Chromosomal Basis of Inheritance Chapter 15 I Mendels

Chromosomal Basis of Inheritance Chapter 15

I. Mendel’s Work was Cool!! A. Mendel’s Laws 1. Segregation The two alleles for each gene separate during gamete formation 2. Independent Assortment Each sex cell gets one of each chromosome (mom’s or dad’s)

I. Mendel’s Work was Cool!! B. Chromosomal Theory of Inheritance 1. Mendelian genes have specific loci on chromosomes 2. Chromosomes undergo segregation and independent assortment.

II. Another Geneticist. . Mr. Morgan A. Morgan’s Experiments 1. Used Fruit Flies • Breed Fast, only four pair of chromosomes, cute • Wild Type. . Most common Red-eyes, normal wings, gray body • Mutant Forms. . Alternates to wild type

A. Morgan’s Experiments 2. Mated a red-eyed female to a whiteeyed male • Resulted in. . . All red-eyed flies 3. Selfed these and got. . • 3 Red-eyed to 1 White-eyed BUT. . • All the girls were red-eyed • Half the boys were RED-eyed • Half the boys were WHITE-eyed

B. Morgan’s Conclusions 1. Specific genes are on specific chromosomes 2. Eye color in flies is linked to sex of the fly. 3. SEX-LINKED genes: Genes that are carried on the X sex chromosome

III. Sex Chromosomes and their Cool Linkage Issues • Human Female = XX • Human Male = XY • Grasshopper Female =X • Grasshopper Male =XX • Bird Females =XY • Bird Males = XX

A. Sex-linkage Problems • Traits travel only on the X!! • Always cross xx and xy • Sex-linked recessive disorders to know. . • Colorblindness • Duchenne Muscular Dystrophy • Hemophilia

• 1. A woman is a hemophiliac, but her husband is not. What are the phenotypic ratios of their potential children? Xh XH X h x X Hy • Genotype results • 1/2 XHXh, 1/2 Xh. Y • Phenotype results • Girls – normal, boys hemophilia Y Xh X HX h X h. Y

• 2. A man is a hemophiliac, but his wife is normal and homozygous for the trait. What are the phenotypic ratios of their potential children? X HX H x X hy XH XH Xh X HX h Y X HY • Genotype results • 1/2 XHXh, 1/2 XHY • Phenotype results • All normal

B. X-Inactivation • In humans, one X chromosome usually becomes inactive during embryo formation. • A Methyl group -CH 3 attaches to the chromosome • This is called a BARR Body. • The X that becomes the Barr body is random (could be from mom or dad) • Stays active in the ovaries of females

IV. The Exception to Mendel’s Laws • Every chromosome has hundreds or thousands of genes. A. Linked Genes • Genes located on the same chromosome tend to be inherited together (chromosome is passed on as a unit)

B. The Example. . • • • Fruit Fly Traits: b+ Gray (wild) vg+ Normal wings (wild) b Black (mutant) vg Vestigial wings (mutant)

• P 1 = b+ b+ vg+ x b b vg vg • F 1 = all b+ b vg+ vg • Next test cross. . . • b+ b vg+ vg x b b vg vg • According to Mendel, we get. . • Equal Numbers of all genotypes • Oops. . he got. . • 965 • 944 b+ b vg+ vg 206 b+ b vg vg b b vg vg 185 b b vg+

C. The Conclusions 1. These genes must be on the same chromosome and inherited together 2. Why did we get any of the recombinants then? • CROSSING OVER

D. Terms. . . êGenetic Recombination: General term for production of offspring with new combinations of traits inherited from two parents êParental Types: Look like the parents êRecombinants: Look different from the parents

E. Fun Problems 1. Frequency of Recombination (crossover frequency) ê Take the number of recombinants and divide by total offspring ê Multiply by 100 to get percent

2. Let’s do our flies from before. . • Total Recombinants… 391 • Total Offspring. . 2300 • Frequency of Recombination. . . 17% 3. When two genes are on different chromosomes, they have a 50% recombination frequency.

F. What is a Linkage Map? 1. Genetic map based on recombination frequencies. 2. The farther apart two genes are on a chromosome, the more likely they are to cross over. 3. Map Units: 1% recombination frequency. These are NOT physical distances! 4. Genes that are really far apart on chromosomes show a 50% recombination frequency just like unlinked genes.

G. Making a Linkage Map 1. Determine the recombination frequencies of each set of genes using the equation. 2. Using the frequencies, try to map out the space on the chromosomes Example. . . • b-cn = 9% • b-vg = 17 % • cn-vg = 9. 5%

C. Genetic Disorders caused by Chromosome Problems 1. Nondisjunction: • • Homologous pairs do not separate during Meiosis I Sister Chromatids do not separate during Meiosis II

2. Aneuploidy: Abnormal chromosome number • Trisomic: 3 copies of a chromosome EX. Down’s Syndrome: Trisomy 21 • Monosomic: 1 copy of a chromosome EX. Turner’s Syndrome: XO, missing a sex chromosome.

Problems with non-disjunction in Sex Chromosomes • XO – Turner’s Syndrome • 1 in 5000 females • No mental impairment • Sterile

• XXY – Klinefelter’s Syndrome • 1 in 2000 • Male sex organs, but small testis • Sterile • Have female characteristics (breast enlargement) • XYY – no name • Tend to be taller than average

3. Polyploidy: More than 2 complete chromosome sets • (more on this weirdness later)

D. Alterations in Chromosome Structure 1. Deletion: • Fragment of a chromosome is lost

2. Duplication: • Fragment joins a homologous chromosome, duplicate genes result (ugh!)

3. Inversion: • Fragment reattaches in reverse order

4. Translocation: • Fragment joins a non homologous chromosome

E. Genetic Imprinting. . . A. A gene on one chromosome is silenced, while the other allele is left to be expressed. B. The same alleles may have different effects on offspring, depending on which parent they came from

E. Genetic Imprinting. . . • Example. . . Fragile X Syndrome: • -Abnormal X chromosome • -Worse if you get the abnormal X from mom • -Worse in boys because they can only get their X from mom.

C. Extranuclear Genes 1. Remember. . MITOCHONDRIA and CHLOROPLASTS have DNA. 2. This DNA gets passed on to children, but not by Mendel’s laws 3. Mom’s egg has the cytoplasm!! Lots of it!! Dad’s sperm was shaved to be fluid dynamic. 4. You get your mom’s mitochondrial DNA

Practice Problems 1. A man with hemophilia (a recessive, sexlinked condition) 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? • • What is the probability that a son of this mating will be a hemophiliac? Show your work:

Cross = XHXh x XHY XH Genotype Results XH 1/4 XHXH, 1/4 XHXh 1/4 XHY, 1/4 Xh. Y Phenotype Results Girls – all normal Boys – 1/2 normal, 1/2 hemophilia Y Xh X HX H X HX h X HY X h. Y

• If the couple has four sons, what is the probability that all four will be born with hemophilia? • 1/2 x 1/2 = 1/16

More Practice Problems 2. Pseudohypertophic muscular dystrophy is a disorder that causes gradual deterioration of the muscles. It is seen only in boys born to apparently normal parents and usually results in death in the early teens. Is this disorder caused by a dominant or a recessive allele? Explain. • Is the inheritance autosomal or sex-linked? Explain. • Why is this disorder seen only in boys and never in girls?

• Recessive – if it were dominant, at least one parent would be affected • Sex-linked – if it were autosomal, girls would be impacted as well • For a girl to get it, she would have to inherit from both parents, but boys don’t live past teen years

Even More Practice 3. Red-green color blindness is caused by a sex-linked recessive allele. A color-blind man marries a woman with normal vision whose father was color-blind. What are the phenotypic ratios of the potential children. Don’t forget to specify the sex. Show your work below.

X b. Y x X B X b XB Xb • Genotype results • 1/4 XBXb, 1/4 Xb. Xb, • 1/4 XBY, 1/4 Xb. Y Y Xb X BX b X b X BY X b. Y Phenotype results • 1/2 girls normal, 1/2 girls colorblind • 1/2 boys normal, 1/2 boys colorblind

• Using the information above, what is the probability of having a colorblind daughter? • 1/2 chance of girl, 1/2 chance of colorblind = 1/4 • What is the probability that their FIRST son will be colorblind? • 1/2 chance boy is colorblind

• 4. A wild-type fruit fly (heterozygous for gray body color and normal wings) is mated with a black fly with vestigial wings. The offspring have the following phenotypic distribution. . Wild Type. . . 778 Black Vestigial. . . 785 Black Normal. . . 158 Gray Vestigial. . . 162 • What is the recombination frequency between these genes for body color and wing type? Show your work.

• • 158 + 162 = 320 recombinants 1883 total offspring 320/1883 17%

• 5. What pattern of inheritance would lead a geneticist to suspect that an inherited disorder of cell metabolism is due to a defective mitochondrial gene? • If the disorder is always inherited from the mother

• 6. Determine the sequence of genes along a chromosome based on the following recombination frequencies: • A-B. . 8 map units • A-C. . . 28 map units • A-D. . . 25 map units • B-C. . . 20 map units • B-D. . . 33 map units

• 7. Assume that genes A and B are linked and are 50 map units apart. An animal heterozygous at both loci is crossed with one that is homozygous recessive at both loci. What percentage of the offspring will show phenotypes resulting from crossovers? If you did not know that genes A and B were linked, how would you interpret the results of this cross?

• 50% of the results would be from crossing over • This is the same as you would get if the genes were on different chromosomes

• 8. A space probe discovers a planet inhabited by creatures who reproduce with the same hereditary patterns seen in humans. Three phenotypic characters are height (T=tall, t=short), head appendages (A=Antenna, a=no antennae), and nose morphology (N=upturned snout, n=downturned snout). Since the creatures are not “intelligent”, Earth scientists are able to do some controlled breeding experiments using various heterozygotes in testcrosses. Remember. . . testcrosses are crosses where one parent is totally recessive.

• For a tall heterozygote with antennae, the offspring were: • Tall-antennae: 46 • Short-antennae: 7 • Short-no antennae: 42 • Tall-no antennae: 5 • What is the recombination frequency? • 7+5 = 12/100 = 12%

• For a heterozygote with antennae and an upturned snout, the offspring were: • antennae-upturned: 47 • antennae-downturned: 2 • no antennae-upturned: 3 • no antennae-downturned: 48 • Calculate the recombination frequencies • 3+2 = 5/100 = 5%

• 9. Using the information above in #8, a further testcross is done using a heterozygote for height and nose morphology. The offspring are. . • Tall-upturned: 40 • Short-upturned: 9 • Tall-downturned: 9 • Short-downturned: 42 • Calculate the recombination frequency from these data. • 9+9 = 18/100 = 18%

• 10. Use your answers from 8 and 9 to determine the correct sequence of the three linked genes. 12 5 • T---------A----N 18