Chapter 3: Mendelian Genetics

Question 1

What factors were crucial to the success of Gregor Mendel’s experiments?

Answer 1

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.

Question 2

Which of the following would be true of a plant heterozygous for a single gene controlling flower color? Assume complete dominance.

Answer 2

If crossed with another heterozygous plant, the majority of progeny will have the dominant flower color.

Question 3

Does the 3:1 phenotypic ratio observed among progeny of an F1 X F1 cross requires random union of gametes?

Answer 3

TRUE

The 1:2:1 genotypic ratio represents relative probabilities of gamete combinations based on the assumption that gamete union is random.

Question 4

What would be the best way to determine which of 2 alleles of a gene is dominant to the other?

Answer 4

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.

Question 5

In an individual that is heterozygous for a particular trait, is the recessive allele expressed?

Answer 5

NO

Only the dominant allele is expressed in an individual that is heterozygous for a particular trait.

Question 6

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?

Answer 6

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.

Question 7

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?

Answer 7

25%

The proportion of rr is ¼, so the next offspring has a 25% chance of being wrinkled.

Question 8

Which two genotypes in an individual would be expressed as the same phenotype in a diploid organism?

Answer 8

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.

Question 9

Mendel’s unit factors in pairs are most accurately known to be _______.

Answer 9

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.

Question 10

Homolog

Answer 10

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.

Question 11

Mendel’s Law of Segregation is supported by a 1:1 testcross ratio.

Answer 11

TRUE

Question 12

To test Mendel’s Law of Segregation, the experimenter needs a minimum of two contrasting forms of a gene.

Answer 12

True

Question 13

A 1:1 phenotypic ratio is expected from a monohybrid testcross with complete dominance.

Answer 13

TRUE

Question 14

Does the law of independent assortment states that one gene in a pair is always dominant to the other?

Answer 14

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.

Question 15

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?

Answer 15

3⁄16

This is the probability of offspring that are either green with round seeds or yellow with wrinkled seeds.

Question 16

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?

Answer 16

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.

Question 17

Which of the following phenomena is a consequence of Independent Assortment?

Answer 17

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.

Question 18

The 9:3:3:1 ratio exhibited in the F2 generation of a dihybrid cross is a genotypic ratio.

Answer 18

FALSE

Question 19

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?

Answer 19

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).

Question 20

How many different kinds of gametes can be produced by an individual with the genotype AABbCCddEeFf ?

Answer 20

8 gametes

1 x 2 x 1 x 1 x 2 x2

Question 21

Which of the following statements is true regarding a trihybrid cross between two true-breeding homozygous individuals with contrasting phenotypes?

Answer 21

The least frequent F2 phenotypic class is recessive for all three traits.

1/64 of the F2 phenotypes is triply recessive.

Question 22

Which of the following statements is an example of independent assortment?

Answer 22

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.

Question 23

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?

Answer 23

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).

Question 24

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.

Answer 24

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).

Question 25

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?

Answer 25

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.

Question 26

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?

Answer 26

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).

Question 27

Is Mendel’s postulate of independent assortment supported by a 1:1:1:1 testcross ratio?

Answer 27

YES

Question 28

Which phenotypic ratio is likely to occur in crosses of two completely dominant, independently segregating gene pairs when both parents are fully heterozygous?

Answer 28

9:3:3:1

Question 29

How could the botanist best determine whether the genotype of the green-pod plant is homozygous or heterozygous?

Answer 29

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.

Question 30

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?

Answer 30

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.

Question 31

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.)

Answer 31

meiosis I, anaphase

Alleles separate from one another during anaphase of meiosis I, when the homologous pairs of chromosomes separate.

Question 32

Independent assortment explains _______.

Answer 32

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.

Question 33

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)?

Answer 33

NO

Even though there would be no genotypic differences in the products of such meioses, random alignment of chromosomes would still have occurred.

Question 34

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

Answer 34

AaBC

This gamete contains two copies of gene “A”. Proper segregation would have separated A from a and allowed only one copy per gamete.

Question 35

How did Mendel’s work support the chromosomal theory of inheritance?

Answer 35

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.

Question 36

True or False?

A gene can have a maximum of two alleles.

Answer 36

FALSE

Question 37

Chi-square analysis can help us to decide whether _______.

Answer 37

our observations of an event differ from our expectations.

This is exactly the purpose of statistical tests.

Question 38

What is the probability of flipping a penny and a nickel and obtaining one head and one tail?

Answer 38

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.

Question 39

A couple has two boys and is expecting a third child. What is the probability that the third child will be a boy?

Answer 39

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.

Question 40

What is the probability of a family with six children having three boys and three girls?

Answer 40

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
???????

Question 41

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?

Answer 41

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.

Question 42

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?

Answer 42

3 degrees of freedom

Question 43

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.

Answer 43

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.

Question 44

For autosomal dominant conditions:

Answer 44

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.

Question 45

For autosomal recessive conditions:

Answer 45

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.

Question 46

Two normal parents have five children, four normal and one albino. What are the most likely genotypes of the parents?

Answer 46

Aa x Aa

Question 47

A normal male and an albino female have six children, all normal. What are the most likely genotypes of the parents?

Answer 47

male: AA
female: aa

Question 48

A normal male and an albino female have six children, three normal and three albino. What are the most likely genotypes of the parents?

Answer 48

male: Aa
female: aa

Question 49

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?

Answer 49

4 normal

4 albino

Question 50

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.

Answer 50

The differences between the observed and expected counts are too large to be attributed to chance.

It is on this basis (p

Question 51

True or False?

The larger the number of total individual counts in a chi-square test, the higher the degrees of freedom (df).

Answer 51

FALSE

df increases with increasing n (number of categories), but not necessarily with increasing number of subjects.

Question 52

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?

Answer 52

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.

Question 53

Proband

Answer 53

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.

Question 54

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.

Answer 54

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.

Question 55

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?

Answer 55

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.

Question 56

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?

Answer 56

There is no difference between the observed values and expected values for a monohybrid cross; the data fits a 3:1 ratio.

Question 57

Independent Events

Answer 57

Chance of event 1 occurring is independent of event 2

Question 58

Product Law

Answer 58

multiply the probability of each event to determine probability of both events

Question 59

Heads and Tails

Answer 59

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

Question 60

Sum Law

Answer 60

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

Question 61

Random events —–> independent assortment

Meoisis

Answer 61

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

Question 62

Genetic material?

Answer 62

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

Question 63

Mendel’s experiments

Answer 63

Used common garden pea

Followed discrete traits

Kept detailed records

Determined units of inheritance

Basis of Mendelian or transmission genetics

Question 64

Mendel’s postulates

Answer 64

Unit factors exist in pairs (alleles)

One unit factor is DOMINANT, the other is recessive

Paired unit factors segregate randomly

Question 65

Monohybrid Cross

Answer 65

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

Question 66

Phenotype

Answer 66

physical expression of a trait

Question 67

Genotype

Answer 67

allelic makeup of a trait

Question 68

Homozygote / homozygous

Answer 68

identical alleles

eg. DD or dd

Question 69

Pure breeding lines

Answer 69

homozygous

Question 70

Heterozytoge / heterozygous

Answer 70

different alleles

eg. Dd

Question 71

Genes are parts of chromosomes

Answer 71

nonhomologous chromosomes assort independently.

all possible gametic combination are formed with equal probability

Question 72

Backcross

Answer 72

crossing of an F1 individual to either one of the P1 parents (or parent genotype)

Question 73

Testcross

Answer 73

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.

Question 74

Mendel’s seven monohybrid crosses of the garden pea

Answer 74

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.

Question 75

Dihybrid cross

Answer 75

Follow 2 pairs of contrasting traits

Independent genes

Independent events (can use product law)

Probability calculation (forked line method)

Question 76

’s of genotypes & phenotypes

Answer 76

First, determine # of heterozygous gene pairs in cross (n)

2n = # of different gametes
2n = # of different phenotypes
3n = # of different genotypes

Question 77

Trihybrid cross

Answer 77

of phenotypes = 2n
(n= 3 of heterozygous gene pairs)
2^3 =
8

Question 78

True breeding

Answer 78

homozygous

Question 79

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

Answer 79

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

Question 80

Binomial Theorem

Answer 80

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

Question 81

2 males & 2 females in 4 children

Answer 81

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)

Question 82

Cystic fibrosis question

2 heterozygous parents for CF

They have 5 (n) children, what is probability that 3 will be normal?

Answer 82

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

Question 83

Goodness of Fit tests

Answer 83

Use statistical test to determine if observed data “fits” expected

Question 84

Chi-Square Analysis

Answer 84

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

Question 85

Human genetics

Answer 85

Can’t do large scale crosses…

So we reconstruct genotypes from analysis of family histories

Pedigree analysis

Question 86

Human pedigree symbols

Answer 86

!!!!!

Question 87

Lethal alleles

Answer 87

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

Question 88

Autosomal recessive traits

Answer 88

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)

Question 89

Autosomal dominant traits

Answer 89

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

Question 90

Huntington’s Disease

Answer 90

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

Question 91

Dominant Human traits

Answer 91

Achondroplasia
Brachydactyly
Congenital night blindness
Ehler-Danlos syndrome
Hypotrichois
Huntington disease
Hypercholesterolemia
Marfan syndrome
Neurofibromatosis
Phenylthiocarbamide (PTC) tasting
Porphyria (some)
Widow’s peak

Question 92

Recessive Human traits

Answer 92

Albinism
Alkaptonuria
Ataxia telangiectasia
Color blindness
Cystic fibrosis
Duchenne MD
Galactosemia
Hemophilia
Lesch-Nyhan syndrome
Phenylketonuria
Sickle-cell anemia
Tay-Sachs disease

Question 93

Penetrance

Answer 93

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%

Question 94

Expressivity

Answer 94

Observed range of the mutant phenotype

Eg., trisomy 21 (Down Syndrome) has variable expressivity in observed phenotype

Question 95

Phenylketonuria

Answer 95

Autosomal recessive disorder

Inability to metabolize phenylalanine

Leads to mental retardation

Question 96

X- Linkage

Answer 96

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

Question 97

Drosophila X-linked example

Answer 97

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

Question 98

hemizygous

Answer 98

Having a gene present in a single dose in an otherwise diploid cell.

Usually applied to genes on the X chromosome in heterogametic males.

Question 99

X-linkage in humans

Answer 99

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)

Question 100

Ishihara color-blindness chart

Answer 100

Red-green color-blind individuals see a 3, rather than an 8 visualized by those with normal color vision.

This colorblindness is x-linked