Genetics Week 1-5

a. The ability to taste the chemical phenylthiocarbamide is an autosomal dominant phenotype, and the inability to taste it is recessive. If a taster woman with a nontaster father marries a taster man who in a previous marriage had a nontaster daughter, what is the probability that their first child will be (1) A nontaster girl (2) A taster girl (3) A taster boy b. What is the probability that their first two children will be tasters of either sex?

- Set up a pedigree to help visualize what is occurring - Let T= taster , t= nontaster - cross Tt x Tt progenies: 3 (taster) : 1 (non taster) a. p (taster)= 3/4 p (non taster) = 1/4 p(boy) = 1/2 p(girl) = 1/2 p(non taster girl) 1/4 x 1/2 = 1/8 p( taster girl) = 3//4 x 1/2= 3/8 p(taster boy) = 3/4 x 1/2 = 3/8 b. p(first two children tasters) = 3/4 x 3/4 = 9/16

The plant Haplopappus gracilis is diploid and 2n = 4. There are one long pair and one short pair of chromosomes. The diagrams below (numbered 1 through 12) represent anaphases ("pulling apart" stages) of individual cells in meiosis or mitosis in a plant that is genetically a dihybrid (A/a ; B/b) for genes on different chromosomes. The lines represent chromosomes or chromatids, and the points of the V's represent centromeres. In each case, indicate if the diagram represents a cell in meiosis I, meiosis II, or mitosis. If a diagram shows an impossible situation, say so.

1) impossible: the alleles of the same genes are on non homologous chromosomes 2) Meiosis II 3) Meiosis II 4) Meiosis II 5) Mitosis 6) Impossible: it looks like it is in mitotic phase, but alleles of sister chromatids are not identical 7) Impossible: too many chromosomes 8) Impossible: too many chromosomes 9) Impossible: too many chromosomes 10) Meiosis I 11) Impossible: It appears to be meiosis of homozygous a/a; B/B 12) Impossible: the alleles of the same gene are on non homologous chromosomes

haploid

1N and are nonhomologous c'somes

diploid

2N and are organisms that have homologous pairs, 2 versions of c'somes derived from mom and dad

In a diploid organism of 2n = 10, assume that you can label all the centromeres derived from its female parent and all the centromeres derived from its male parent. When this organism produces gametes, how many maleand female-labeled centromere combinations are possible in the gametes?

2^5: find number of haploids and do 2^n power: 32

In what ways does the second division of meiosis differ from mitosis?

As cells divide mitotically, each chromosome consists of identical sister chromatids that are separated to form genetically identical daughter cells. Although the second division of meiosis appears to be a similar process, the "sister" chromatids are likely to be different from each other. Recombination in earlier meiotic stages will have swapped regions of DNA between sister and non-sister chromosomes such that the two daughter cells of this division are typically not genetically identical.

If children obtain half their genes from one parent and half from the other parent, why aren't siblings identical?

Because the "half: inherited is very random, the chances of receiving exactly the same half is vanishingly small. Ignoring recombination and focusing just on which chromosomes are inherited from one parent, there are a 2^23 possible combinations

Holstein cattle are normally black and white. A superb black-and-white bull, Charlie, was purchased by a farmer for $100,000. All the progeny sired by Charlie were normal in appearance. However, certain pairs of his progeny, when interbred, produced red-and-white progeny at a frequency of about 25 percent. Charlie was soon removed from the stud lists of the Holstein breeders. Use symbols to explain precisely why

Charlie was heterozygous for the red gene.Genotypically, he was Bb instead of BB. So when he mated females that were Bb, there was a 25% chance that his progeny would be bb, and a 50% chance that his other offspring, also out of Bb dams, were Bb or heterozygous black. When interbred, the F2 generation produced 75% phenotypical black-and-white progeny, but 25% phenotypical red-and-white progeny, a big no-no with purebred Holstein breeders, because breed standards for Holsteins account for only those females and males that are B&W; a R&W female or male does not meet breed standards, and if these are found, the papers are traced back to the sire and thus the sire is removed because he was hetero black, not homozygous black. a) it is a case of dominant and recesive gene, of the gene which is responsible for skin colour in the Holstein cattle. The allele which causes red and white is recessive and the allele which causes black and white is dominant over this gene and so it is expressed in most of the cases. b) the trait studied is that of skin colour and pattern of Holstein cattle i.e black and white and the other being black and red. c) this is single gene (but in different forms or alleles) control this trait d) two alleles are involved. Let us represent allele which causes red and white is recessive by b and the allele which causes black and white is dominant over this gene, let it be represented by B e) B is dominant over b f) BB cross Bb (1st generation) BB, Bb, BB, Bb (2nd generation) When Bb will be inbred i.e (Bb will be crossed with Bb), the following genotypes will be resulted; BB, Bb, bB, bb g) The phenotypes for BB and Bb genotypes will be black and white, but the phenotype for bb genotype will be red and white. i) If the female, with which Charlie was crossed was truly a pure bred, its genotype is BB, and in this case Charlie is the carrier for the recessive gene, and responsible for producing red and white cows. This is the reason for the farmers' decision for removing Charlie for Black and white Holstein breed, however only maximum 25% of cows could be red and white if a heterozygous cattle is bred with a pure bred black and white catlle.

If you had a fruit fly (drosophila melanogaster ) that was of phenotype A, what test would you make to determine if the fly's genotype was A/A or A/a?

Do a test- cross ( cross to a/a). If the fly was A/A, all the progeny will be phenotypically A; of the guy was A/a, half of the progeny will be A, and half will be a.The use of of a/a ( recessive individual known as the tester) means that it can be ignored because all of its gametes are recessive and do not contribute to the phenotypes of the progeny.

Design a different nuclear-division system that would achieve the same outcome as that of meiosis.

It is pretty hard to beat several billion years of evolution, but it might be simpler if DNA did not replicate prior to meiosis. The same events responsible for halving the DNA and producing genetic diversity could be achieved in a single cell division if homologous chromosomes paired, recombined, randomly aligned during metaphase, and separated during anaphase. However, you would lose the chance to check and repair DNA that replication allows

In examining a large sample of yeast colonies on a petri dish, a geneticist finds an abnormal-looking colony that is very small. This small colony was crossed with wild type, and products of meiosis (ascospores) were spread on a plate to produce colonies. In total, there were 188 wild-type (normal-size) colonies and 180 small ones. a. What can be deduced from these results regarding the inheritance of the small-colony phenotype? (Invent genetic symbols.) b. What would an ascus from this cross look like?

Since there is equal segregation within one meiocyte occurring in yeasts, the cross produced two meiotic products which are normal (s+ genotype) and small colony (s) phenotypes. The mutation in a single gene is supposed to yield 50/50

In the plant Arabidopsis thaliana, a geneticist is interested in the development of trichomes (small projections). A large screen turns up two mutant plants (A and B) that have no trichomes, and these mutants seem to be potentially useful in studying trichome development. (If they were determined by single-gene mutations, then finding the normal and abnormal functions of these genes would be instructive.) Each plant is crossed with wild type; in both cases, the next generation a. What do these results show? Include proposed genotypes of all plants in your answer. b. Under your explanation to part a, is it possible to confidently predict the F_1F1​ from crossing the original mutant A with the original mutant B?

The F2 progenies suggested that the two mutant (A and B) are homozygous recessive. When crossed with a wild type, all progenies (F1) are normal. Selfing F1 plants yielded 3 normal and 1 mutant phenotype ratio. Let T= normal; t = mutant Parents (cross 1): TT x tt F1 progenies: All Tt Cross 2: Tt x Tt F2 progenies= 1 TT: 2 Tt: 1 tt 3 normal: 1 mutant (no trichomes) The F1 progenies cannot be surely determined after crossing mutant A and B. The mutation should be considered whether if they belong to the same or different genes since both scenarios will yield different progenies

Normal mitosis takes place in a diploid cell of genotype A/a; B/b. Which of the following genotypes might represent possible daughter cells? a. A;B b. a;b c. A;b d. a;Be. A/A; B/B f. A/a; B/b g. a/a; b/b

The genotype of the daughter cells will be identical with that of the original cell: (f) A/a; B/b.

In nature, the plant Plectritis congesta is dimorphic for fruit shape; that is, individual plants bear either wingless or winged fruits, as shown in the illustration. Plants were collected from nature before flowering and were crossed or selfed with the following results: Interpret these results, and derive the mode of inheritance of these fruit-shaped phenotypes. Use symbols. What do you think is the nongenetic explanation for the phenotypes marked by asterisks in the table?

The results suggest that winged is dominant to wingless. The crosses then become:1. WW x WW2. Ww x Ww3. ww x ww4. WW x ww5. Ww x ww6. WW x W/-7. WW x W/- The unusual plants are most likely due to either to human error on classification or to contamination. Alternatively, they could result from environmental effects on development

This pedigree is for Huntington disease, a late-onset disorder of the nervous system. The slashes indicate deceased family members a. Is this pedigree compatible with the mode of inheritance for Huntington disease mentioned in the chapter? b. Consider two newborn children in the two arms of the pedigree, Susan in the left arm and Alan in the right arm. Study the graph in Figure 2-24 and form an opinion on the likelihood that they will develop Huntington disease. Assume for the sake of the discussion that parents have children at age 25.

Yes, the pedigree is consistent It is a autosomal dominant trait. In first generation, The affected parent is Hh and unaffected is hh. Thus 50% of its progeny can be Aa (affected). Likewise, III-3&4 are unaffected (hh) and thus none of the child in the lineage is affected thereafter. Thus Susan and Alan have no likelihood of disease because they carry the recessive alleles

In a fungus with four ascospores, a mutant allele lys-5 causes the ascospores bearing that allele to be white, whereas the wild-type allele lys-5+ results in black ascospores. (Ascospores are the spores that constitute the four products of meiosis.) Draw an ascus from each of the following crosses: a. lys-5 × lys-5+ b. lys-5 × lys-5 c. lys-5+ × lys-5+

You expect two lys-5+ (black) spores and two lys-5 (white) spores. You expect all lys-5 (white) spores. You expect all lys-5+ (black) spores

Human cells normally have 46 chromosomes. For each of the following stages, state the number of nuclear DNA molecules present in a human cell: a. Metaphase of mitosis b. Metaphase I of meiosis c. Telophase of mitosis d. Telophase I of meiosis e. Telophase II of meiosis

a. Metaphase of mitosis: 46 chromosomes, each with two chromatids= 92 chromatids b. Metaphase I of meiosis : 46 chromosomes, each with two chromatids= 92 chromatids c. Telophase of mitosis: 46 physically separate chromosomes in each of two about to be formed cells d. Telophase I of meiosis: 23 chromosomes in each of two about to be formed cells, each with 46 chromatids e. Telophase II of meiosis: 23 chromosomes in each of two about to be formed cells

A rare recessive allele inherited in a Mendelian manner causes the disease cystic fibrosis. A phenotypically normal man whose father had cystic fibrosis marries a phenotypically normal woman from outside the family, and the couple consider having a child. a. Draw the pedigree as far as described. b. If the frequency in the population of heterozygotes for cystic fibrosis is 1 in 50, what is the chance that the couple's first child will have cystic fibrosis? c. If the first child does have cystic fibrosis, what is the probability that the second child will be normal?

a. first child: 1/ 200 b. since the first child has CF then the probability of the other getting CF is 3/4

Assume independent assortment and start with a plant that is dihybrid A/a ; B/b: a. What phenotypic ratio is produced from selfing it? b. What genotypic ratio is produced from selfing it? c. What phenotypic ratio is produced from testcrossing it? d. What genotypic ratio is produced from testcrossing it?

a: 9: A/_ ; B/_3: A/_ ; bb3: aa ; B_1: aa ; bb b: 9:3:3:1 c: 1:1:1:1 d: 1: Aa ; Bb1: Aa ; bb1: aa ; Bb1: aa ; bb

The plant blue-eyed Mary grows on Vancouver Island and on the lower mainland of British Columbia. The populations are dimorphic for purple blotches on the leaves—some plants have blotches and others don't. Near Nanaimo, one plant in nature had blotched leaves. This plant, which had not yet flowered, was dug up and taken to a laboratory, where it was allowed to self. Seeds were collected and grown into progeny. One randomly selected (but typical) leaf from each of the progeny is shown in the accompanying illustration. a. Formulate a concise genetic hypothesis to explain these results. Explain all symbols and show all genotypic classes (and the genotype of the original plant). b. How would you test your hypothesis? Be specific

about 3:1 leaves are blotched which means blotched is dominantIf parents were Bb (heteros) the ratio of blotched to unblotched progeny would be 3:1 as seen

phenotype

all observable traits of an organism

c'some segregation in meiosis

alleles of a locus assort independently during meiosis

law of segregation

alleles of a locus segregate at random into gametes during meiosis equal segregation

mutant allele

an allele differing from the allele found in standard or the wild type

mutant

an organism or cell carrying a mutation

genome

an organism's full DNA sequence or the DNA or RNA sequence of a virus

genetic variation

arises from mutations, recombination, selection and epigenetic necessary for process of evolution

metaphase

c'somes align on the equatorial plane.

prophase I

c'somes condense and replicated c'somes become visable homologous c'somes pair or synapse: where recombination can occur nuclear membrane starts to disappear and the spindles start to form

prophase

c'somes condensed and replicated so c'somes become visible c'somes attach to spindle fiber via kinetochore nucleolus and nuclear membrane start to disappear

telophase I

cell division nuclear membrane may appear

telophase

cell division starts and reformation of nucleus and nucleolus. ​

anaphase II

centromeres divide sister c'tids seperate

epigenetics

changes in phenotypes in the absence of any underlaying change in DNA sequences

genetic crosses

controlled mating of 2 individuals - to obtain desired genotype - to deduce genotypes of parents

testcross

cross the genotype in question with a tester - a tester is always the homozygous recessive genotype

In the pedigree below, the black symbols represent individuals with a very rare blood disease. If you had no other information to go on, would you think it more likely that the disease was dominant or recessive? Give your reasons

dominant and the father is heterozygous. In the pedigree, the black symbols represent individuals with a very rare blood disease. The disease is more likely to be dominant and the genotypes of the parents: the father is heterozygous. Explanation: - The disease being followed in this pedigree is very rare. - If the allele that results in this disease is recessive, then the father would have to be homozygous and the mother would have to be heterozygous for this allele. - On the other hand, if the trait is dominant, then all that is necessary to explain the pedigree is that the father is heterozygous for the allele that cause the disease. - This is the better choice as it is more likely, given the rarity of the disease

law of independent assortment

during gamete formation, segregation of alleles of a gene is independent of segregation of alleles of other genes

recombination

exchange of genetic material

selection

favoring particular combinations of genes in a given environment different genotypes contribute alleles to the next generation in proportion to their selective advantage

gene

genomic sequence corresponding to a unit of inheritance

Name two key functions of meiosis

halve the DNA content and to reshuffle the genetic content of the organism to generate genetic diversity among the progeny

recombinant type

have different combination of alleles from those in the gametes that formed the F1

parental type

have same alleles as the gametes that formed the F1

mutations

heritable changes in genetic material may be spontaneous or induced

metaphase I

homologous c'somes pair at midplane facing each other nuclear membrane disappear spindle forms

anaphase I

homologous c'somes separate

mitosis

no change in genetic material or c'some number end product is identical to the initial cell

prophase II

not much change

telophase II

nuclear envelope cells start to divide

ploidy

number of sets of c'somes

meiosis

occurs in germ cells and change the c'somes number

allele

one of the different forms of a gene that can exist in a particular locus

genetic map

show whether genes with related functions are on the same chromosome and are useful for cloning and genome sequencing

anaphase

sister chromatids separate & migrate to opposite poles. ​

gene

specific sequences of nucleotides that pass traits from parents to offspring

metaphase II

spindles form c'somes align at midline no pairing of c'somes

pedigrees

study pattern of distribution of a trait or disease in multi generation families

genotype

the complete genetic makeup of an organism

wild type

the genotype or the phenotype that is found in nature or a standard laboratory stock in a given organism

locus

the location of a gene on the physical map

In a possible future scenario, male fertility drops to zero, but, luckily, scientists develop a way for women to produce babies by virgin birth. Meiocytes are converted directly (without undergoing meiosis) into zygotes, which implant in the usual way. What would be the short- and long-term effects in such a society?

the offspring would be identical to the mother and not be differentiated from each other. moreover there will be no more male linked chromosome and the male sex will be extinct

sum rule

the probability of any of several mutually exclusive events occurring, is the sum of their individual probabilities (or rule)

product rule

the probability that two independent events will occur at the same time and is the result of their individual probabilities (and rule)

gene expression

the process by which a gene produces its product and the product carries out its function

probability

the ratio of number of times a particular even is expected to happen to the number of trials during which the event could have happened

Name the key function of mitosis

to generate two daughter cells that are genetically identical to the original parent cell

autosomal dominant

usually does not skip a generation male and female are both affected the abnormal (mutant) allele is dominant lethal genotypes affect the frequency of the lethal alleles in a population

autosomal recessive

usually skips generation or generations males and females are both affected if the parent of an affected individual do not exhibit the trait then they must be heterozygous