UNIT VI - MENDELIAN GENETICS

UNIT VI - MENDELIAN GENETICS
Baby Campbell – Ch 9 Big Campbell – Ch 14, 15 Hillis – Chp 8

I. MENDEL Mendel’s Experiments Worked with ______________
Eliminated ________________________ and controlled ____________________ P Generation - True-breeding pea plants with one trait X true-breeding pea plants with another trait Produced hybrids also known as F1 ___________________________ F1 Phenotype = F1 Genotype = F1 X F1 → F2 F2 Phenotype Ratio = F2 Genotype Ratio = pea plants self-pollination cross-pollination TT x tt 4 tall : 0 short 0 TT : 4 Tt : 0 tt Tt x Tt 3 tall : 1 short 1 TT : 2 Tt : 1 tt

I. MENDEL, cont Mendel’s Principles
Alternative versions of genes known as ______________ account for variations in inherited characters. Organisms inherit _____ alleles for each trait If alleles at a locus differ; that is; if the genotype is _______________, the allele that shows is known as the ________________ allele. Law of Segregation – two alleles for a heritable character segregate during meiosis Law of Independent Assortment – each pair of alleles segregates independently of each other pair of alleles during meiosis alleles 2 heterozygous dominant

If dependent assortment… then gametes = YR yr
Independent Assortment or Dependent Assortment??? P generation = YYRR x yyrr Gametes = YR x yr F1 Generation = YyRr Predict F2 offspring… If YyRr x YyRr…. If dependent assortment… then gametes = YR yr If Independent Assortment… Gametes are ____________

II. ANALYZING PROBABILITY OF TRAIT INHERITANCE
Test Cross Organisms with dominant phenotype (PP or Pp) crossed with ______________________(pp)__ to determine genotype Punnett Square – tools used to predict the possible outcomes of meiosis and fertilization. Multiplication Rule States the probability of 2 or more independent events occurring together can be calculated by multiplying individual probabilities For example, Determine the probability of a homozygous recessive short plant produced from F1 X F1 Cross = Tt x Tt Probability of egg carrying t = ½ Probability of sperm carrying t = ½ Probability of tt offspring = ¼ homozygous recessive

II. ANALYZING PROBABILITIES, cont
Addition Rule States that the probability of 2 or more mutually exclusive events occurring can be calculated by adding together their individual probabilities For example, Determine the probability of a heterozygous plant produced from F1 X F1 Tt x Tt Chance of egg carrying T = ½ Chance of sperm carrying t = ½ Chance of sperm carrying T = ½ Chance of egg carrying t = ½ Probability of Tt offspring = ¼ + ¼ = ½

Crosses Involving Multiple Characters Determine the genotype ratios of the offspring for the cross BbCCDD X BBccDd

Crosses Involving Multiple Characters Determine the phenotype ratios of the offspring for the cross YyRrSS X yyRrss

Crosses Involving Multiple Characters In the cross, PpYyRr X Ppyyrr, what is the probability of offspring that are purple, green, & round? P= purple, p = white Y = yellow, y = green R = round, r = wrinkled

Probability Practice Makes Perfect! 
In pea plants, long stems are dominant to short stems purple flowers are dominant to white, and round seeds are dominant to wrinkled. A plant that is heterozygous for all three loci self-pollinates and 2048 progeny are examined. How many of the resulting plants would you expect to be long-stemmed with purple flowers, producing wrinkled seeds? 10

III. VARIATIONS IN INHERITANCE
Co-Dominance Both alleles affect phenotype in separate & distinguishable ways Often designated with 2 different “big letters” For example, __In chickens, BB = black WW = white

III. VARIATIONS IN INHERITANCE
Incomplete Dominance Neither allele is dominant; heterozygotes show a blend of two homozygous phenotypes One allele designated with “big letter’, the other with “big letter prime”; for example T T’ For example, in snapdragons, RR = RED R’R’ = WHITE

III. VARIATIONS IN INHERITANCE, cont
Multiple Alleles Many genes have more than 2 alleles Example, ABO blood groups in humans Three alleles Rh Factor Phenotype Genotype A B AB O A woman with O positive blood has a child with Type A negative blood. The man she claims is the father has AB positive blood. Is it possible that he is the father of this child?

Polygenic Inheritance For example, AABBCC = very dark skin; aabbcc = very light skin. Intensity based on units; in other words, AaBbCc and AABbcc individuals would have the same pigmentation

Epistasis Gene at one locus alters phenotypic expression of a gene at a second locus For example, A dominant allele, P causes the production of purple pigment; pp individuals are white. A dominant allele C is also required for color production; cc individuals are white. What proportion of offspring will be purple from a ppCc x PpCc cross?

Pleiotropy – gene can have multiple phenotypic effects. EX: the gene for flower color also affects the gene for seed coat color.

Environmental Impact on Phenotypes Phenotype of a trait depends on environment as well as genotype. EX: size and shape of leaves on a tree also depends on sun, wind, water, etc. in addition to alleles. EX: sun on skin, exercise on build, nutrition on height, etc.

IV. SEX-LINKED INHERITANCE
First recognized by Thomas Hunt Morgan Drosophila melanogaster Fruit flies Excellent organism for genetic studies Prolific breeding habits Simple genetic make-up; 4 pairs of chromosomes → 3 pairs of autosomes, 1 pair of sex chromosomes Crossed true-breeding wild-type females with true-breeding mutant males Mutant trait showed up in ½ male F2 offspring ; was not seen in F2 females Determined mutant allele was on X-chromosome; thus inherited differently in males versus females In females, In males,

IV. SEX-LINKED INHERITANCE, cont
Red-green colorblindness is caused by a sex-linked recessive allele. A color-blind man marries a woman with normal vision whose father was colorblind. What is the probability she will have a colorblind daughter?

The gene for amber body color in Drosophila is sex-linked recessive. The dominant allele produces wild type body color. The gene for black eyes is autosomal recessive; the wild type red eyes are dominant. If males with amber bodies, heterozygous for eye color are crossed with females heterozygous for eye color and body color, calculate the expected phenotype ratios in the offspring.

X Inactivation in Females During embryonic development, one X chromosome in female cells is inactivated due to addition of methyl group to its DNA Dosage compensation Inactive X chromosome condenses; known as Barr body

IV. SEX-LINKED IN INHERITANCE, cont
Occurs randomly Females will have some cells where “Dad’s copy” of X is inactivated, some where “Mom’s copy” is inactive Therefore, females are a mosaic of cells Preserved in mitosis In ovaries, Barr body chromosome is reactivated for meiosis and oogenesis

X Inactivation Calico coloration in female cats

V. PEDIGREE ANALYSIS

Recessive Trait Dominant Trait

VI. CHROMOSOMAL BASIS OF INHERITANCE
Chromosomal Theory of Inheritance States genes occupy specific loci on chromosomes During meiosis, chromosomes undergo segregation & independent assortment Linked Genes During Thomas Morgan’s work with Drosophila, he recognized Two genes located on same chromosome were linked; that is, inherited together However, offspring phenotypes showed this wasn’t always true

VI. CHROMOSOMAL BASIS OF INHERITANCE, cont
Linked Genes, cont In fruit flies, normal wild-type phenotype is gray body, normal wings – both genes are located on same chromosome G = wild-type (gray) body; g = black body W = wild-type wings; w = mutant wings True-breeding wild type flies X true-breeding mutants F1 showed all F1 X test cross Counted 2300 offspring Should have counted

965 GWgw 944 gwgw 206 Gwgw 185 gWgw Morgan realized variation in probabilities due to

Linkage Maps In crossing over, the further apart two genes are, the higher the probability that a crossover will occur between them and therefore, the higher the recombination frequency.

Linkage Maps, cont Recombination Frequency = # recombinants__ X 100 total # offspring One map unit = 1% recombination frequency Morgan’s Data … F1 X test cross GW/gw X gw/gw Counted 2300 offspring Should have counted 965 GWgw 944 gwgw 206 Gwgw 185 gWgw

In a genetics experiment with fruit flies, P generation true-breeding flies are crossed. The female parent is brown and wingless while the male parent is black with normal wings. All of the F1 flies are brown with normal wings. In the next cross, F1 females are test-crossed with black, wingless males. The following offspring are counted: 85 brown winged flies 728 black winged flies 712 brown wingless flies 75 black wingless flies How many map units apart are the two genes?

The genes for vestigial wings, black body color, and cinnabar eyes are linked genes. In controlled crosses . .. The gene for vestigial (vg) wings and body color (b) have a 17% crossover rate. The gene for eye color (cn) and body color (b) have a 9% crossover rate. The gene for eye color (cn) and vestigial wings (vg) have a 9 ½% crossover rate. Draw the chromosome.

VII. MUTATIONS Change in the nucleotide sequence of DNA
May be spontaneous mistakes that occur during replication, repair, or recombination May be caused by mutagens; for example, x-rays, UV light, carcinogens Classification Gene Chromosomal

VIII. INHERITED GENETIC DISORDERS
Most often due to gene mutations Classified according to type of chromosome on which affected gene is located Autosomal Disorders Affects 1st 22 pair of chromosomes Sex-linked Disorder Affects last pair of chromosomes (X) Grouped according to path of inheritance Recessive Co-Dominant Dominant

VIII. INHERITED GENETIC DISORDERS
Autosomal Recessive Disorders Albinism Cystic Fibrosis PKU Tay-Sachs

VIII. INHERITED GENETIC DISORDERS, cont
Autosomal Co-Dominant Disorders Sickle Cell Anemia

Autosomal Dominant Disorders Huntington’s Disease

Autosomal Dominant Disorders Marfan Syndrome

Autosomal Dominant Disorders Achondroplasia

Hypercholesteremia

Sex-Linked Disorders Vast majority are recessive Affect mostly males Hemophilia

Colorblindness Duchenne Muscular Dystrophy

IX. CHROMOSOMAL MUTATIONS
Due to abnormal chromosome structure or abnormal number of chromosomes Abnormal Chromosome Structure Often due to mistakes in

IX. CHROMOSOMAL MUTATIONS, cont
Abnormal Number of Chromosomes Most commonly due to nondisjunction Results in aneuploid gametes Detected with __karyotype_______

IX. CHROMOSOMAL MUTATIONS, cont
Chromosomal Number Disorders Examples Down Syndrome

X. CHROMOSOMAL DISORDERS, cont
Turner Syndrome

X. CHROMOSOMAL DISORDERS, cont
Klinefelter Syndrome

XI. TESTING FOR INHERITED GENETIC DISORDERS
Identification of Carriers/Genetic Counseling Tests are available for Tay-Sachs, sickle cell, cystic fibrosis, Huntington’s, PKU, & many others Fetal Testing Gene Testing Amniocentesis Performed between 14th-16th weeks of pregnancy Cells collected, tested

XI. GENETIC TESTING, cont
Chorionic Villus Sampling (CVS) Narrow tube inserted through mother’s vagina, cervix Small tissue sample suctioned from placenta (organ that transmits nutrients, removes wastes from fetus) Testing may be done earlier in pregnancy but not suitable for all types of testing Karyotype

XI. GENETIC TESTING, cont
Newborn Screening PKU PGD Pre-Implantation Genetic Diagnosis

UNIT VI - MENDELIAN GENETICS

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