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Biology Notes

[Last Updated 30 March 2008]


General Biology
Classification of Animals
Mendelian Genetics


General Biology

PROBLEM

Estimate the number of atoms in the human body.

SOLUTION

A human body contains, by mass, 61.43% oxygen, 22.86% carbon, 10.00% hydrogen, 2.57% nitrogen, 1.43% calcium, 1.11% phosphorus, and less than 1.00% other elements. For a 70 kg individual, the mass of the top six elements (Emsley 1998) is as shown in column [5] of the table below. The number of moles of atoms of each element is obtained by dividing the number of grams of the element by the molar mass (Giancoli 1998). The number of atoms is obtained by multiplying the number of moles by Avogadro's number, NA = 6.02 x 1023.

[1]
Element
[2]
Symbol
[3]
Atomic Number
[4]
Atomic Weight (u)
[5]
Mass (kg)
[6]
Moles
[7]
Atoms
oxygen
O
8
15.9994
43
2687.601
1.61794 x 1027
carbon
C
6
12.011
16
1332.112
8.01932 x 1026
hydrogen
H
1
1.00794
7
6944.858
4.1808 x 1027
nitrogen
N
7
14.00674
1.8
128.5096
7.73628 x 1025
calcium
Ca
20
40.078
1
24.95134
1.50207 x 1025
phosphorus
P
15
30.97376
0.78
25.18261
1.51599 x 1025
Total
69.58
6.708 x 1027

Summing the values in column [7] for the individual elements, we get 6.708 x 1027. This is approximately the number of atoms in a 70 kg person. For a baby, the number of atoms would be roughly one tenth of this or 6.708 x 1026. So the number of atoms in a human body is ~ 1027.

Emsley, John 1998, The Elements, Third Edition (New York: Oxford University Press).

Giancoli, Douglas C. 1998, Physics: Principles with Applications, Fifth Edition (Upper Saddle River, New Jersey: Prentice Hall).

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Classification of Animals

Vertebrates [having a backbone]
  Warm Blooded
    Mammals Mammalia
      Primates
        Anthropoids Anthropoidea
          Hominids
            Humans Homo sapiens
            Cro-Magnon Man Homo sapiens sapiens [extinct; lived 10,000-40,000 years ago; discovered in Cro-Magnon, Les Eyzies, Dordogne, France, in 1868]
          Apes
            Chimpanzees Pan troglodytes [equatorial Africa]
          Old World Monkeys
            Baboons Papio ursinus [Africa, Arabia; largest members of the monkey family]
            Macaques/Rhesus Monkeys Macaca mulatta [India]
          New World Monkeys
      Giraffes Giraffa camelopardalis [Africa]
      Elephants Elephantidae
        African Elephants Loxodonta africana [Africa]
        Indian Elephants Elephas maximus [India]
        Mastodons/Mammoths Mammut americanum [extinct; 11000 to 3.75 million years ago; North America]
      Equines Equus
        Horses Equus caballus
        Zebras [Africa]
        Eohippus [extinct]
      Felines
        Saber Toothed Tigers Smilodon fatalis [extinct]
    Birds Aves
      Parakeets
        Monk Parakeets Myiopsitta monarchus
  Cold Blooded
    Reptiles Reptilia
      Dinosaurs [extinct]
        Sauropods
          Brachiosaurus
          Apatosaurus/Brontosaurus [extinct; Jurassic; 150 million years ago; found in Utah, Colorado, Wyoming; correct name is Apatosaurus]
          Diplodocus [extinct; late Jurassic; Colorado, Montana, Wyoming]
        Pterosaurs/Pterodactyls [extinct; Triassic to Cretaceous; 65-228 million years ago]
    Amphibians Amphibia
    Fish
      Jawless Fish
      Cartilaginous Fish
      Bony Fish
        Esox
          Northern Pike Esox lucius [carnivorous; brackish water and freshwater]
        Lutjanus
          Red Snappers Lutjanus campechanus [Gulf of Mexico; saltwater]
        Scombridae
          Bluefin Tuna [saltwater]
        Xiphias
          Swordfish Xiphias gladius [saltwater]
Invertebrates [without a backbone]

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Mendelian Genetics


PROBLEM

In tomatoes, red fruit color is dominant to yellow. Suppose a tomato plant homozygous for red is crossed with one homozygous for yellow. Determine the appearance of
(a) the F1 tomatoes
(b) the F2 tomatoes
(c) the offspring of a cross of the F1 tomatoes back to the red parent.
(d) the offspring of a cross of the F1 tomatoes back to the yellow parent.

[from Russell, Peter J. 2005, iGenetics: A Mendelian Approach, Second Edition (Benjamin Cummings), problem 11.1]

SOLUTION

(a) Let R = red and r = yellow. Then the genotype of the homozygous red plant is RR and that of the yellow plant is rr. The F1 tomatoes are the first filial (F1) offspring, which are the offspring of the homozygous red and the yellow. Construct a Punnett diagram corresponding to this cross:

r r
R
Rr
Rr
R
Rr
Rr

The F1 tomatoes are all heterozygous (Rr) and appear red.

(b) The F2 tomatoes are offspring of two of the F1 plants. Construct a Punnett diagram:

R r
R
RR
Rr
r
Rr
rr

1/4 of the F2 tomatoes are homozygous red (RR) and appear red. 1/2 of the F2 tomatoes are heterozygous (Rr) and also appear red. 1/4 of the F2 tomatoes are homozygous yellow (rr) and appear yellow. Thus, 3/4 of the F2 tomatoes appear red and 1/4 appear yellow.

(c) If an F1 plant is crossed with the homozygous red parent, the results are:

R R
R
RR
RR
r
Rr
Rr

All offspring appear red.

(d) If an F1 plant is crossed with the yellow parent, the results are:

r r
R
Rr
Rr
r
rr
rr

1/2 of the offspring appear red, and the other 1/2 appear yellow.

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PROBLEM

In maize, a dominant allele A is necessary for seed color, as opposed to colorless (a). Another gene has a recessive allele wx that results in waxy starch, as opposed to normal starch (Wx). The two genes segregate independently. An Aa WxWx plant is testcrossed. What are the phenotypes and relative frequencies of offspring?

[from Russell, Peter J. 2005, iGenetics: A Mendelian Approach, Second Edition (Benjamin Cummings), problem 11.2]

SOLUTION

Let A = colored, a = colorless, W = normal starch, and w = waxy starch. Then the "Aa WxWx" plant can be represented more concisely as AaWW. In a testcross, the plant is crossed with one which is homozygous for both recessive alleles, so it is crossed with an aaww plant. The AaWW can contribute AW or aW to the offspring, while the aaww can contribute only aw. The corresponding Punnett diagram is therefore:

aw
AW
AaWw
aW
aaWw

1/2 of the offspring are AaWw and appear colored, and the other 1/2 are aaWw and are colorless. All offspring have Ww and produce normal starch.

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PROBLEM

F2 plants segregate 3/4 colored:1/4 colorless. If a colored plant is picked at random and selfed, what is the probability that both colored and colorless plants will be seen among a large number of its progeny?

[from Russell, Peter J. 2005, iGenetics: A Mendelian Approach, Second Edition (Benjamin Cummings), problem 11.3]

SOLUTION

Let C = colored and c = colorless. Then the F2 plants are 1/4 CC, 1/2 Cc, and 1/4 cc. Consider what happens if each of these is crossed with itself.

If a CC is crossed with itself, the results are 100% CC. All offspring are colored.

If a Cc is crossed with itself, the results are 1/4 CC, 1/2 Cc, and 1/4 cc. 3/4 of the offspring are colored and 1/4 are colorless.

If a cc is crossed with itself, the results are 100% cc. All offspring are colorless.

Thus, the only way that both colored and colorless plants will be seen among a large number of offspring is if a Cc is picked. The probability that this will happen if only colored plants are picked from is 2/3, because 2/3 of the colored plants are Cc, and 1/3 of the colored plants are CC.

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PROBLEM

In guinea pigs, rough coat (R) is dominant over smooth coat (r). A rough-coated guinea pig is bred to a smooth one, giving eight rough and seven smooth progeny in the F1 generation.
(a) What are the genotypes of the parents and their offspring?
(b) If one of the rough F1 animals is mated to its rough parent, what progeny would you expect?

[from Russell, Peter J. 2005, iGenetics: A Mendelian Approach, Second Edition (Benjamin Cummings), problem 11.4]

SOLUTION

(a) The rough-coated parent could be either RR or Rr. The smooth parent must be rr. If RR is crossed with rr, the results are all going to be rough. So the rough-coated parent must be Rr. The rough offspring are Rr and the smooth offspring are rr.

(b) If a rough-coated offspring is mated with the rough-coated parent, an Rr is being crossed with an Rr. The result would be 1/4 RR, 1/2 Rr, and 1/4 rr. 3/4 of the offspring would be rough-coated, and 1/4 would be smooth.

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PROBLEM

In cattle, the polled (hornless) condition (P) is dominant over the horned (p) phenotype. A particular polled bull is bred to three cows. Cow A, which is horned, produces a horned calf; polled cow B produces a horned calf; and horned cow C produces a polled calf. What are the genotypes of the bull and the three cows, and what phenotypic ratios do you expect in the offspring of these three matings?

[from Russell, Peter J. 2005, iGenetics: A Mendelian Approach, Second Edition (Benjamin Cummings), problem 11.5]

SOLUTION

The bull is either PP or Pp. Cow A must be pp because it is horned. Cow B is either PP or Pp. Cow C must be pp because it is horned.

If crossing the bull (PP or Pp) with cow A (pp) produced a horned calf (pp), the bull must be Pp. If crossing the bull (Pp) with cow B (PP or Pp) produced a horned calf (pp), cow B must be Pp.

The results of crossing the bull (Pp) with cow A (pp) are 1/2 Pp and 1/2 pp. 1/2 of the offspring are polled and 1/2 are horned.

The results of crossing the bull (Pp) with cow B (Pp) are 1/4 PP, 1/2 Pp, and 1/4 pp. 3/4 of the offspring are polled and 1/4 are horned.

The results of crossing the bull (Pp) with cow C (pp) are 1/2 Pp and 1/2 pp. 1/2 of the offspring are polled and 1/2 are horned.

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PROBLEM

In jimsonweed, purple flowers are dominant to white. Self-fertilization of a particular purple-flowered jimsonweed produces 28 purple-flowered and 10 white-flowered progeny. What proportion of the purple-flowered progeny will breed true (i.e., produce only purple-flowered offspring)?

[from Russell, Peter J. 2005, iGenetics: A Mendelian Approach, Second Edition (Benjamin Cummings), problem 11.6]

SOLUTION

Let P = purple and p = white. If self-fertilization of a particular jimsonweed produces both purple and white flowered offspring, the parent must be Pp.

1/4 of the offspring are PP, 1/2 are Pp, and 1/4 are pp. The PP and Pp are both purple, whle the pp are white. Only the PP will breed true, so 1/3 of the purple-flowered offspring will breed true.

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PROBLEM

Two black female mice are crossed with the same brown male. In a number of litters, female X produced 9 blacks and 7 browns, and female Y produced 14 blacks. What is the mechanism of inheritance of black and brown coat color in mice? What are the genotypes of the parents?

[from Russell, Peter J. 2005, iGenetics: A Mendelian Approach, Second Edition (Benjamin Cummings), problem 11.7]

SOLUTION

Let B = black and b = brown. Assume that black is dominant and brown is recessive. Then the brown male is bb, female X is Bb, and female Y is BB. This explains the results of the crossings.

Alternatively, let B = brown and b = black. Assume that brown is dominant and black is recessive. Then the brown male is either BB or Bb. Brown females X and Y are both either BB or Bb. There is no way the brown male could be crossed with brown female Y to produce all blacks, so the first assumption that we made above, that black is dominant, is the correct one.

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PROBLEM

Bean plants may have different symptoms when infected with a virus. Some show local lesions that do not seriously harm the plant; others show general systemic infection. The following genetic analysis was made:

P local lesions x systemic infection
F1 all local lesions
F2 785 local lesions:269 systemic infection

What is the likely genetic basis of this difference in beans? Assign gene symbols to all the genotypes occurring in the genetic analysis. Design a testcross to verify your assumptions.

[from Russell, Peter J. 2005, iGenetics: A Mendelian Approach, Second Edition (Benjamin Cummings), problem 11.8]

SOLUTION

The difference in beans is caused by a gene with two alleles, one for local lesions and the other for systemic infection. The allele for local lesions is dominant.

Let S = local lesions and s = systemic infection. Then, in the analysis described, a plant which is homozygous for local lesions (SS) is crossed with a plant which is homozygous for systemic infection (ss). The F1 plants are all Ss and exhibit local lesions only. The F2 plants are 1/4 SS, 1/2 Ss, and 1/4 ss. 3/4 exhibit local lesions and 1/4 exhibit systemic infection.

The analysis described can be considered a testcross where the parents were obtained by identifying plants which breed true for local lesions and for systemic infection. If the systemic infection allele was dominant, the F1 plants would have all exhibited systemic infection rather than local lesions.

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PROBLEM

A normal Drosophila (fruit fly) has both brown and scarlet pigment granules in its eyes, which appear red as a result. Brown (bw) is a recessive allele on chromosome 2 that, when homozygous, results in brown eyes because of the absence of scarlet pigment granules. Scarlet (st) is a recessive allele on chromosome 3 that, when homozygous, results in scarlet eyes because of the absence of brown pigment granules. Any fly homozygous for recessive alleles at both genes produces no eye pigment and has white eyes. The following results were obtained from crosses:

P brown-eyed fly x scarlet-eyed fly
F1 red eyes (both brown and scarlet pigment present)
F2 9/16 red:3/16 scarlet:3/16 brown:1/16 white

(a) Assign genotypes to the P, F1, and F2 generations.
(b) Design a testcross to verify the F1 genotype, and predict the results.

[from Russell, Peter J. 2005, iGenetics: A Mendelian Approach, Second Edition (Benjamin Cummings), problem 11.9]

SOLUTION

(a) Let B = dominant on chromosome 2, b = turns off scarlet on chromosome 2, S = dominant on chromosome 3, and s = turns off brown on chromosome 3.

The eye colors are determined using the following rules:
1. Brown is turned off only if "ss" is present.
2. Scarlet is turned off only if "bb" is present.
3. If brown and scarlet are both turned on, the color is red.
4. If brown is turned off but scarlet is turned on, the color is scarlet.
5. If brown is turned on but scarlet is turned off, the color is brown.
6. If brown and scarlet are both turned off, the color is white.

Then the genotype of the brown-eyed parent is either bbSS or bbSs, and the genotype of the scarlet-eyed parent is either BBss or Bbss.

The fact that the F1 generation are all red-eyed can be used to narrow down the possible genotypes of the parents. Consider the four possible crosses between a brown-eyed parent and a scarlet-eyed parent:

Cross Possible Outcomes Comment
bbSS x BBss
bS x Bs = BbSs (red)
possible
bbSS x Bbss
bS x (Bs or bs) = BbSs (red) or bbSs (brown)
not possible
bbSs x BBss
(bS or bs) x Bs = BbSs (red) or Bbss (scarlet)
not possible
bbSs x Bbss
(bS or bs) x (Bs or bs) = BbSs (red), bbSs (brown), Bbss (scarlet), or bbss (white)
not possible

The only combination that can produce only red-eyed offspring is the first one, so the brown-eyed parent is bbSS, and the scarlet-eyed parent is BBss. The F1 generation are all BbSs.

The genotypes of the F2 generation can be determined using a Punnett diagram. Each F1 can contribute BS, Bs, bS, or bs to the F2 generation. Thus, the Punnett diagram has the following appearance:

BS Bs bS bs
BS
BBSS
red
BBSs
red
BbSS
red
BbSs
red
Bs
BBSs
red
BBss
scarlet
BbSs
red
Bbss
scarlet
bS
BbSS
red
BbSs
red
bbSS
brown
bbSs
brown
bs
BbSs
red
Bbss
scarlet
bbSs
brown
bbss
white

9/16 of the F2 generation are red-eyed, 3/16 are brown-eyed, 3/16 are scarlet-eyed, and 1/16 are white-eyed.

(b) An F1 Drosophila can be crossed with a white-eyed Drosophila (bbss). If the F1 Drosophila is really BbSs, we would expect the following results (the BbSs can contribute BS, Bs, bS, or bs to the offspring, while the bbss can only contribute bs to the offspring):

bs
BS
BbSs
red
Bs
Bbss
scarlet
bS
bbSs
brown
bs
bbss
white

There should be equal numbers of red-eyed, brown-eyed, scarlet-eyed, and white-eyed offspring.

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PROBLEM

In Drosophila, vestigial wings (vg) is recessive to normal wings (vg+), and the gene for this trait is located on an autosome (i.e., a non-sex chromosome). Yellow body color (y) is recessive to normal body color (y+) and sex-linked.
(a) If a homozygous yellow vestigial female is crossed to a normal male (i.e., a male which has only alleles for normal wings and normal body color), what will be the genotypes and phenotypes of the F1 offspring?
(b) If you interbreed these F1 offspring, what will be the genotypes and phenotypes of the F2 offspring? Indicate the frequency of each genotypic and phenotypic combination; please show a Punnett square or a branch diagram.

[from Biology 22a Exam 1, Spring 2007, Brandeis University, problem 1]

SOLUTION

(a) Assume body color is x-linked (i.e., on the x sex chromosome). Let vg = vestigial wings, vg+ = normal wings, y = yellow body color (on x sex chromosome), y+ = normal body color (on x sex chromosome), and Y = y sex chromosome. Then the homozygous yellow vestigial female can be represented as yy vg vg, and the normal male can be represented by y+Y vg+vg+.

The female parent can contribute only y vg to the F1 generation. The male parent can contribute y+ vg+ or Y vg+. The F1 generation will have the genotypes and phenotypes indicated in the following Punnett diagram:

y+ vg+
male
Y vg+
male
y vg
female
y+y vg+vg
female
normal wings
normal color
yY vg+vg
male
normal wings
yellow

The females in the F1 generation will have genotype y+y vg+vg, normal wings, and normal body color. The males in the F1 generation will have genotype yY vg+vg, normal wings, and yellow body color.

(b) If a male and female in the F1 generation are interbred, the male can contribute y vg+, y vg, Y vg+, or Y vg, and the female can contribute y+ vg+, y+ vg, y vg+, or y vg. The offspring in the F2 generation will have the genotypes and phenotypes indicated in the following Punnett diagram:

y vg+, male y vg, male Y vg+, male Y vg, male
y+ vg+, female
y+y vg+vg+, female
normal wings
normal color
y+y vg+vg, female
normal wings
normal color
y+Y vg+vg+, male
normal wings
normal color
y+Y vg+vg, male
normal wings
normal color
y+ vg, female
y+y vg+vg, female
normal wings
normal color
y+y vg vg, female
vestigial
normal color
y+Y vg+vg, male
normal wings
normal color
y+Y vg vg, male
vestigial
normal color
y vg+, female
yy vg+vg+, female
normal wings
yellow
yy vg+vg, female
normal wings
yellow
yY vg+vg+, male
normal wings
yellow
yY vg+vg, male
normal wings
yellow
y vg, female
yy vg+vg, female
normal wings
yellow
yy vg vg, female
vestigial
yellow
yY vg+vg, male
normal wings
yellow
yY vg vg, male
vestigial
yellow

The F2 phenotypes can be summarized as follows:

8/16 female:
  3/16 normal wings, normal color
  3/16 normal wings, yellow
  1/16 vestigial wings, normal color
  1/16 vestigial wings, yellow

8/16 male:
  3/16 normal wings, normal color
  3/16 normal wings, yellow
  1/16 vestigial wings, normal color
  1/16 vestigial wings, yellow

The F2 genotypes can be summarized as follows:

8/16 female (without "Y"):
  1/16 y+y vg+vg+
  2/16 y+y vg+vg
  1/16 y+y vg vg
  1/16 yy vg+vg+
  2/16 yy vg+vg
  1/16 yy vg vg

8/16 male (with "Y"):
  1/16 y+Y vg+vg+
  2/16 y+Y vg+vg
  1/16 y+Y vg vg
  1/16 yY vg+vg+
  2/16 yY vg+vg
  1/16 yY vg vg

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