Chapter 3: Mendelian Genetics Flashcards
What factors were crucial to the success of Gregor Mendel’s experiments?
He kept detailed quantitative records.
He observed only one or very few traits in any given experiment.
He chose traits that were not greatly influenced by the environment.
Mendel often crossed different individuals in his experiments.
Which of the following would be true of a plant heterozygous for a single gene controlling flower color? Assume complete dominance.
If crossed with another heterozygous plant, the majority of progeny will have the dominant flower color.
Does the 3:1 phenotypic ratio observed among progeny of an F1 X F1 cross requires random union of gametes?
TRUE
The 1:2:1 genotypic ratio represents relative probabilities of gamete combinations based on the assumption that gamete union is random.
What would be the best way to determine which of 2 alleles of a gene is dominant to the other?
Observe the relevant phenotype in the progeny that result from a cross between individuals from two different pure-breeding lines.
All progeny will be heterozygous for the trait in question and will display the phenotype that corresponds with the dominant allele.
In an individual that is heterozygous for a particular trait, is the recessive allele expressed?
NO
Only the dominant allele is expressed in an individual that is heterozygous for a particular trait.
Round (R) seed shape is dominant to wrinkled (r) seed shape in pea plants. If an RR plant is crossed with an rr plant, what is the frequency of phenotypes in the F2 generation?
3 round seeds, 1 wrinkled seed
The F1 generation would have the genotype Rr, so crossing two heterozygotes would result in 3 plants with round seeds and 1 plant with wrinkled seeds.
If the first three F2 offspring grown from the cross described above are round, what is the probability that the next F2 offspring will be wrinkled?
25%
The proportion of rr is ¼, so the next offspring has a 25% chance of being wrinkled.
Which two genotypes in an individual would be expressed as the same phenotype in a diploid organism?
homozygous dominant and heterozygous
Both of these genotypes would express the dominant phenotype. The homozygous dominant individual would have both copies of the dominant allele, and the heterozygous individual would have one copy of the dominant allele and one copy of the recessive allele, but only the dominant phenotype would be observed.
Mendel’s unit factors in pairs are most accurately known to be _______.
two alleles on paternal and maternal homologs
Mendel observed patterns of inheritance in offspring that stemmed from traits found in the parents. Mendel explained this phenomenon as the passage of “unit factors” that are transmitted from the parents to the offspring.
Homolog
A gene related to a second gene by descent from a common ancestral DNA sequence. The term, homolog, may apply to the relationship between genes separated by the event of speciation or to the relationship betwen genes separated by the event of genetic duplication.
One member of a chromosome pair.
Mendel’s Law of Segregation is supported by a 1:1 testcross ratio.
TRUE
To test Mendel’s Law of Segregation, the experimenter needs a minimum of two contrasting forms of a gene.
True
A 1:1 phenotypic ratio is expected from a monohybrid testcross with complete dominance.
TRUE
Does the law of independent assortment states that one gene in a pair is always dominant to the other?
NO
The law of independent assortment states that during gamete formation, segregating pairs of unit factors assort independently of each other; the law says nothing about dominance.
If a yellow pea plant with round seeds that has the genotype GgWw is crossed to itself, what proportion of the offspring will be green with round seeds?
3⁄16
This is the probability of offspring that are either green with round seeds or yellow with wrinkled seeds.
The following phenotypic ratios are determined for a trihybrid cross in which the gametes assort independently:
A 1 : A 2 = ¼ : ¾
B 1 : B 2 = ¼ : ¾
C 1 : C 2 = ¾ : ¼
What is the probability that the F2 offspring will have the phenotype A 1 B 1 C 1?
3⁄64
This is the probability that the offspring will be
A 1 B 1 C 1,
A 1 B 2 C 2,
or A 2 B 1 C 2.
Which of the following phenomena is a consequence of Independent Assortment?
The phenotypic ratio among phenotypes produced from an F1 X F1 dihybrid cross is 9:3:3:1.
The 9:3:3:1 ratio requires that the four gamete classes produced occur with equal frequency. This requires independent assortment.
The 9:3:3:1 ratio exhibited in the F2 generation of a dihybrid cross is a genotypic ratio.
FALSE
A pea plant is heterozygous for two genes; one controlling height, one controlling color. The genotype is written PpTt. Based on the Law of Independent Assortment, approximately what proportion of the pollen produced by this plant should have the genotype PT?
1/4
There are 4 different possible genotypes. Independent Assortment assumes that all are equally likely. Therefore, each of the 4 should occur with roughly equal frequency (1/4).
How many different kinds of gametes can be produced by an individual with the genotype AABbCCddEeFf ?
8 gametes
1 x 2 x 1 x 1 x 2 x2
Which of the following statements is true regarding a trihybrid cross between two true-breeding homozygous individuals with contrasting phenotypes?
The least frequent F2 phenotypic class is recessive for all three traits.
1/64 of the F2 phenotypes is triply recessive.
Which of the following statements is an example of independent assortment?
A tall pea plant is no more or less likely to have round seeds than a dwarf pea plant.
Traits that are not on the same chromosome are inherited independently of each other.
One character in peas that Mendel studied was yellow versus green seeds.
A cross between a homozygous yellow line (GG) and a homozygous green line (gg) will result in F1 plants that are heterozygous (Gg) for this trait and produce yellow seeds.
Cross between homozygous yellow and green pea plants.
When an F1 plant undergoes meiosis, what gamete types will it produce, and in what proportions?
Mendel’s Law of Segregation states that allele pairs segregate equally into gametes during meiosis. This means that a gamete will have only one allele of any given gene, and that the probability of a gamete having one allele or the other is equal (and therefore ½, or 50%, for either allele).
This Punnett square shows the results of a Gg x Gg cross to form F2 progeny.
Use your understanding of Mendel’s Law of Segregation and the rules of probability to complete the Punnett square for this cross.
Identify the gametes. Identify the male and female gamete types and the gamete frequencies.
Then identify the F2 progeny. Identify the progeny genotypes and the progeny frequencies.
A Punnett square is a convenient method for representing Mendel’s Law of Segregation in a visual form. Using a Punnett square allows one to easily see gamete types and frequencies, as well as the genotypes and frequencies of progeny formed by random gamete fusion.
The genotype frequencies inside a Punnett square are calculated using the product rule: The probability of two independent events occurring simultaneously is the product of their individual probabilities.
In this example, the genotype frequencies inside the square (¼) are the product of the gamete frequencies that led to their formation (½ x ½ = ¼).
Gg x Gg cross to form F2 progeny.
What is the probability that three F2 seeds chose at random will include one green seed and two yellow seeds?
Use all of the probabilities of getting one green and two yellow seeds in the bunch. So: G, Y, Y = 1/4 X 3/4 X 3/4 = 9/64
Y, G, Y = 3/4 X 1/4 X 3/4 = 9/64
Y, Y, G = 3/4 X 3/4 X 1/4 = 9/64
Add all the probabilities together.
9/64 + 9/64 + 9/64 = 27/64.
27/64 is the probability that three F2 seeds chosen at random will include one green seed and two yellow seeds.
Gg x Gg cross to form F2 progeny.
What is the probability that three f2 seeds chosen at random will include at least one yellow seed?
63/64 all you have to do is add up all the possibilities of drawing at least 1 yellow seed YYY=.75x.75x.75=27/64 YYG=.75x.75x.25=9/64 YGY=.75x.75x.25=9/64 GYY=.75x.75x.25=9/64
totaling up to 63/64!!!!!
63/64 there are seven possible groupings that have at least one yellow seed (shown in blue in the table); the sum of their individual probabilities yields the final answer (63/64). Note that you could have also calculated this answer by subtracting the probability of the one category that doesn’t fit the criterion (Green, Green, Green) from the total probability (1 - 1/64 = 63/64).
Is Mendel’s postulate of independent assortment supported by a 1:1:1:1 testcross ratio?
YES
Which phenotypic ratio is likely to occur in crosses of two completely dominant, independently segregating gene pairs when both parents are fully heterozygous?
9:3:3:1
How could the botanist best determine whether the genotype of the green-pod plant is homozygous or heterozygous?
Cross the green-pod plant with a yellow-pod plant.
A cross between a plant of unknown genotype and one that is known to be homozygous recessive is called a test cross because the recessive homozygote tests whether there are any recessive alleles in the unknown. Because the recessive homozygote will contribute an allele for the recessive characteristic to each offspring, the second allele (from the unknown genotype) will determine the offspring’s phenotype.
Suppose that the botanist carried out the test cross described in Parts A and B and determined that the original green-pod plant was heterozygous (Gg). Which of Mendel’s findings does her test cross illustrate?
law of segregation
The law of segregation states that the two alleles for a gene separate during gamete formation, and end up in different gametes. In the case of the heterozygous green-pod plant (Gg), one gamete will receive the dominant allele (G), and the other gamete will receive the recessive allele (g). The law of segregation accounts for the prediction that 50% of the offspring of the test cross will have green pods and 50% will have yellow pods.
During which part of meiosis (meiosis I or meiosis II) do the two alleles of a gene separate? During which phase does the separation occur? (Assume no recombination.)
meiosis I, anaphase
Alleles separate from one another during anaphase of meiosis I, when the homologous pairs of chromosomes separate.
Independent assortment explains _______.
why a collection of meiotic products that includes gametes of genotype Ab will also include gametes of genotype AB in roughly the same proportion.
Because alignment of the chromosomes is random, the alignment that produces Ab is just as likely as the alignment that produces AB. Therefore, the two genotypes should both occur with equal frequency.
Is it true that Independent Assortment occurs only in cells that are heterozygous for two genes (AaBb) and not in cells that are completely homozygous (AABB or aabb)?
NO
Even though there would be no genotypic differences in the products of such meioses, random alignment of chromosomes would still have occurred.
Which of the following genotypes represents a gamete produced by a failure to segregate during meiosis in a cell of genotype AaBbCC?
ABC aBC abC AaBC
AaBC
This gamete contains two copies of gene “A”. Proper segregation would have separated A from a and allowed only one copy per gamete.
How did Mendel’s work support the chromosomal theory of inheritance?
The behavior of chromosomes during meiosis, as observed under a microscope, correlated with Mendel’s principles of inheritance.
The observation that chromosomes segregate independently and randomly during meiosis fit very nicely with Mendel’s postulates. Mendel’s work supported the idea that the genetic information was carried on the chromosomes.
True or False?
A gene can have a maximum of two alleles.
FALSE
Chi-square analysis can help us to decide whether _______.
our observations of an event differ from our expectations.
This is exactly the purpose of statistical tests.
What is the probability of flipping a penny and a nickel and obtaining one head and one tail?
1/2
You can have S={HH,HT,TH,TT}
Two options have one head and one tail out of the four outcomes.
Therefore, the probability of obtaining one head and one tail is 2/4 = 1/2.
A couple has two boys and is expecting a third child. What is the probability that the third child will be a boy?
The probability of having a boy is ½, and the probability of having a girl is ½. The sex of the previous children does not influence the sex of the third child.
What is the probability of a family with six children having three boys and three girls?
20⁄64
The term of interest in the expansion (a + b)^6 is the middle term: (a^3)(b^3). Using Pascal’s triangle, the coefficient of this term is 20, so the probability of three boys and three girls is 20⁄64.
a=girls=.5 chance
b=boys=.5 chance
^3 for three each
20 = 6! / 3!3!
2^6 = 64
???????
If three differently colored dice are rolled at the same time, what is the probability that a 6 will be the outcome on each one?
1⁄216
Each die is a different color, and thus the probability of rolling a 6 on all three is the product of the individual probabilities: 1⁄6 × 1⁄6 × 1⁄6.
If one is testing a goodness of fit to a 9:3:3:1 ratio, how many degrees of freedom would be associated with the Chi-square analysis?
3 degrees of freedom
If an individual has a genetic condition that neither parent has, then that condition must be recessive. Dominant conditions require that every affected individual have at least one affected parent.
In situations where the inheritance mode of a rare condition cannot be definitely determined, the most likely mode is the one that requires the fewest unrelated individuals to have the condition-causing allele.
For autosomal dominant conditions:
Unaffected individuals are homozygous for the recessive, wild-type allele.
Affected individuals with only one affected parent are heterozygous.
Affected individuals with any unaffected children are heterozygous.
Affected individuals with two affected parents may be homozygous dominant or heterozygous.
For autosomal recessive conditions:
Affected individuals are always homozygous recessive.
Unaffected children of an affected parent are always carriers (heterozygous).
Both parents of affected individuals must have at least one recessive allele.
If both parents are carriers, their unaffected children may be carriers or homozygous for the dominant, wild-type allele.
Two normal parents have five children, four normal and one albino. What are the most likely genotypes of the parents?
Aa x Aa
A normal male and an albino female have six children, all normal. What are the most likely genotypes of the parents?
male: AA
female: aa
A normal male and an albino female have six children, three normal and three albino. What are the most likely genotypes of the parents?
male: Aa
female: aa
Suppose that one of the normal children Aa and one of the albino children aa become the parents of eight offspring. What would be the expected number of normal and albino offspring?
4 normal
4 albino
You would like to know whether the progeny of a pair of mated fruit flies are distributed among the resulting four phenotypic classes in a 1:1:1:1 ratio. You perform a chi-square test and obtain a p value of 0.04. Assuming you have done the test correctly, select the best interpretation of this result.
The differences between the observed and expected counts are too large to be attributed to chance.
It is on this basis (p
True or False?
The larger the number of total individual counts in a chi-square test, the higher the degrees of freedom (df).
FALSE
df increases with increasing n (number of categories), but not necessarily with increasing number of subjects.
The expected ratio of phenotypes among the progeny of a test cross is 1:1:1:1. Out of 200 total resulting progeny, 48 occur in one of the four phenotypic classes. Given this information, which of the following must also be true?
The value of observed - expected for this cell = -2.
With a total of 200, the expected number in each cell when the predicted ratio is 1:1:1:1 = 50. The observed number is 48.
Proband
Term that describes the individual in a pedigree whose phenotype was first brought to the attention of a medical researcher.
The proband is the affected individual who is first brought to the attention of a medical researcher; usually the pedigree is constructed around this individual.
Which of the following statements about autosomal recessive traits is true?
If an individual expresses the trait, then all of his or her offspring will also express the trait. If an individual expresses the trait, none of his or her offspring will express the trait. Only females can pass the trait to their offspring. If neither parent expresses the trait, but the offspring does, both parents must be heterozygous for the trait.
If neither parent expresses the trait, but the offspring does, both parents must be heterozygous for the trait.
For an autosomal recessive trait to be expressed, the individual must be homozygous for the recessive allele. The only way (excluding new mutations) for homozygous recessive offspring to be produced from parents who do not express the trait is if both parents are heterozygous for the trait.
What is the most likely mode of transmission for a trait that is not expressed in parents but is expressed by one half of the sons?
Rare X-linked recessive
This trait is X-linked, and since only half the sons are affected, it is a recessive trait and the mother must have been heterozygous.
The results of two of Mendel’s crosses are shown. You want to calculate the χ2 value and determine the p value for both crosses.
What is the null hypothesis to be tested using χ2 analysis?
There is no difference between the observed values and expected values for a monohybrid cross; the data fits a 3:1 ratio.
Independent Events
Chance of event 1 occurring is independent of event 2
Product Law
multiply the probability of each event to determine probability of both events
Heads and Tails
P(Heads) on a single flip of the coin = 1/2
P(Tails) on a single flip of the coin= 1/2
P(H)P(T) = 1/2 x 1/2 = 1/4 P(H)P(H) = 1/2 x 1/2 = 1/4 P(T)P(H) = 1/2 x 1/2 = 1/4 P(T)P(T) = 1/2 x 1/2 = 1/4 P(T)P(T)P(T) = 1/2 x 1/2 x 1/2 = 1/8
2 or more events happening simultaneously = product of their individual probabilities
Sum Law
Outcome can be accomplished in more than one way
Sum the probabilities of each way of obtaining the outcome
toss 2 coins, what is the “outcome” of one head & one tail: PH x PT + PT x PH 1/2 x 1/2 + 1/2 x 1/2 = 1/4 + 1/4 = 1/2
this accounts for the different order in which you can achieve the outcome
Random events —–> independent assortment
Meoisis
1st meiotic anaphase, homologs separate independently
Possible combinations = 2n
Humans: n = 23, therefore 2^23 = 8 x 10^6
8 million gametes possible just because of random assortment of homologous chromosomes!
Meiosis 1 features the independent assortment of homologs
Genetic material?
Schwann proposed cell theory around 1830
Darwin’s Origin of Species 1859. Darwin’s proposed pangenesis included gemmules to describe physical units representing body parts etc.
Lamarck: “use it or lose it” theory of heredity
Mendel’s experiments
Used common garden pea
Followed discrete traits
Kept detailed records
Determined units of inheritance
Basis of Mendelian or transmission genetics
Mendel’s postulates
Unit factors exist in pairs (alleles)
One unit factor is DOMINANT, the other is recessive
Paired unit factors segregate randomly
Monohybrid Cross
Trait transmission from one generation to the next
P1 = parental generation F1 = first filial generation F2 = second filial generation
Unit factors = genes
2 categories: df= 2-1 = 1
Phenotype
physical expression of a trait
Genotype
allelic makeup of a trait
Homozygote / homozygous
identical alleles
eg. DD or dd
Pure breeding lines
homozygous
Heterozytoge / heterozygous
different alleles
eg. Dd
Genes are parts of chromosomes
nonhomologous chromosomes assort independently.
all possible gametic combination are formed with equal probability
Backcross
crossing of an F1 individual to either one of the P1 parents (or parent genotype)
Testcross
crossing of an F2 offspring to homozygous recessive individual to determine genotype of the F2 individual
Cross to homozygous recessive to determine genotype of unknown.
Mendel’s seven monohybrid crosses of the garden pea
In each case, pollen derived from plants exhibiting one trait was used to fertilize the ova of plants exhibiting the other trait. In the F1 generation, one of the two traits, (dominant) was exhibited by all plants. The contrasting trait (recessive) then reappeared in approximately 1/4 of the F2 plants.
Dihybrid cross
Follow 2 pairs of contrasting traits
Independent genes
Independent events (can use product law)
Probability calculation (forked line method)
’s of genotypes & phenotypes
First, determine # of heterozygous gene pairs in cross (n)
2n = # of different gametes 2n = # of different phenotypes 3n = # of different genotypes
Trihybrid cross
of phenotypes = 2n
(n= 3 of heterozygous gene pairs)
2^3 =
8
True breeding
homozygous
2 true breeding parents crossed
P1 Round (WW) terminal (tt) Violet (VV) constricted (cc)
P2 wrinkled (ww) Axial (TT) white (vv) Full (CC)
Round, Axial, Violet, Full are Dominant
F1 Cross yields WwTtVvCc
F2 Most common phenotype: Round Axial Violet Full
Ww Tt Vv Cc x Ww Tt Vv C
3/4 x 3/4 x 3/4 x 3/4 = 81/256
Least likely phenotype in F2 ?
All recessive ww tt vv cc
= 1/4 x 1/4 x 1/4 x 1/4 = 1/256
Binomial Theorem
Used when 2 outcomes are possible in any one trial
Use to quickly calculate specific number of outcomes among large number of events
Eg., calculate number of 2 males & 2 females in a family of 4 children
2 alternative outcomes
Probability based on # of trials
Can calculate the probability of # of outcomes of one type in n trials
p= (n!/s!t!) x (a^s)(b^t)
p= probability; n= # of trials; a=prob of event a; b=prob of event b; s= # of times of outcome a; t=# of times of outcome b
2 males & 2 females in 4 children
Queston: P(2 males & 2 females in 4 trials)
p= (n!/s!t!) x (a^s)(b^t) p= probability; n= # of trials; a=prob of event a; b=prob of event b; s= # of times of outcome a; t=# of times of outcome b
n = 4; a = .5; b = .5; s = 2; t = 2
P= (4!/2!2!) x (.52)(.52) P= 24/4 x (.25)(.25)
Cystic fibrosis question
2 heterozygous parents for CF
They have 5 (n) children, what is probability that 3 will be normal?
Cystic fibrosis: autosomal recessive disorder: Cf-/Cf- affected
Pa = normal = 3/4 = 0.75 Pb = afflicted = 1/4 = 0.25
P3normal = 5!/3!2! x (.75^3)(.25^2) P3normal = 10 x .0264 = 0.264
Goodness of Fit tests
Use statistical test to determine if observed data “fits” expected
Chi-Square Analysis
Tests observed vs. expected results
Null hypotheses
“Reject” or “fail to reject” the null hypothesis
Significance level, p values
Degrees of freedom (# of categories - 1)
x= ∑ (o-e)^2/e = d^2/e
Human genetics
Can’t do large scale crosses…
So we reconstruct genotypes from analysis of family histories
Pedigree analysis
Human pedigree symbols
!!!!!
Lethal alleles
Can be recessive or dominant lethal
2 copies of mutant present is lethal (homozygous): recessive lethal allele
1 copy of mutant is lethal (heterozygous): dominant lethal, eg. Huntingtons disease
Autosomal recessive traits
of males = # of females
Affected usually produce unaffected
Most affected have unaffected parents
Parents of affected sometimes related
Individual with affected sibling has 25% chance of being affected
Trait may skip generations but still be carried through carriers (heterozygous)
Autosomal dominant traits
of male = # of females
Affected has usually affected parents
Individual with affected parent has 50% chance of being affected
Does not skip generations
Example: Huntington’s Disease
Huntington’s Disease
Huntington protein aggregates,
leading to degeneration of brain tissue
Mapped to Chromosome 4
Autosomal dominant
Disease caused by increase in CAG polynucleotide repeat
in the gene, leads to repeat of glutamine in the protein
Mouse model of disease now being studied
Dominant Human traits
Achondroplasia Brachydactyly Congenital night blindness Ehler-Danlos syndrome Hypotrichois Huntington disease Hypercholesterolemia Marfan syndrome Neurofibromatosis Phenylthiocarbamide (PTC) tasting Porphyria (some) Widow’s peak
Recessive Human traits
Albinism Alkaptonuria Ataxia telangiectasia Color blindness Cystic fibrosis Duchenne MD Galactosemia Hemophilia Lesch-Nyhan syndrome Phenylketonuria Sickle-cell anemia Tay-Sachs disease
Penetrance
percentage of individuals that show some degree of expression of the mutant genotype
Eg., In Huntingtions disease, all having the dominant mutant gene get the disease: Penetrance is 100%
Expressivity
Observed range of the mutant phenotype
Eg., trisomy 21 (Down Syndrome) has variable expressivity in observed phenotype
Phenylketonuria
Autosomal recessive disorder
Inability to metabolize phenylalanine
Leads to mental retardation
X- Linkage
Many animals and some plants have one sex contains a pair of “unlike” chromosomes
These chromosomes involved in sex determination
This pair must contain some region of homology for pairing
However, one chromosome contains genes not found on the other
In human and Drosophila, this is the case for males, which have one X and one Y chromosome
Genes present on X, but absent on Y are said to be “X-linked” genes
In males, whatever alleles are on the X are expressed
Males are neither homozygous or heterozygous for X-linked genes
Males are hemizygous for these X-linked genes
Drosophila X-linked example
white eye mutation
Normal color is red
Results were different depending on which parent exhibited the recessive mutant trait
Reciprocal crosses did not yield identical results
Conclusion was that the white locus was on the X chromosome
hemizygous
Having a gene present in a single dose in an otherwise diploid cell.
Usually applied to genes on the X chromosome in heterogametic males.
X-linkage in humans
Identified in pedigrees by crisscross pattern of inheritance: all sons exhibit mutant trait inherited from the mother
Examples: certain types of color blindness, Hemophilia A, B
Duchenne Muscular Dystrophy (DMD)
Ishihara color-blindness chart
Red-green color-blind individuals see a 3, rather than an 8 visualized by those with normal color vision.
This colorblindness is x-linked
