Reyes, A Module 3 - Lecture notes 1, 3-7 PDF

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HUMAN GENETICS WITH CYTOGENETICSMODULEREYES, ALLESANDRAZELENA R.PRE-Listen to this song and take note of five (5) important terms and concepts that are all necessary for understanding Mendelian Genetics. After which define or describe your noted terms or concepts. One dominant, one recessive - When ...

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PRE-

MODULE REYES, ALLESANDRA ZELENA R.

HUMAN GENETICS WITH CYTOGENETICS

Listen to this song and take note of five (5) important terms and concepts that are all necessary for understanding Mendelian Genetics. After which define or describe your noted terms or concepts. 1. One dominant, one recessive - When an organism is heterozygous at a specific locus and carries one dominant and one recessive allele, the organism will express the dominant phenotype. 2. Traits are controlled by genes - An organism's traits are controlled by the alleles it inherits from its parents 3. We can use punnett squares to show possible crossing pairs - A Punnett square shows all the possible genotypes that can result from a given cross. It is also used to predict possible phenotypes in offspring. 4. Inheritance of traits are alleles - Although an individual gene may code for a specific physical trait, that gene can exist in different forms, or alleles. One allele for every gene in an organism is inherited from each of that organism's parents. 5. Offspring will be the dominant - One out of the four boxes of the Punnett square holds the dominant homozygote

L2 Check-in Activity Seatwork 1

RY

Ry

rY

ry

Seatwork 2

Complete the punnet square and determine the PR of the cross. Given: P= RrYy x RrYy

RYRY

RYRy

RYrY

RYry

RyRY

RyRy

RyrY

Ryry

rYRY

rYRy

rYrY

rYry

ryRY

ryRy

ryrY

ryry

PR = 9 : 3 : 3 : 1

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In humans, brown eyes (B) are dominant over blue (b)*. A brown-eyed man marries a blue-eyed woman and they have three children, two of whom are browneyed and one of whom is blue-eyed. Draw the Punnett square that illustrates this marriage. What is the man’s genotype? What are the genotypes of the children?

(* Actually, the situation is complicated by the fact that there is more than one gene involved in eye color, but for this example, we’ll consider only this one gene.) •

Suppose you counted 79 R_ and 33 rr. The total number of individuals you counted, N, is 112. You expect 3/4 to be R_ (84) and 1/4 to be rr (28). Are your results close enough to these ratios for you to accept the null hypothesis—that there is no real difference? •

Observe Expected d = o - d2 d2/e d (e) e (o) R_ 79 .75 x 112 = 84 -5 25 .30 rr 33 .25 x 112 = 28 5 25 .89 Total 112 112 0 1.19 Χ2 = ∑ (observed-expected)2 / (expected). This means add up the values in the last column. Phenotypes

You can compare the chi-square sum, 1.19, with the numbers in a table of critical values to decide whether to accept the null hypothesis—that the observed results are so close to expected results that there is no difference, and our original hypothesis is accepted.

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An ear of corn has a total of 381 grains, including 216 Purple & Smooth, 79 Purple & Shrunken, 65 Yellow & Smooth, and 21 Yellow & Shrunken. This ear of corn was produced by a dihybrid cross (PpSs x PpSs) involving two pairs of heterozygous genes resulting in a theoretical (expected) ratio of 9:3:3:1. Test your hypothesis using chi square.

GRAIN PHENOTYPE (COLUMN 1)

OBSERVED NUMBER (COLUMN 2)

OBSERVED RATIO (COLUMN 3)

EXPECTED RATIO (COLUMN 4)

EXPECTED NUMBER (COLUMN 5)

[OBS NO. – EXP NO.]2 / Expected

PURPLE & SMOOTH PURPLE & SHUNKREN YELLOW & SMOOTH YELLOW & SHUNKREN TOTAL NUMBER:

216

10.3

9

79

3.8

3

65

3.1

3

21

1.0

1

381

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381 x 9/16 =214 381 x 3/16 = 71 381 x 3/16 = 71 381 x 1/16 = 24 Chi Square Value:

No. (COLUMN 6) 4/214 = 0.019 64/71 = 0.901 36/71 = 0.507 9/24 = 0.375 1.80

L4 Check-in Activity For an ice breaker, I want all of you to take the proprofs quiz that will make your brain cells work and spike your adrenaline. But don’t you worry this will be an additional grade for you to have. Just click the site and after you took the quiz, there is a big revelation that I know you will be very happy to have. So what are you waiting for, start to render your best ability. Good luck everyone!

M3 POSTTASK We come to the last post-task of this module, what I would like everyone to do is to read and understand the following problems and answer them showing your solutions. Again, upload them in the reply box found on this page. For 20 points. 1. An individual whose genotype is AABbCc is crossed with an individual who is heterozygous for all three of these genes. List the gametes that the AABbCc parent could produce. 

ABc, Abc, AbC, ABc

2. Perform the Fork-lined method of the given alleles. A male having a characteristic of TtSsYy and a female TTSsYY ....