genomic Imprinting
Genomic imprinting is the epigenetic phenomenon by which certain genes are expressed in a parent-of-originspecific manner. If the allele inherited from the father is imprinted, it is thereby silenced, and only the allele
from the mother is expressed. If the allele from the mother is imprinted, then only the allele from the father is
expressed.
It is an epigenetic process that involves DNA methylation and histone methylation without altering the genetic
sequence. These epigenetic marks are established ("imprinted") in the germline (sperm or egg cells) of the parents
and are maintained through mitotic cell divisions in the somatic cells of an organism.
19. Common Human Chromosomal Disorders
a. Karyotype – autosomes, sex chromosomes
Autosomes = pairs 1-22. Males and females inherit autosomal characteristics equally.
Sex = 23 pair; Males = XY, Females = XX
I.
II.
III.
IV.
b. Downs, description, effect
This affects about 1 in 700 births.
This individuals possess an extra 21 Autosome (A.K.A. Trisomy 21)
General syndrome features – mental retardation, flat face, squinted eyes, small.
Mainly the result of women of advanced age having babies.
c. Cri du Chat, description, effect
Rare genetic disorder due to deletion of part of chromosome #5.
The syndrome gets its name from the characteristic cry of affected infants, which is similar to that of a meowing
kitten, due to problems with the larynx and nervous system. About 1/3 of children lose the cry by age 2. Other
symptoms of cri du chat syndrome may include:
feeding problems because of difficulty swallowing and sucking;
low birth weight and poor growth;
severe cognitive, speech, and motor delays;
behavioral problems such as hyperactivity, aggression, tantrums, and repetitive movements;
unusual facial features which may change over time;
excessive drooling;
small head and jaw;
wide eyes;
skin tags in front of eyes.
d. Klinefelter’s, description, effect
Klinefelter’s Syndrome (XXY)
b. Still genetically a male (remember Y makes the guy), but show an increase in feminine
characteristics.
c. General characteristics: Increased breast development, smaller testicular size, decreased beard
growth, narrow shoulders, wide hips.
e. Turners, description, effect
Turners Syndrome (X_) – Only have 1 X chromosome (no additional X or Y)
a. General characteristics: These individuals appear fairly normal. They just are lacking the
“sexual” characteristics we normally see in individuals. The “sexual” traits are missing because
there is no second sex chromosome to help create those traits. These individuals are usually
raised as females.
20. Test Cross
a. Purpose?
b. Description of the process.
If all of the OFFSPRING ARE DOMINANT then it’s assumed the parent is HOMOZYGOUS DOMINANT.
If any of the OFFSPRING ARE RECESSIVE then the parent phenotype is HETEROZYGOUS.
21. Chi-square analysis
Chi-squared is always used to determine if your observed results support the expected results (hypothesis). A fancy
way of saying it is “Is there a significant statistical difference in the observed and expected results?”
a. Purpose in genetics?
To determine if the expected offspring (offspring actually born) are equal to the ratios expected (determined by
Punnett Squares/Genetics)
If they do not match then the inheritance pattern does not follow simple Mendelian Genetics. For example, the
inheritance may be LINKED, EPISTASIS, PLEIOTROPY, ETC.
b. Formula
i. Degrees of Freedom, Critical Value, Chi-squared, How to determine expected results in genetic
problems? What would null hypothesis be?, Accept or Reject hypothesis


Degrees of freedom = number of choices – 1
Critical Value = Found by using a probability of p=0.05 and the degrees of freedom of your experiment.
This states you are 95% certain in your support/rejection of the null hypothesis.





Chi-Squared = this is the actual value found in the math equation. Compare to the critical value to
determine whether you should support/reject
The null hypothesis states “There is no difference between the expected and the observed”
Determine expected results through Punnett Square
Accept null hypothesis if chi-squared is less than the critical value (Reject if chi-squared is greater than chisquared).