Genetics Final

Front 1

___ hold homologous chromosomes together during meitoic metaphase I.

Back 1

Chiasmata

Front 2

Exchange of DNA b/t nonsister chromatids occurs during ____ of meiosis.

Back 2

Prophase I

Front 3

When a human female is born, all of her oocytes are arrested in ____.

Back 3

Prophase of Meiosis I.

Front 4

Individuals with blood type B make antibodies against _____.

Back 4

A Antigen

Front 5

The relationship between I^a and I^b allele of human ABO bloodtype system is that of ____.

Back 5

Codominance

Front 6

___ is cause by autosomal anueplody

Back 6

Downsyndrome

Front 7

Humans with 47XXY

Back 7

males with Kleinfelter syndrome

Front 8

individuals with testicular ferminization ____.

Back 8

have testes, produce testosterone, but cells do not respond to the testosterone.

Front 9

Fruitflies with chromosome composition XXY + 4A have a total of ___ chromosomes in each of their somatic cells.

Back 9

15 (1+1+1)+ 4(3) = 15

Front 10

reactivation of inactivated X chromosomes occur _____.

Back 10

Prior to meiosis.

Front 11

In humans, the ___ gene is expressed only from inactive X chromosomes.

Back 11

XIST

Front 12

Caused by aneuploidy

Back 12

Down Syndrome

Front 13

defective cell surface receptors

Back 13

hypercholesterlemia

Front 14

can be caused by a nondisjunction during a mitotic division

Back 14

true hermaphroditism

Front 15

defective blood clotting factor

Back 15

Hemophilia A

Front 16

Disease with very low penetrance

Back 16

Hemochromatosis

Front 17

defective in amino acid metabolism

Back 17

phenylketonuria

Front 18

caused by abberant recombination of the male parent.

Back 18

XX male

Front 19

caused by random X chromosome inactivation

Back 19

calico cat coloration pattern

Front 20

Fruit flies with the chromosomal composition XXY + 3A have a total of ___ chromosomes in each of their somatic cells.

Back 20

12! (1+1+1) + 3(3) = 12

Front 21

Nondisjunction during ____ could explain the formation of an XYY zygote

Back 21

meiosis II in the father

Front 22

Nondisjunction during ____ could explain the formation of Turner syndrome in a child

Back 22

Meiosis I and II in mother
Meiosis I and II in father

Front 23

The recombination frequency between genes on different chromosomes is ___%

Back 23

50%

Front 24

In a chi-square analysis, a pvalue of 0.035 ____

Back 24

causes you to reject the null hypothesis.

Front 25

In a chi-square analysis, a pvalue of 0.085 ____

Back 25

supports but does not prove the null hypothesis.

Front 26

A single crossover between two circular DNA molecules generates _____.

Back 26

A single circular molecule

Front 27

The E.Coli F factor is a ____.

Back 27

conjugative episome

Front 28

Specialized transduction occurs because of errors made during _____.

Back 28

exicision of the phage chromosome from the bacterial chromosome.

Front 29

If adenine makes up 20% of the nitrogenous base composition of a certain species of mammals, guanine should make up approximately ___% of that organisms nitrogenous bases.

Back 29

30% (30+20=50%)

Front 30

___ provided evidence that DNA was injected into bacteria during bacteriophage infection.

Back 30

Hershey and Crick

Front 31

____ is not a standard component of any of the naturally occuring amino acids.

Back 31

Phosphorous

Front 32

DNA absorbs UV light maximally at a wavelength of 260nm. Proteins absorb at 280nm. The most MUTAGENIC wavelength of light is ___.

Back 32

260nm

Front 33

Peptide bonds are formed between ____.

Back 33

one amino group and one carboxyl group

Front 34

Alpha helices in polypeptide chains are held together primarily by

Back 34

regularly spaced hydrogen bonds between components of peptide bonds.

Front 35

Transcription factors are ___.

Back 35

proteins that help regulate gene expression.

Front 36

An unusual 5' to 5' linkage is

Back 36

used to attach the 7-methyl guanosine cap to eukaryotic mRNA's

Front 37

One way eukaryotes regulate gene expression is to regulate the rate of shortening of the ____ in the cytoplasm.

Back 37

Poly A tails

Front 38

The ____ binds to the prokaryotic Pribnow box

Back 38

sigma factor.

Front 39

The prokaryotic binding ribosome site binds to ____ during the initiation of protein synthesis.

Back 39

the 3' end of 16S rRNA

Front 40

Isoaccepting tRNA's are groups of tRNA's that have ____

Back 40

different anticodons but are recognized by the same aminoacyl tRNA synthetase.