Sex Determination Seminar Notes
Sex Determination
The mechanisms that control whether a given animal will develop into a
male or develop into a female, or in some kinds of animal into a
hermaphrodite
In humans and other mammals, development into a male is
Quiz 4 review on Bones and Cartilage review Notes
artilage and Bone
Vertebrate skeletons are made of cartilage and bone.
Bone is a composite of type I collagen (protein fibers) and calcium phosphate
crystals.
Cartilage is a composite of type II collagen a
Exam 2 review (Part 1)
Specific Examples of apoptosis:
1) Self destruction of the tail of tadpoles.
2) Self destruction of the cells that were between gums and lips.
3) Self destruction of cells between the fingers (otherwise webbing, like duck
feet).
4)
Exam 1 Review Questions
1) If you can take a wall-paper pattern and either move it 2 inches to the right,
or to the left, or move it 4, 6, 8 etc inches to either the right or left, and it would
look the same, then what kind of symmetry does it have? (Let'
Quiz 1 Review Notes
Epithelia (Epithelium, singular)
Seal off spaces from each other;
Have distinct apical surfaces and baso-lateral surfaces
(incidentally, "endothelium" is a special kind of epithelium that lines blood vessels)
During embryonic developme
Midterm Study Guide
* indicates more difficult questions.
What combination of properties would two diffusible chemicals, A and B, need to have in
order to generate regular spatial patterns of lines and dots. Hint, B needs to diffuse faster
than A, and the
Exam 2 Practice Key
Each question was worth 5 points total.
1) Circle which 3 of the following have the same symmetries A N I X Z H % * @ &
There are two valid answers: NZ% or IXH
2) What is this symmetry? How many fold, what symmetry?NZ% have two-fold
ro
Solutions to Problem Set 9
1
1
15.6.2 Since 1 + e Z A(U )
= 1 + e Z A0 (U )
almost surely, we then
Z
0 Z
have e A(U ) = e A0 (U ) almost surely. Because A0 is continuous with density bounded above and below, we can get e Z dA(U ) =
dA
e0 Z dA0 (U ). Thus,
Solutions to Problem Set 7
P (| | > u)du. However, this is not norm,
10.5.1 (a) Recall that 2,1 = 0
because it does not satisfy the triangle inequality.
(b)For the rst inequality, if we can prove E ( 2 ) = 2 0 P (| | > u)udu
2 2,1 2 , then we have 2 = E
Solutions to Problem Set 6
9.6.6 C1 can pick out O(nV1 1 ) subsets from X1 , then C1 X2 can pick out
O(nV1 1 ) from X1 X2 , similarly, C2 X1 can pick out O(nV2 1 ) subsets from X1 X2
(C1 X2 ) (C2 X1 ) = (C1 C2 ) (X2 X1 ) can pick out O(nV1 +V2 1 )
subset
Solutions to Problem Set 3
6.5.8 Denote t as the coordinate projections x x(t) on C [a, b], then
t (x) = x(t);
First prove that t is continuous: cfw_x C [a, b], x x, |t x
t x| = |x (t) x(t)| sup|x (t) x(t)| = x (t) x(t) , Hence t is
continuous.
Therefor
Solutions to Problem Set 1
2.4.1 Let F (s, t) P 1cfw_X s, Y t, F1 (s) = F (s, ), and F2 (t) =
F (, t). For each > 0, choose = s1 < s2 < < sk1 = and
= t1 < t2 < < tk2 = such that F1 (sj ) F1 (sj 1 ) < , for
all 1 < j k1 , and F2 (tj ) F2 (tj 1 ) < , for a
Lecture 27 Notes
Lee, Kosorok and Fine (2005) proposed inference based on sampling from a posterior
distribution based on the profile likelihood.
The quadratic expansion of the previous section can generate confidence sets for by
inverting the log-likelih
Lecture 26 Notes
The most common approach to efficient estimation is based on modifications of
maximum likelihood estimation that lead to efficient estimates.
These modifications, which we will call likelihoods, are generally not really
likelihoods (produ
Lecture 12 Notes
The purpose of the remainder of this section is to discuss more substantive preservation
results such as the following, which is a minor modification of Theorem 3 of van der
Vaart and Wellner (2000):
THEOREM 1. Suppose that F1, . . . , Fk
Lecture 1 Notes
An empirical process is a function or other quantity computed from a data sample.
Empirical process methods are a set of specialized techniques and ways of thinking that
enable statistical inference and problem solving for empirical proces
Solutions to Problem Set 11
22
19.5.3 H 2 = cfw_h2 D [0, ] : function h2 with bounded total variation; is
a linear operator from H 2 to H 2, and
22
=
(1 + )e Z Y (s)h2 (s) Z
e
1 + e Z A(u)
Dene g (s) = P
(1+ )e0 Z Y (s)h2 (s)
1+e0 Z A0 (u)
Y (u)h2 (u)dA