Chapter 15 - The Chromosomal Basis of Inheritance 15...The offspring had the following phenotypic...

Chapter 15 - The Chromosomal Basis of Inheritance

A. Bergeron+AP BiologyPCHS

Do Now - Predicting Unpredictable Genotypes

Devise a strategy for determining the unknown genotype of both parental tomato plants using as few resources as possible. Remember, I want to maximize the yield of fruit from all of the future generations of tomato plants.

As an inexperienced (albeit precocious) gardener, I am always looking to maximize my yield of fruits and vegetables during the summer.

Last summer, I identified two tomato plants that produced an inordinate amount of fruit. I contemplated breeding these two plants (either through cross-pollination or self-fertilization) but was unsure of either plant’s genotype.

Do Now - Sex-linked InheritanceThe table below shows the results of a cross of two red-eyedfruit flies. Some of the offspring have “ruby”-colored (deep pink) eyes instead of the usual red-colored eyes.

Eye Color! ! Numbers of Offspring! ! ! Males!! FemalesRed eyes! ! 77! ! 151“Ruby” eyes! ! 75! ! 0

b. How do the results of the cross indicate that ruby eye color is sex-linked?

a. Using XR for the dominant allele and Xr for the recessive allele, write the genotypes of the two parents.

TestcrossUsed to determine the genotype of a “parent” with the dominant phenotype but an unknown genotype

Dominant parent “crossed/mated” with a homozygous recessive parent

Genotype of the dominant parent can be determined based onthe offspring produced from the cross

If offspring with ONLY the dominant phenotype are produced, the parent must have the homozygous dominant genotype

If offspring with the dominant and the recessive phenotype areproduced, the parent must have the heterozygous genotype

Sample Testcross

Reading Quiz

1. The most common phenotype in a population of organisms is called the ____________.

2. _______ genes have loci located on the same chromosome.

3. A recombination frequency greater than 50% suggests that two genes are (linked or unlinked)?

4. A father carries a recessive allele for hemophilia on his X chromosome. What is the probability that he will transmit the hemophilia allele to a son?

5. In females, one of the two X-chromosomes is inactivated early in embryonic development. What term is used to describe this inactivated chromosome?

Chromosomal Basis of Mendelian Inheritance

Do Now - Sex-Linked Genetics Practice Problems1. A man with hemophilia (a recessive, sex-linked disease) has a daughter of normal phenotype. She marries a man who is normal for the trait. What is the probability that a daughter of this mating will be a hemophiliac? That a son will be a hemophiliac? If the couple has four sons, what is the probability that all four sons will be born with hemophilia?

2. Red-green color blindness is caused by a sex-linked recessive allele. A color-blind man marries a woman withnormal vision whose father was color blind. What is the probability that they will have a color-blind daughter? What is the probability that their first son will be color-blind?

Chromosomal Basis of Mendelian Inheritance

Mendel’s laws of segregation and independent assortmentassumed that the factors (i.e. alleles) that are responsible for each of the observed traits were on different chromosomes ! -Remember that Mendel did not know of the existence! of genes, chromosomes, or even DNA.

Chromosomal Theory of Inheritance! -Mendelian factors (genes/alleles) are located on ! chromosomes! -Chromosomes segregate and assort independently

Morgan’s Discovery of Sex-linked InheritanceParental Cross: White-eyed male (XwY) x red-eyed (XRXR) female fruit flies (Drosophila)F1 generation (males and females) had red eyesF2 generation had (expected) 3 (red):1(white) ratio of eye color phenotypes but only males had whiteeyes!What could explain this peculiar result?Answer: Sex-linked inheritance - the locus of some genes is located exclusively on one of the sex c’somes

Parental Cross: White-eyed male (XwY) x red-eyed (XRXR)

female fruit flies (Drosophila)

F1 generation (males and females) had red eyes

F2 generation had (expected) 3 (red):1(white) ratio of eye color phenotypes

But only males had white eyes

How Did Morgan Know that the Gene for Red Eyes Was on a Sex Chromosome and Not an

Autosome?

1. Since Morgan hypothesized that the allele for white eyes was recessive, a white-eyed female would need to inherit two (2) X chromosomes with the recessive allele! -This was not observed in the F2 females

2. A white-eyed male would not have a wild-type (normal) allele to “mask” the presence of the recessive allele! -Since the white-eye allele was proposed to be on the ! X chromosome, a single copy of the allele would ! guarantee the white eye phenotype in male flies

Sex-Linked Inheritance (Three Possibilities...)

X-Chromosome Inactivation in Female MammalsExample - Female calico cats

Inactivated X c’somes are called Barr bodies b/c the inactivated c’some appearsas a dark mass of DNA under the microscope

In females, one of the two X chromosomes is inactivatedto prevent multiple copies of a gene from being expressed

The inactivation is random.Either chromosome could beinactivated.

Evidence for Linked Genes in Drosophila

Linked genes do NOT assort independently during meiosis.

Genes located on the same chromosome are considered to be linked

Variations away from expected results of a particular cross are indicative of a non-Mendelian inheritance pattern

Parental = offspring that have the same phenotype as one or more of the parentsRecombinant = offspring whose phenotypes are different from the parents! Recombinant gametes are often the products of ! crossing over

Important Genetics Vocabulary

What would be the expected phenotypic ratio in the offspringif the genes for body color and wing type were unlinked

(i.e. assorted independently)?

(Assume that Morgan produced approximately 2,300 offspring in his cross)

Mendel’s Principle of Independent Assortment

MomDad

A aB b

A a

Bb

A

b

a

B

Ab

Ab

aB

aB

AB

ab

AB

AB

ab

ab

Sex cells

What would be the expected phenotypic ratio in the offspringif the genes for body color and wing type were linked

(i.e. genes located on the same chromosome that donot assort independently)?

Crossing Over

Calculating Recombination Frequency# Recombinants

Total # offspringX 100%

1. A wild-type fruit fly (heterozygous for gray body color and normal wings) was mated with a black fly with vestigialwings. The offspring had the following phenotypicdistribution:! ! 778 wild-type! ! 785 black, vestigial! ! 158 black, normal! ! 162 gray, vestigial

What is the recombination frequency between these genesfor body color and wing type?

Using Recombination Data to Calculate the Distance Between Genes on a Chromosome

A recombination frequency that is greater than or equal to 50% indicates that genes are unlinked (i.e. on separate chromosomes) OR are so far apart from each other on the same chromosome that they behave as if they are on separate chromosomes (frequent cross overs)

Crossover events are more likely to occur between genes thatare further apart on a chromosome

Creating a Genetic MapDistance between genes on achromosome is measuredin centimorgans (cM) in honor of Thomas Hunt Morgan

1 centimorgan = 1 map unit = 1% recombination frequency

One “map unit” is defined as the distance within which a crossover event is expected to occur in an average of 1%of gametes

Genetics Practice Problems

1. Determine the sequence of genes along a chromosome based on the following recombination frequencies:! ! A-B: 8 cM! ! A-C: 28 cM! ! A-D: 25 cM! ! B-C: 20 cM! ! B-D: 33 cM

1 centimorgan (cM) = 1% recombination frequency

2. A fruit fly of genotype BR/br is test-crossed to br/br. (Note: Alleles grouped on either side of the slash are locatedon the same chromosome) In 84% of the meioses, there are no chiasmata between the linked genes; in 16% of themeioses, there is one chiasmata between the genes. Whatproportion of the progeny will be Br/br?

Why Shouldn’t I Include the Parental Gametes Produced as a Result of the Crossover?

Recall…# Recombinants

Total # offspringX 100%% Rec. =

The “formula” for calculating % recombination does NOT take the parental gametes into account

Thus, the probability of producingrecombinant gametes is equivalent to the % recombination

Don’t forget that there is a 50%chance that either of the two gametes will fertilize an egg!

Genetics Practice Problems3. In the plant Arabidopsis thaliana the loci for pod length (L = long, l = short) and fruit hairs (H = hairy, h = smooth) are located 20 cM apart on the same chromosome. The following crosses were made:! ! L H/LH X lh/lh --> F1! ! Lh/Lh X lH/lH --> F1

If the F1’s from above are crossed, what proportion of the progeny are expected to be lh/lh?

4. Using the same information from #3 on the previous slide, if the F1’s are crossed, what proportion of the progeny are expected to be Lh/lh?

Do Now

Genetics Practice Problems5. In Drosophila, the gene for white-eyes and the gene that produces “hairy” wings have both been mapped to the same chromosome and have a crossover frequency of 1.5%. A geneticist notices that in a particular stock of flies, these two genes assorted independently; that is, they behaved as though they were on separate chromosomes. What explanation can you offer for this observation?

Errors and Exceptions to “Normal” Chromosomal Inheritance

Deletion

Duplication

Inversion

Translocation

6. In mice, black coat is dominant to brown and intense pigment is dominant to diluted. One mouse with dilute black pigment is crossed to another with intense brown pigment. The progeny are:


a. What are the genotypes of parents and progeny?

b. Draw the alleles involved on their respective chromosomes.

7. The following testcross of a plant is made:


If the two loci are 15 m.u. apart, what proportion of progeny will be (a) AB/ab? (b) ab/ab? (c) Ab/ab?

8. The A locus and the D locus are so tightly linked that no recombination is ever observed between them. If Ad/Ad is crossed to aD/aD, and the F1 is intercrossed, what genotypes will be seen in the F2 and in what proportions?

A female animal with genotype A/a.B/b is crossed with a double-recessive male a/a.b/b. Their progeny include 442 A/a.B/b, 458 a/a.b/b, 46 A/a.b/b, and 54 a/a.B/b. Explain these proportions and draw the chromosomes of the dihybrid parent showing the positions of the genes and alleles.

Do Now - Genetics Practice Problems

Three-point Testcross

How can we find the order and relative map distance between genes on a chromosome?

Assume you have the following alleles at three different loci:v (vermillion eyes) vs. v+ (red eyes)cv (absence of crossvein on wing) vs. cv+ (presence of crossvein)ct (snipped wing edges) vs. ct+ (smooth wing edges)

Red (left) vs. Vermillion (right)


Available parental stocks: v+/v+ . cv/cv . ct/ct.v/v . cv+/cv+ . ct+/ct+

What is the F1?

Testcross F1 females to male homozygotes

Progeny PhenotypeProgeny PhenotypeProgeny Phenotype # Fliesv cv+ ct+ 580v+ cv ct 592v cv ct+ 45v+ cv+ ct 40v cv ct 89v+ cv+ ct+ 94v cv+ ct 3v+ cv ct+ 5


1. What is the map distance between all three genetic loci?2. Draw the gene map for these three genes.

9. In corn, a triple heterozygote was obtained carrying the mutant alleles s (shrunken), w (white aleurone), and y (waxy endosperm), all paired with their normal wild-type alleles. This triple heterozygote was testcrossed, and the progeny contained 116 shrunken, white; 4 fully wild type; 2538 shrunken; 601 shrunken, waxy; 626 white; 2708 white, waxy; 2 shrunken, white, waxy; and 113 waxy.


a. Determine if any of these three loci are linked, and, if so, show map distances.

BONUS: Show the allele arrangement on the chromosomes of the triple heterozygote used in the testcross.

10. If A/A.b/b is crossed to a/a.B/B, and the F1 is testcrossed, what percent of the testcross progeny will be a/a.b/b if the two genes are (a) unlinked, (b) completely linked (no crossing-over at all), (c) 12 map units apart, (d) 24 map units apart?


Genes of unknown linkage are shown separated by a dot, A/a.B/b

11. You have a Drosophila line that is homozygous for autosomal recessive alleles a, b, and c, linked in that order. You cross females of this line with males homozygous for the corresponding wild-type alleles. You then cross the F1 heterozygous males with their heterozygous sisters. You obtain the following F2 phenotypes (where letters denote recessive phenotypes and pluses denote wild-type phenotypes):

1364 + + +, 365 a b c, 87 a b +, 84 + + c, 47 a + +, 44 + b c, 5 a + c, and 4 + b +.


(a) What is the map distance between genes a and b? b and c? a and c?

Genetics Quiz - Chapter 151. A strain of Neurospora (common bread mold) with the genotype H.I is crossed with a strain with the genotype h.i. Half the progeny are H.I and half are h.i. Briefly explain how this is possible.

a. What pattern of inheritance is shown by each condition in this pedigree?

Genetics Practice Problems12. The following pedigree shows a family with two rare abnormal phenotypes: blue sclerotic (a brittle bone defect), represented by a black-bordered symbol, and hemophilia, represented by a black center in a symbol. Individuals represented by completely black symbols have both disorders.

Genetics Practice ProblemsBlue sclerotic: Autosomal dominantHemophilia: Sex (X)-linked recessive

Pedigree Analysis - Important Symbols

Important Points to Consider When Performing a Pedigree Analysis

If a trait is recessive:! Individuals with the trait must be homozygous! Unaffected parents of children with the trait will be ! heterozygous! Children may exhibit the trait even if their parents do not

If a trait is dominant:! Any child with the trait will have at least one parent ! with the trait! A trait caused by a dominant inheritance pattern! will show up in every generation

Inheritance Patterns & Phenotypic CharacteristicsSimple (Autosomal) Recessive -Males and females are equally likely to have the trait -Either parent can transmit the trait to offspring of either gender *The trait often skips generations -Often, both parents of offspring who have the trait are heterozygous; they carry at least one copy of the allele -Only homozygous recessive individuals have the trait -The trait may appear in siblings w/o appearing in parents -If one parent has the trait, those offspring who do not have it are carriers of the trait

Pedigree Analysis

Alkaptonuria - Is the disease caused by a dominant or recessive allele? What are the possible genotypes of

individuals 1, 2, and 3?

1

2

3

AutosomalRecessive

Pedigree Analysis

Familial Hypercholesterolemia - Is disease caused by a dominant or recessive allele? What are the possible

genotypes of individuals 1 and 2?

1

2

AutosomalDominant

Do Now - Pedigree Analysis

Determine the mode of inheritance of the trait diagrammed inthe above pedigree.

Hint: If one than one mode of inheritance is possible, determine which is most probable.

Pedigree Analysis

Sickle cell disease - Is disease caused by a dominant or recessive allele? What are the possible genotypes of

individuals 1, 2, and 3?

AutosomalRecessive

1

2 3


Pedigree Analysis

Determine the mode of inheritance of the trait diagrammed inthe above pedigree.


Pedigree Analysis

Hint: If one than one mode of inheritance is possible, determine which is most

probable.

Inheritance Patterns & Phenotypic CharacteristicsX-linked Recessive -All daughters of a male who has the trait are carriers -There is no male-to-male transmission -The trait is far more common in males than in females -The son of a female carrier has a 50% chance of having the trait -Mothers of males who have the trait are either carriers or homozygous and have the trait -The daughters of a female carrier have a 50% of being a carrier

Inheritance Patterns & Phenotypic CharacteristicsX-linked Dominant -All daughters of a male who has the trait will also have the trait -There is no male-to-male transmission -A female who has the trait may or not pass the gene for that trait to her son or daughter

Diseases Caused by Recessive Genotypes

1. Cystic fibrosis

2. Tay-Sachs disease

3. Sickle-cell disease

Diseases Caused by Dominant Genotypes1. Achondroplasia

2. Huntington’s disease

*Be familiar with the cause of the disease (dominant or recessive allele/s) and disease symptoms*

Fly Lab Instructions

1. I will give each group 2 petri dishes with a different fly phenotype. Observe each phenotype under the microscope. Be sure to observe each phenotype!

2. Practice separating male flies from female flies. You can use any phenotype to practice.

3. Set up labeled culture vials with Drosophila medium.! Add 1 scoop of tap water to 1 scoop of food flakes! Sprinkle a few grains of yeast on top of the culture.! Wait 4-5 minutes before adding flies

4. Add 4-5 mating pairs of F1 flies from each cross to your labeled culture vials. The flies should be asleep! They will wake up later! I promise…

1) After reviewing the class data (which I will provide for you), you should be able to write a hypothesis that explains the mode of inheritance acting in each cross.

Example - Monohybrid/dihybrid (autosomal) cross! Monohybrid/dihybrid (sex-linked) cross

2) You should assume that chance has been responsible for the independent assortment of alleles that produced the F1 and F2 generations of flies.

ANY variation of your observed results when compared tothe expected results is due to CHANCE

3) In this experiment, variation due to chance alone is called the “null hypothesis”

4) Can the class data be used to support the null hypothesis?

Chi-Square (Χ2) Statistical AnalysisΧ2 is used to determine how well observed ratios fit expected ratios

Chi-square formulaΧ2 = Σ (o-e)2

! -------!! e

You will have observed values from the class data setThe expected values will come from your expected phenotyperatios! For example, if you predict that the offspring should exist in a 3:1 ! ratio and there are 400 offspring, 300 offspring should have one ! phenotype and 100 offspring should have the other phenotype

The experimental Χ2 value is compared to actual Χ2 values from a statistical table to determine significance

Critical Values of the Chi-Square Distribution

20.518.516.313.810.80.001 (0.1%)

15.113.211.39.216.640.01 (1%)

11.19.497.825.993.840.05 (5%)

54321Prob. due to chance

Degrees of Freedom

Degrees of freedom = (# phenotypes - 1)

Unless told otherwise, use Χ2 values for the 5% probability value

Practice ProblemA true-breeding corn plant with purple corn kernels is crossed with a corn plant with yellow corn kernels. An F1 offspring is “selfed” to produce the F2 generation. The results of the F1 cross are shown in the table below:

Phenotype # Observed (o) # Expected (e) (o-e) (o-e)2 (o-e)2/2Yellowkernels

149

Purple kernels

388

Total 537

Are differences between the observed and expected values due to chance alone or is there an underlying cause of the variation? Perform a Χ2 analysis to support your answer.

Chi-Square AnalysisPhenotype # Observed (o) # Expected (e) (o-e) (o-e)2 (o-e)2/2Yellowkernels

149

Purple kernels

388

Total 537

Chi-square formula:Χ2 = Σ (o-e)2

! -------!! e

Null hypothesis: There is no significant difference between the expected and observedresults. The difference/s are dueto chance.

Chi-Square AnalysisPhenotype # Observed (o) # Expected (e) (o-e) (o-e)2 (o-e)2/eYellowkernels

149 134 15 225 1.68

Purple kernels

388 403 -15 225 .56

Total 537 537 2.24

Chi-square formula:Χ2 = Σ (o-e)2

! -------!! e

Null hypothesis: There is no significant difference between the expected and observedresults. The difference/s are dueto chance.

Probability due to chance

1 2 3 4 5

0.05 (5%) 3.84 5.99 7.82 9.49 11.1

0.01 (1%) 6.64 9.21 11.3 13.2 15.1

0.001 (0.1%) 10.8 13.8 16.3 18.5 20.5

Critical Values of the Chi-Square Distribution

Degrees of FreedomDegrees of freedom:(# phenotypes - 1)

Making Conclusions Using Chi-SquareIf the calculated X2 value is less than the critical value for agiven probability of variation due to chance, you must accept your null hypothesis.! -There is no significant difference between the ! expected and observed values.! -Any difference between the expected and observed ! values probably occurred due to chance.

If the calculated X2 value is greater than the critical valuefor a given probability of variation due to chance, youmust reject your null hypothesis! -There is a significant difference between the expected! and observed values! -There is an underlying reason why a difference exists ! between the expected and observed values

Practice Problem - Chi Square AnalysisSuppose that a scientist crossed pink-flowered snapdragonsand obtained 236 offspring. 66 of the offspring grew into plantswith red flowers, 115 with pink flowers, and 55 with white flowers.

a. What are the expected phenotypic ratios of the offspring in the F1 generation?

b. What are the expected number of offspring with each type of flower color?

c. Are the differences between the observed and expected results significant enough to conclude that the results did NOT occur by chance alone?

Chapter 15 - The Chromosomal Basis of Inheritance 15...The offspring had the following phenotypic...

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