BIOS 1500 Lecture Notes - Lecture 1: Cumulative Incidence

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lemoncheetah669

29 Apr 2017

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A ratio is the result of dividing one quantity by another. Specifically if a and b are two numbers, then the ratio of a to b is. A p(cid:396)opo(cid:396)tio(cid:374) is a t(cid:455)pe of (cid:396)atio (cid:449)he(cid:396)e the (cid:374)u(cid:373)e(cid:396)ato(cid:396) is also a pa(cid:396)t of the de(cid:374)o(cid:373)i(cid:374)ato(cid:396). If (cid:449)e ha(cid:448)e t(cid:449)o (cid:374)u(cid:373)(cid:271)e(cid:396)s a a(cid:374)d (cid:271) the p(cid:396)opo(cid:396)tio(cid:374) that (cid:494)a(cid:495) is to the total (cid:894)a+(cid:271)(cid:895) is the (cid:396)atio: a / (cid:894)a+(cid:271)(cid:895). Note that p(cid:396)opo(cid:396)tio(cid:374)s (cid:449)ill al(cid:449)a(cid:455)s fall (cid:271)et(cid:449)ee(cid:374) (cid:1004) a(cid:374)d (cid:1005) like p(cid:396)o(cid:271)a(cid:271)ilities. A rate is a (cid:396)atio i(cid:374)di(cid:272)ati(cid:374)g ho(cid:449) (cid:395)ui(cid:272)kl(cid:455) so(cid:373)e (cid:858)p(cid:396)o(cid:272)ess(cid:859) is o(cid:272)(cid:272)u(cid:396)(cid:396)i(cid:374)g. it can be defined as the amount of change in one variable relative to the amount of change in another variable. Rate(y:x) = dy/ dx (in units of y per unit of x) A linear relationship between two quantities say y and x is given by the equation:

Related Questions

___1. Nationally, the most common cause of Down Syndrome is
a. nondisjunction of Chromosome 21 during meiosis
b. translocation of Chromosome 21 onto another chromosome
c. inversion of a large section of Chromosome 21
d. deletion of the Down Syndrome gene
___2. In humans, sex is determined by
a. The presence or absence of the Y chromosome
b. The ratio of X chromosomes to sets of autosomes
c. The temperature during fetal development
d. Which parent is stronger at the time of conception
___3. A translocation is
a. a chromosomal abnormality in which a piece of one chromosome breaks off and attaches to a different chromosome
b. an abnormality in which the chromosomes are translocated out of the nucleus
c. a condition in which the chromosome has broken in two places and come back together with the middle segment backward
d. an abnormality in which there is an extra copy of one chromosome
___ 4. In snapdragons, a red-flowered plant crossed with a white-flowered plant produces all pink-flowered offspring. But crossing two pink-flowered plants produces red, pink, and white in a 1:2:1 ratio. This is a good example of
a. Incomplete dominance
b. Dominance of the red allele over the white allele
c. Lethal effects of the pink allele
d. Sex-linkage of the flower-color alleles
___ 5. If a person with type AB blood marries a person with Type O blood, what blood types are possible in their children? (Assume this trait is a single-gene trait, not complicated by interaction with other genes).
a. A, B, AB, or O
b. A or B only
c. AB or O only
d. AB only 2
___6. In what phase of meiosis are haploid nuclei first formed? a. Anaphase I b. Metaphase II c. Telophase II d. Metaphase I e. Telophase I
___7. What is the phenotype of a heterozygous tall pea plant? a. tall b. Tt c. TT d. tt e. dominant
___ 8. If genes A and B are far apart on a chromosome, and A and C are close together, which of the following observations is expected?
a. Crossing over will occur more often between A and C than between A and B
b. Crossing over will occur at equal frequencies between A and C, and between A and B
c. Crossing over will occur more often between A and B than between A and C
___9. Sometimes an organismâs genotype for one gene masks the effect of a different gene. This phenomenon is called
a. Dominance
b. Incomplete dominance
c. Co-dominance
d. Epistasis
e. Variable expressivity
___10. The dominant allele of a gene is
a. The one that is expressed most often in a population
b. The one that is expressed most often in any particular cross
c. The one that is expressed in an individual that has two different alleles
d. The one that is most healthy to have, that gives the greatest fitness
___11. In human cells, DNA replication occurs
a. In S phase of interphase
b. In G1 phase of interphase
c. In G2 phase of interphase
d. In prophase of mitosis
e. Whenever the cell needs to read the DNA to make proteins
___ 12. A process in which recipient cells take up genes from DNA molecules floating free in the surrounding medium is called
a. conjugation
b. specialized transduction
c. generalized transduction
d. transformation 3
___ 13. Most calico cats are female. However, a few rare males are observed. What is the genotype of a male calico cat?
a. X BX 0
b. X OY
c. X BY
d. X BYY
e. X BX OY
___ 14. A type of red-green color-blindness is a sex-linked (X-linked) recessive trait in humans. Can a woman have this type of color-blindness?
a. Yes, a woman could be color-blind if her mother and father were both heterozygous.
b. Yes, a woman could be color-blind but only if she has an abnormal number of sex chromosomes.
c. No, women can't show sex-linked traits because they lack a Y chromosome.
d. No, women can't show sex-linked traits because they have two X chromosomes.
e. Yes, a woman could be color-blind if her father was color-blind and her mother was a carrier (heterozygous).

CASE ONE

The study is conducted to investigate the association betweenchewing tobacco and oral leukoplakia (a precancerous lesion) amongcurrently active professional baseball players in the southeasternUnited States. A roster of all active players is obtained (n=500).All potential study subjects agree to participate. Each subject hasan interview regarding current use of chewing tobacco and has hismouth examined by a dentist. Of the 500 subjects, 125 subjects chewtobacco and 375 do not chew tobacco. Of the chewers, 25 haveevidence of oral leukoplakia. Of the non-chewers, 15 have evidenceof oral leukoplakia. All 500 players were followed for a period of5 years. Of those who had evidence of oral leukoplakia, 18 died ofsome type of oral cancer.

Draw a 2 x 2 table demonstrating the relationship betweenchewing tobacco and oral leukoplakia. In drawing this table, putthe exposure variable on the columns and the health outcomevariable on the rows.
Draw a second 2 x 2 table demonstrating the relationshipbetween chewing tobacco and oral leukoplakia, but this time, putthe exposure variable on the rows and the health outcome variableon the columns.
Using the table drawn in part a, compute the prevalence ratioand the prevalence odds ratio of oral leukoplakia for chewerscompared to non-chewers. Are these two estimates close toone-another? Why are these prevalence measures and not incidencemeasures?
Using the table drawn in part b, compute the prevalence ratioand the prevalence odds ratio of oral leukoplakia for chewerscompared to non-chewers. Are these estimates equal to theircorresponding estimates computed using the table drawn in part a?Explain.
Ignoring the issue of statistical inference and the control ofother variables, what do these results say about the relationshipbetween chewing tobacco and the presence of oral leukoplakia?
Calculate the case-fatality rate (actually, a risk) in thisstudy. Why is this a measure of risk?
Based on the information provided, why canât you evaluatewhether tobacco chewers have a higher case-fatality risk thannon-chewers?

yellowcheetah313

In this study you will learn about a modification of Mendelian inheritance which is also an important source of genetic variation.

You are about to perform what looks like a classic dihybrid testcross ("A" is the gene for eye color and "B" is the gene for wing color.)

A allele for red eyes

a allele for white eyes

B allele for clear wings

b allele for spotted wings

(AaBb) X (aabb) Note the genotypes and phenotypes.

Question #1: Predict the expected phenotypic ratio of this P X P cross.

Observed results:

500 red eyes clear wings

500 white eyes spotted wings

Question #2: How do the expected results compare with the observed results?

Question #3: Considering Mendel's second law (regarding dihybrid crosses), suggest an explanation for the discrepancy between the observed results and those predicted.

Genes are located at specific positions on a chromosome. The distance between genes is measured in terms of "mapping units". In this exercise the gene for wing phenotype âBâ will be moved from its original position of zero to different positions on the chromosome. You will observe the effect of this gene movement.

First, gene "B" (wing color) is moved from position zero to position 10 on the same chromosome. Gene "A" (eye color) stays at position 0.

Observed Results: (position 10)

450 red eyes clear wings

50 red eyes spotted wings

50 white eyes clear wings

450 white eyes spotted wings

Question #4: How do these results differ from those predicted by Mendel's second law?

Question #5: From those in the first cross?

Question #6: Can you suggest a hypothesis to explain these new results?

Gene "B" (wing color) is now moved to position 20 on the same chromosome. Gene "A" (eye color) stays at position 0.

Observed Results: (position 20)

400 red eyes clear wings

100 red eyes spotted wings

100 white eyes clear wings

400 white eyes spotted wings

Question #7: How do the proportions of the phenotypic classes in this cross differ from the proportions you obtained in the earlier crosses?

The wing color gene "B" is now moved farther away (positions 30, 40, 50, and 60).

Observed Results: (position 30)

350 red eyes clear wings

150 red eyes spotted wings

150 white eyes clear wings

350 white eyes spotted wings

Observed Results: (position 40)

292 red eyes clear wings

220 red eyes spotted wings

195 white eyes clear wings

293 white eyes spotted wings

Observed Results: (position 50)

250 red eyes clear wings

250 red eyes spotted wings

250 white eyes clear wings

250 white eyes spotted wings

Observed Results: (position 60)

250 red eyes clear wings

250 red eyes spotted wings

250 white eyes clear wings

250 white eyes spotted wings

Question #8: Prepare a line graph plotting the % of recombinant offspring (Y-axis) vs. difference in position between the two genes (X-axis). (Hint: your graph should be one line with labeled points at map positions 0, 10, 20, 30, 40, 50, & 60).

Question #9: A genetic definition of "1 map unit" is the distance between 2 genes that gives 1% recombinants out of the total number of progeny. Using this definition, how many map units separate the "A" and "B" genes when the eye color gene "A" is at position 0 and the wing color gene "B" is at position 40? (Show how you determined the results.)

Question #10: How does the distance between two genes on the same chromosome affect the proportion of recombinants? Why?

Question #11: Maximum crossover frequency can be defined as the point at which no more recombination can be detected no matter how far apart the genes are located. Do your results indicate that you are approaching the maximum crossover frequency?

Question #12: Show (using a Punnett Square) the expected phenotypic rations if the two genes were located on different chromosomes.

Question #13: Which of your simulated cross or crosses do the data from the Punnett Square most closely fit? Why?

Question #14: A wild-type fly (heterozygous for gray body color and normal wings - b⁺ vg⁺/b vg) was mated to a black fly with vestigial wings (b vg/b vg). The F1 had the following phenotypic distribution: wild type, 850; black body - vestigial wings, 899; black body - normal wings, 79; gray body - vestigial wings, 85. What is the recombination frequency (RF) between these genes for body color and wing type? (show work)

Question #15: A cross produced 915 offspring with normal pigment and 310 with albinism. Conclusion?

a. One of the parents was homozygous for albinism.

b. Both parents were heterozygous.

c. One parent was homozygous for normal pigmentation.

d. Both parents were albinos. e. 605 albino zygotes must have failed to develop.

NOTE: Please show all work to support your answers.

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BIOS 1500 Lecture Notes - Lecture 1: Cumulative Incidence

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