Mendelian Genetics More Genetics Problems Linked Genes Chromosomal

  • Slides: 51
Download presentation
Mendelian Genetics More Genetics Problems Linked Genes Chromosomal Disorders Bio Trivia 5 pt 5

Mendelian Genetics More Genetics Problems Linked Genes Chromosomal Disorders Bio Trivia 5 pt 5 pt 10 pt 10 pt 15 pt 15 pt 20 pt 20 pt 25 pt 25 1 pt

If a man with Huntington’s marries a normal female, what percentage of the offspring

If a man with Huntington’s marries a normal female, what percentage of the offspring are likely to have Huntington's? I 5 2

If a man with Huntington’s I 5 a marries a normal female, what percentage

If a man with Huntington’s I 5 a marries a normal female, what percentage of the offspring are likely to have Huntington's? 50% 3

I 10 The ability of one gene to control the expression of another gene

I 10 The ability of one gene to control the expression of another gene is known as ______. 4

I 10 a The ability of one gene to control the expression of another

I 10 a The ability of one gene to control the expression of another gene is known as episatsis. 5

I 15 In the cross: Rr. Tt x Rrtt, what is the probability of

I 15 In the cross: Rr. Tt x Rrtt, what is the probability of getting an offpring that is dominant for both traits? 6

I 15 a In the cross: Rr. Tt x Rrtt, what is the probability

I 15 a In the cross: Rr. Tt x Rrtt, what is the probability of getting an offpring that is dominant for both traits? ¾ x ½ = 3/8 7

I 20 What is the probability of flipping a tails three times in a

I 20 What is the probability of flipping a tails three times in a row? 8

I 20 a What is the probability of flipping a tails three times in

I 20 a What is the probability of flipping a tails three times in a row? ½ x ½ = 1/8 9

An AABb individual is mated I 25 with another AABb individual. The possible number

An AABb individual is mated I 25 with another AABb individual. The possible number of genetically different kinds of offspring is _____. 10

An AABb individual is mated I 25 a with another AABb individual. The possible

An AABb individual is mated I 25 a with another AABb individual. The possible number of genetically different kinds of offspring is 3. 11

If one parent. IIis 5 blood type AB and the other is type O,

If one parent. IIis 5 blood type AB and the other is type O, what fraction of their offspring will be blood type A? 12

If 5 a one parent is blood type II AB and the other is

If 5 a one parent is blood type II AB and the other is type O, what fraction of their offspring will be blood type A? ½ 13

IIThe 10 result of the following cross indicates the orange eyes are _____ black

IIThe 10 result of the following cross indicates the orange eyes are _____ black eyes. 14

IIThe 10 aresult of the following cross indicates the orange eyes are recessive to

IIThe 10 aresult of the following cross indicates the orange eyes are recessive to black eyes. 15

II 15 If a parent is Aa. BBCcdd. Ee, how many different types of

II 15 If a parent is Aa. BBCcdd. Ee, how many different types of gametes can it produce? 16

II 15 a If a parent is Aa. BBCcdd. Ee, how many different types

II 15 a If a parent is Aa. BBCcdd. Ee, how many different types of gametes can it produce? 2 x 1 x 2=8 17

IIA 20 woman is A+ and has Bchild. A father with this blood type

IIA 20 woman is A+ and has Bchild. A father with this blood type could NOT be the father: A. A+ B. BC. AB + D. AB 18

woman is A+ and has BIIA 20 a child. A father with this blood

woman is A+ and has BIIA 20 a child. A father with this blood type could NOT be the father: A. A+ 19

An organism homozygous for red II 25 color was crossed with an organism homozygous

An organism homozygous for red II 25 color was crossed with an organism homozygous for blue color. The offspring were green. Two of the green organisms were crossed. What is (are) the expected phenotypes in this second group of offspring? 20

An organism homozygous for red IIcolor 25 a was crossed with an organism homozygous

An organism homozygous for red IIcolor 25 a was crossed with an organism homozygous for blue color. The offspring were green. Two of the green organisms were crossed. What is (are) the expected phenotypes in this second group of offspring? 1 Red: 2 green: 1 blue 21

If two genes are linked _____. III 5 a. they are on different chromosomes

If two genes are linked _____. III 5 a. they are on different chromosomes b. they assort independently c. they code for the same protein d. they are on the same chromosome 22

If two genes are linked _____. III 5 a d. they are on the

If two genes are linked _____. III 5 a d. they are on the same chromosome 23

If a color-blind woman marries III a 10 man with normal vision, the female

If a color-blind woman marries III a 10 man with normal vision, the female offspring will be _____, and the male offspring will be _____. 24

If a color-blind woman marries III a 10 a man with normal vision, the

If a color-blind woman marries III a 10 a man with normal vision, the female offspring will be normal, and the male offspring will be color-blind. 25

Four genes (A, B, C, and D) are on the III same 15 chromosome.

Four genes (A, B, C, and D) are on the III same 15 chromosome. The crossover frequencies below are obtained. What is the correct sequence of genes? A-B 19% A-D 21% B-C 14% C-D 16% A-C 5% B-D 2% 26

Four genes (A, B, C, and D) are on the same chromosome. The III

Four genes (A, B, C, and D) are on the same chromosome. The III 15 a crossover frequencies below are obtained. Which is the correct sequence of genes? A-B 19% A-D 21% B-C 14% C-D 16% A-C 5% B-D 2% ACBD or DBCA 27

III A 20 father (whose father was color-blind) and a normal woman can produce:

III A 20 father (whose father was color-blind) and a normal woman can produce: A. a color-blind son B. a color-blind daughter C. no color-blind children D. sons that have a 50% chance of being color-blind. 28

IIIA 20 a father (whose father was color-blind) and a normal woman can produce:

IIIA 20 a father (whose father was color-blind) and a normal woman can produce: C. no color-blind children 29

In a particular species of mammal, black III 25 hair (B) is dominant to

In a particular species of mammal, black III 25 hair (B) is dominant to green hair (b), and red eyes (R) are dominant to white eyes (r). When a Bb. Rr individual is mated with a bbrr individual, the observed offspring are: black-red 1, 070; black-white 177; green-red 180; green-white 1, 072. How far apart are the B and R genes? 30

In a particular species of mammal, black hair (B) is dominant to green III

In a particular species of mammal, black hair (B) is dominant to green III 25 a hair (b), and red eyes (R) are dominant to white eyes (r). When a Bb. Rr individual is mated with a bbrr individual, the observed offspring are: black-red 1, 070; black-white 177; green-red 180; green-white 1, 072. How far apart are the B and R genes? 14. 3 MU 31

IV 5 A chart of chromosomes in which they are arranged by number is

IV 5 A chart of chromosomes in which they are arranged by number is called a _________. 32

IV 5 a A chart of chromosomes in which they are arranged by number

IV 5 a A chart of chromosomes in which they are arranged by number is called a karyotype. 33

IV 10 A male with a Barr body has what disorder? 34

IV 10 A male with a Barr body has what disorder? 34

IV 10 a A male with a Barr body has what disorder? Klinefelter’s Syndrome

IV 10 a A male with a Barr body has what disorder? Klinefelter’s Syndrome XXY 35

IV 15 The exchange of parts between nonhomologous chromosomes is called _____. 36

IV 15 The exchange of parts between nonhomologous chromosomes is called _____. 36

IV 15 a The exchange of parts between nonhomologous chromosomes is called translocation. 37

IV 15 a The exchange of parts between nonhomologous chromosomes is called translocation. 37

IV 20 The only known viable human monosomy is: 38

IV 20 The only known viable human monosomy is: 38

IV 20 a The only known viable human monosomy is Turner’s Syndrome. 39

IV 20 a The only known viable human monosomy is Turner’s Syndrome. 39

If non-disjunction occurs in IV 25 Meiosis I, what types of gametes will be

If non-disjunction occurs in IV 25 Meiosis I, what types of gametes will be produced? A. n and n + 1 B. n and n-1 C. n + 1 and n – 1 D. n, n + 1, and n - 1 40

If non-disjunction occurs in IV 25 a Meiosis I, what types of gametes will

If non-disjunction occurs in IV 25 a Meiosis I, what types of gametes will be produced? C. n + 1 and n – 1 41

Who am I? V 5 a. Mendel c. Kleinfelter b. Morgan d. Turner 42

Who am I? V 5 a. Mendel c. Kleinfelter b. Morgan d. Turner 42

Who am I? V 5 a b. Thomas Hunt Morgan 43

Who am I? V 5 a b. Thomas Hunt Morgan 43

Ophioglossum reticulatum, a type of V 10 has the most chromosomes of fern, any

Ophioglossum reticulatum, a type of V 10 has the most chromosomes of fern, any organism. How many chromosomes does each cell have? a. 110 b. 256 c. 1260 d. 2448 44

 • Polyploidy is a common conduction in 10 a plants, but seemingly taken

• Polyploidy is a common conduction in 10 a plants, but seemingly taken to its V limits in the Ophioglossum reticulatum. This fern has roughly 630 pairs of chromosomes or 1260 chromosomes per cell 45

V 15 What is this condition called? 46

V 15 What is this condition called? 46

V 15 a Polydactyly 47

V 15 a Polydactyly 47

20 read 5 bases/second, how If Vyou long would it take you to read

20 read 5 bases/second, how If Vyou long would it take you to read the entire human genome (without stopping)? a. 7 years b. 19 years c. 48 yearsd. 87 years 48

VIt 20 a would take 19 years to read aloud without stopping, at 5

VIt 20 a would take 19 years to read aloud without stopping, at 5 bases per second, the entire sequence of the genome within the nucleus of the human embryo 49

V 25 Why was Aberforth Dumbledore prosecuted? 50

V 25 Why was Aberforth Dumbledore prosecuted? 50

V 25 a For practicing inappropriate charms on a goat 51

V 25 a For practicing inappropriate charms on a goat 51